SECTION 1.5: PIECEWISE-DEFINED FUNCTIONS; …(Section 1.5: Piecewise-Defined Functions; Limits and...

(Section 1.5: Piecewise-Defined Functions; Limits and Continuity in Calculus) 1.5.1

SECTION 1.5: PIECEWISE-DEFINED FUNCTIONS;

LIMITS AND CONTINUITY IN CALCULUS

LEARNING OBJECTIVES

• Know how to evaluate and graph piecewise-defined functions.

• Know how to evaluate and graph the greatest integer (or floor) function.

• Preview limits and continuity from calculus.

PART A: DISCUSSION

• A piecewise-defined function applies different rules, usually as formulas, to

disjoint (non-overlapping) subsets of its domain (subdomains).

• To evaluate such a function at a particular input value, we need to figure out

which rule applies there.

• To graph such a function, we need to know how to graph the pieces that

correspond to the different rules on their subdomains.

• The greatest integer (or floor) function and its graph, seen in calculus and

computer science, exhibit similar features.

• We will take a peek into calculus and preview the related topics of one- and two-

sided limits and continuity. Piecewise-defined functions appear frequently in these

sections of a calculus course.


PART B: THE ABSOLUTE VALUE FUNCTION

Let f x( ) = x . We discussed the absolute value function f in Section 1.3, Part N.

The piecewise definition of f is given by:

f x( ) = x =x, if x 0

x, if x < 0

• For instance, f 3( ) = 3 = 3, because 3 0 , and we use the top rule, which

applies to the subdomain

0, ) .

• However, f 3( ) = 3 = 3( ) = 3 , because 3< 0 , and we use the

bottom rule, which applies to the subdomain

, 0( ) .

We say that f is continuous at 0. That is, if we trace the graph of f with a pencil in

the vicinity of the point 0, 0( ) , we do not have to lift our pencil.

• We will provide a formal definition of continuity in Part G.

WARNING 1: Piecewise-defined functions are often discontinuous (i.e., they

lose continuity) where they switch rules. §


PART C: EVALUATING PIECEWISE-DEFINED FUNCTIONS

Example 1 (Evaluating a Piecewise-Defined Function)

Let the function f be defined by:

f x( ) =x2 , 2 x < 1

x +1, 1 x 2

• To evaluate f 1( ) , we use the top rule, since 2 1< 1.

f 1( ) = 1( )2

= 1

• To evaluate f 1( ) , we use the bottom rule, since 1 1 2 .

f 1( ) = 1( ) +1

= 2

• f 10( ) , for example, is undefined, because we have no rule for x = 10 .

10 is not in the domain of f , which is

2, 2 . §

PART D: GRAPHING PIECEWISE-DEFINED FUNCTIONS

Example 2 (Graphing a Piecewise-Defined Function with a Jump Discontinuity;

Revisiting Example 1)

Graph the function f from Example 1.

§ Solution

WARNING 2: Clearly indicate any endpoints and whether they are

included in, or excluded from, the graph.

Top rule: We will graph y = x2 on the subdomain

2,1) .

• We obtain a piece of a parabola.

•

2( )2

= 4 , so the left endpoint of this piece is at 2, 4( ) .


• 2 is included in the subdomain

2,1) . Therefore, the endpoint

2, 4( ) is included in this piece, and we plot it as a filled-in circle.

• 1( )

2

= 1, so the right endpoint of this piece is at 1,1( ) .

•• Why do we use the top rule, instead of the bottom rule? See Part F.

• 1 is excluded from the subdomain

2,1) . Therefore, the endpoint

1,1( ) is excluded from the piece, and we plot it as a hollow circle.

Bottom rule: We will graph y = x +1 on the subdomain 1, 2 .

• We obtain a line segment.

• 1( ) +1= 2 , so the left endpoint of this piece is at

1, 2( ) .

• 2( ) +1= 3 , so the right endpoint of this piece is at

2, 3( ) .

• Both 1 and 2 are included in the subdomain 1, 2 . Therefore, the

endpoints 1, 2( ) and

2, 3( ) are included in this piece, and we plot

them as filled-in circles.

The graph of f is below.

In calculus, we say that f has a jump discontinuity at 1 x = 1( ) .

(See Part G.) §


Example 3 (Graphing a Piecewise-Defined Function with a Removable

Discontinuity)

Graph f , if

f x( ) =x + 3, x 3

7, x = 3.

§ Solution

• We graph the line y = x + 3 , except we leave a hole at the point 3, 6( ) ,

since 3 is deleted from the subdomain of the top rule.

• The bottom rule defines f 3( ) to be 7, so we plot the point 3, 7( ) .

• In calculus, we say that there is a removable discontinuity at 3, since

the discontinuity could be removed by simply redefining f 3( ) to be 6.

We would end up with a different function, though. (See Part G.)

• Although f 3( ) = 7 , we will write limx 3

f x( ) = 6 in calculus. This is

because, as x approaches 3, f x( ) approaches 6. (See Part F.) §


PART E: THE GREATEST INTEGER (OR FLOOR) FUNCTION

The greatest integer (or floor) function is defined by f x( ) = x or x ,

the greatest integer that is not greater than x.

• Think: Round x down.

• If x is nonnegative, we simply take the integer part.

•

x = max y y x{ } . “max” takes the greatest element of the set.

Example Set 4 (Evaluating the Greatest Integer or Floor Function)

2.9 = 2

3 = 3

1.1 = 2

2.9 = 2

3 = 3

1.1 = 2

§

The ceiling function is defined by f x( ) = x ,

the least integer that is not less than x.

• Think: Round x up.

• If x is nonpositive, we simply take the integer part.

•

x = min y y x{ } . “min” takes the least element of the set.

Example Set 5 (Evaluating the Ceiling Function)

2.1 = 3 , 3 = 3, and 1.9 = 1. §


We may think of these as piecewise constant functions.

For example, the greatest integer (or floor) function is given by:

f x( ) = x or x =

2, 2 x < 1

1, 1 x < 0

0, 0 x <1

1, 1 x < 2

2, 2 x < 3

Example 6 (Graphing the Greatest Integer or Floor Function)

Looking at its graph below, we can see why the greatest integer function is

called a staircase function.

• Step functions are similar, except that the range must be a finite set; there are

only a finite number of distinct function values for a step function.

TIP 1: Remember that the left endpoints at 0, 0( ) ,

1,1( ) ,

2, 2( ) , etc. are

included in the graph. The right endpoints are excluded.

• In calculus, we say that f has jump discontinuities at all integer values

of x. (See Part G.)


PART F: LIMITS IN CALCULUS

Example 7 (Limits and the Greatest Integer or Floor Function)

Let f x( ) = x or x .

A Left-Hand, One-Sided Limit

Evaluate

limx 2

f x( ) , which is read:

“the limit of f x( ) as x approaches 2 from the left.”

• We want the real number that f x( ) approaches as x approaches 2

from lesser numbers (imagine approaching x = 2 from the left along

the real number line), if such a number exists.

• A table can help:

x 1.9 1.99 1.999 2

f x( ) 1 1 1

1

We see that limx 2

f x( ) = 1, even though f 2( ) = 2 .

WARNING 3: Sometimes,

limx a

f x( ) does not equal f a( ) .

For example, the value of f 2( ) , or whether it exists at all, is

technically irrelevant (though sometimes helpful) when

evaluating limits as x 2 .

• A graph confirms that

limx 2

f x( ) = 1. Along the graph of f below,

as x approaches 2 from the left, y approaches 1.


A Right-Hand, One-Sided Limit

Evaluate

limx 2+

f x( ) , which is read:

“the limit of f x( ) as x approaches 2 from the right.”

• We want the real number that f x( ) approaches as x approaches 2

from greater numbers (imagine approaching x = 2 from the right

along the real number line), if such a number exists.


x 2+ 2.001 2.01 2.1

f x( )

2 2 2 2

We see that

limx 2+

f x( ) = 2 .

• The graph of f below confirms this.

Two-Sided Limits

The two-sided limit limx 2

f x( ) does not exist here, because the

corresponding left-hand and right-hand limits are unequal.

If they existed and were equal, the common value would be the

two-sided limit.

• In Example 3, limx 3

f x( ) = 6 . (See the Exercises.) §


Example 8 (Finding Limits; Revisiting Examples 1 and 2)

Let the function f be defined by:

f x( ) =x2 , 2 x < 1

x +1, 1 x 2

While f 1( ) = 1( ) +1= 2 , the left-hand limit lim

x 1f x( ) = 1.

• When evaluating this left-hand limit, we only care about values of x

that are close to 1 and less than 1. The top rule, f x( ) = x2 , is the only

rule that applies to those x values. Therefore, lim

x 1f x( ) = lim

x 1x2

,

which we can compute by evaluating x2 at 1:

lim

x 1x2

= 1( )2

= 1.

• This explains why our graph approaches the endpoint at 1,1( ) from

the left, even though 1,1( ) itself is excluded from the graph.


x 0.9 0.99 0.999

1

f x( )

x2

0.81 0.9801 0.998001

1

WARNING 4: Tables can sometimes be misleading.

If the limit were 0.999, for instance, that might have been

difficult to detect.

• A graph confirms the limit:

§


PART G: CONTINUITY IN CALCULUS

f is continuous at a, or x = a

1) f a( ) is defined,

2) limx a

f x( ) exists, and

3) limx a

f x( ) = f a( ) .

f is discontinuous at a f is not continuous at a.

(Some sources require that f be defined “around” a.)

Comments

1) ensures that there is literally a point at a.

2) constrains the behavior of f immediately around a.

3) then ensures safe passage through the point

a, f a( )( ) on the graph of f .

In Part B, the absolute value function was continuous everywhere on .

f has a removable discontinuity at a, or x = a

1) limx a

f x( ) exists, but

2) f is still discontinuous at a

Then, the graph of f has a “hole” at the point a, lim

x af x( )( ) .

In Example 3,

f x( ) =x + 3, x 3

7, x = 3. f had a removable discontinuity at 3.


f has a jump discontinuity at a, or x = a

1)

limx a

f x( ) exists, and (call this limit L

1)

2)

limx a+

f x( ) exists, but (call this limit L

2)

3)

limx a

f x( ) limx a+

f x( ) . ( L

1L

2)

Therefore, the two-sided limit limx a

f x( ) does not exist.

It is irrelevant how f a( ) is defined, or if it is at all.

Example 2 and the greatest integer function gave us jump discontinuities.

f has an infinite discontinuity at a, or x = a

lim

x af x( ) = or , or

lim

x a+f x( ) = or

Then, the graph of f has the line x = a as a vertical asymptote (VA).

It is irrelevant how f a( ) is defined, or if it is at all.

For example, if f x( ) =1

x, then f has an infinite discontinuity at 0.

(Section 1.6: Combining Functions) 1.6.1

SECTION 1.6: COMBINING FUNCTIONS

LEARNING OBJECTIVES

• Know how to add, subtract, multiply, and divide functions.

• Be able to identify linear combinations of functions.

• Know how to construct and decompose composite functions.

• Find the domains of the functions that result from these operations.

• Be able to model functions using constraint equations and composite functions.

PART A: DISCUSSION

• In Section 1.5, we constructed functions from pieces of functions on disjoint

(non-overlapping) subdomains.

• In this section, we will combine functions by adding, subtracting, multiplying,

and dividing them over basically the same domain.

• Linear combinations are usually not formally introduced to students until a linear

algebra course, but linear combinations of functions will enable us to take a

broader view of symmetry in Section 1.7.

• We will also combine functions by composing them. This corresponds to the

successive application of functions. We did this when we transformed functions

and graphs in Section 1.4.

• Calculus theorems, including linearity theorems and the Chain Rule for differentiating

composite functions, will use the notation and ideas of this section. Composite functions are also

related to the u-substitution technique of integration.

• In applied calculus problems such as related rates and optimization problems, we will need to

devise functions, compose them, and combine them with constraint equations.


PART B: ARITHMETIC COMBINATIONS OF FUNCTIONS

Let f and g be functions.

If their domains overlap, then the overlap (intersection) Dom f( ) Dom g( )

is the domain of the following functions with the specified rules, with one

possible exception (*):

f + g , where

f + g( ) x( ) = f x( ) + g x( )

f g , where

f g( ) x( ) = f x( ) g x( )

fg , where fg( ) x( ) = f x( )g x( )

f

g, where

f

gx( ) =

f x( )g x( )

(*) WARNING 1:

Domf

g= x Dom f( ) Dom g( ) g x( ) 0{ } .

Example 1 (Adding Functions)

Let f x( ) = x2 and g x( ) = x3 . We will use tables and graphs for f and g to

develop a table and a graph for f + g .

• Observe that the domain is for f , g, and f + g .

x

f x( )

x2

g x( )

x3

f x( ) + g x( )

x2+ x3

3 9

27

18

2 4

8

4

1 1

1 0

0 0 0

0

1 1 1

2

2 4 8

12

3 9 27

36


f x( ) = x2 yields the blue graph below;

g x( ) = x3 yields the red graph.

f x( ) + g x( ) = x2+ x3 yields the brown graph; we add y-coordinates.

§

Example 2 (Subtracting Functions)

Let f x( ) = 4x and

g x( ) = x +

1

x.

Find f g( ) x( ) and Dom f g( ) .

§ Solution

f g( ) x( ) = f x( ) g x( )

= 4x( ) x +1

x

WARNING 2: Use grouping symbols when expanding g x( )

here. Be careful when subtracting an expression with more than

one term.

= 4x x1

x

= 3x1

x

Dom f( ) = . We omit only 0 from

Dom g( ) and also

Dom f g( ) .

Dom f g( ) = \ 0{ } = x x 0{ } = , 0( ) 0,( ) . §


Example 3 (Dividing Functions)

Let f x( ) = 1 and g x( ) =

x

x 2. Find

f

gx( ) and

Domf

g.

§ Solution

f

gx( ) =

f x( )g x( )

=1

x

x 2

=x 2

xx 2( )

WARNING 3: Watch the domain when you take reciprocals. We state

x 2( ) , because that restriction is not implied by the expression x 2

x.

• Dom f( ) = .

• Dom g( ) = \ 2{ } = x x 2{ } .

• g x( ) = 0 x = 0 , so we require: x 0 .

• Therefore, we must exclude both 0 and 2 from

Domf

g.

Domf

g= \ 0, 2{ } in set-difference form.

Also,

Domf

g …

… in set-builder form is: x x 0 and x 2{ } , or

x : x 0 and x 2{ }

… in graphical form is:

… in interval form is: , 0( ) 0, 2( ) 2,( )

§


PART C: LINEAR COMBINATIONS OF FUNCTIONS

Let f and g be functions. Let c and d be real numbers, possibly 0.

cf + dg is then called a linear combination of f and g.

• Its domain is the overlap (intersection) Dom f( ) Dom g( ) .

• More generally, a linear combination of objects is a sum of constant

multiples of those objects.

Example 4 (Linear Combinations)

a) 3 f + 4g ,

1

2f 5g , g , and 0 are linear combinations of f and g.

b) 2 f + 3g 4h and 3.7 f + g are linear combinations of f , g, and h. §

• In Section 1.7, we will see that a linear combination of even functions is even.

The same goes for odd functions.

• WARNING 4: People often erroneously attempt to apply linearity properties in

precalculus. For example, remember that the “square root of a sum” is typically

not equal to the “sum of the square roots.” Think: 1+ 4 1 + 4 .

• In calculus, we will see important linearity theorems for limits, derivatives, and integrals.

For example, let’s say we have a group of functions of x. We find their limits as x a , and all

the limits exist as real numbers. We can then find the limit of any linear combination of those

functions as x a . This is because, if the limits exist, …

•• the limit of a constant multiple of a function equals the constant times the limit of

the function. That is, limx a

cf x( ) = c limx a

f x( ) .

(Constant factors can “move across” the limit operator.)

•• the limit of a sum (difference) of functions equals the sum (difference) of the limits

of the functions. That is,

limx a

f x( ) + g x( ) = limx a

f x( ) + limx a

g x( ) , and

limx a

f x( ) g x( ) = limx a

f x( ) limx a

g x( )

(Limits can be taken term-by-term.)


PART D: COMPOSITION OF FUNCTIONS

We compose functions when we apply them in sequence, as we did in

Section 1.4 on transformations.

Let f and g be functions. The composite function f g is defined by:

f g( ) x( ) = f g x( )( )

Its domain is

x x Dom g( ) and g x( ) Dom f( ){ } .

• The domain consists of the “legal” inputs to g that yield outputs

that are “legal” inputs to f .

x g g x( ) f

f g

f g x( )( )

Think of f g as a “merged” function.

WARNING 5: The function f g applies g first and then f .

Think of pressing a g button on a basic calculator followed by an f button.

WARNING 6: f g may or may not be the same function as g f , which

applies f first. Composition of functions is not commutative the way that, say,

addition is. (See the Exercises.)

Example 5 (Composition of Functions)

Let f u( ) =

1

u and

g x( ) = x 1 .

WARNING 7: Writing f x( ) =

1

x together with

g x( ) = x 1 would

be acceptable, but that may lead to confusion.

Find

f g( ) x( ) and Dom f g( ) .


§ Solution

f g( ) x( ) = f g x( )( )= f x 1( )

Substitute u = x 1 into f u( ) =

1

u.

Conceptually, f takes the reciprocal of its input.

=1

x 1

Find Dom f g( ) .

The only restriction implied by the final expression is: x > 1.

In fact, Dom f g( ) …

… in set-builder form is:

x x > 1{ } , or

x : x > 1{ }


… in interval form is: 1,( )

WARNING 8: As we will see in Example 7, we might not be able to

obtain all the necessary domain restrictions on f g from our final

expression for f g( ) x( ) , as we did here.

• It is typically sufficient to analyze the final expression and the

domain of the first function applied, but there are counterexamples:

• If, instead of f u( ) =

1

u, we had been given

f u( ) =u 2

u u 2( ), then x 5

would be another restriction on top of x > 1. (Check this.)


A more rigorous analysis of Dom f g( ) follows.

xg :

g x( ) = x 1x 1, our u

g x( )

f :

f u( ) =1

u

f g

1

x 1

f g x( )

x > 1x 1 from Dom g( )x = 1 0 is an "illegal" input to f

0 Dom f( )( )

(See the bottom of the diagram in blue.)

• The set of “legal” inputs to g, Dom g( ) = 1, ) .

• The set of outputs from g, Range g( ) = 0, ) , and

• The set of “legal” inputs to f , Dom f( ) = \ 0{ } .

•• The outputs from g (u values) are “legal” inputs to f ,

except 0, which is the output for only the input 1 in Dom g( ) .

x 1 = 0 x = 1.( )

• Therefore, Dom f g( ) = Dom g( ) \ 1{ } = 1,( ) . §


Example 6 (Evaluating Composite Functions; Revisiting Example 5)

Again, let f u( ) =

1

u and g x( ) = x 1 .

Evaluate

f g( ) 3( ) ,

f g( ) 0( ) , and

f g( ) x + h( ) , where x + h > 1.

§ Solution

In Example 5, we developed the formula:

f g( ) x( ) =1

x 1 on

1,( ) .

Evaluate

f g( ) 3( ) . Evaluate

f g( ) x + h( ) .

Observe: 3 1,( ) . We assume: x + h > 1.

f g( ) 3( ) =1

3 1

=1

2

=2

2

f g( ) x + h( ) =1

x + h( ) 1

=1

x + h 1

=x + h 1

x + h 1

Evaluate

f g( ) 0( ) .

Observe: 0 1,( ) .

f g( ) 0( ) is undefined.

We can also evaluate

f g( ) 3( ) and

f g( ) x + h( ) without the formula.

f g( ) 3( ) = f g 3( )( )= f 3 1( )= f 2( )

=1

2

=2

2

f g( ) x + h( ) = f g x + h( )( )

= f x + h( ) 1( )= f x + h 1( )

=1

x + h 1

=x + h 1

x + h 1

§


Example 7 (“Hidden” Domain Restrictions)

Let f x( ) =1

x2 7 and g x( ) = x 3 . Find

f g( ) x( ) and

Dom f g( ) .

§ Solution

Rewriting the rule for f as, say, f u( ) =

1

u2 7, may clarify the solution.

f g( ) x( ) = f g x( )( )= f x 3( )

• If we had kept the formula f x( ) =1

x2 7, we would

informally (and awkwardly) substitute the x in

1

x2 7

with

x 3( ) . Substituting the u in

1

u2 7 with

x 3( )

is less awkward.

=1

x 3( )2

7

=1

x 3( ) 7

=1

x 10x 3; we will explain this( )

Find Dom f g( ) .

It is typically sufficient to analyze the final expression and the

domain of the first function applied, although there are

counterexamples (see Example 5). Here, 1

x 10 imposes the

restriction x 10 , and Dom g( ) imposes the restriction x 3 .


In fact, Dom f g( ) …

… in set-builder form is: x x 3 and x 10{ } , or

x : x 3 and x 10{ }


… in interval form is:

3,10) 10,( )

We write: f g( ) x( ) =1

x 10x 3( ) . The restriction

x 10( ) is

evident.

A more rigorous analysis of Dom f g( ) follows.

xg :

g x( ) = x 3x 3, our u

g x( )

f :

f u( ) =1

u2 7

f g

1

x 10

f g x( )

x 3 from Dom g( ) 7 "illegal" input to f , but not an output from g

x = 10 7 "illegal" input to f

7 Dom f( )( )

• The set of “legal” inputs to g, Dom g( ) = 3, ) .

• The set of outputs from g, Range g( ) = 0, ) , and

• The set of “legal” inputs to f , Dom f( ) = \ 7, 7{ } .

•• This is because: u2 7 = 0 u = ± 7 .

•• The outputs from g (u values) are “legal” inputs to f ,

except 7 , which is the output for only the input 10 in

Dom g( ) .

x 3 = 7 x = 10.( )

• Therefore, Dom f g( ) = Dom g( ) \ 10{ } = 3,10) 10,( ) . §


PART E: “DECOMPOSING” COMPOSITE FUNCTIONS

Example 8 (“Decomposing” a Composite Function)

Find component functions f and g such that

f g( ) x( ) = 3x +1 .

We want to “decompose” f g .

• Neither f nor g may be an identity function.

For example, do not use: g x( ) = x and

f u( ) = 3u +1 .

This would not truly be a decomposition. f does all the work!

xg :

g x( ) = xx, our u

g x( )

f :

f u( ) = 3u +1

f g

3x +1

f g x( )

§ Solution

• We need: f g x( )( ) = 3x +1 .

• We can think of f and g as buttons we are designing on a basic calculator.

We need to set up f and g so that, if x is an initial input to Dom f g( ) , and

if the g button and then the f button are pressed, then the output is 3x +1 .

xg :

g x( ) = ?u = ?

f :

f u( ) = ??

f g

3x +1

f g x( )

• A common strategy is to let g x( ) , or u, be an “inside” expression

(for example, a radicand, an exponent, a base of a power, a denominator, an

argument, or something being repeated) whose replacement simplifies the

overall expression.

• Here, we will let g x( ) = 3x +1 .


• We then need f to apply the square root operation. We will let f u( ) = u .

The use of u is more helpful in calculus, but f x( ) = x is also

acceptable. However, f u( ) = x is not acceptable.

Possible Answer: Let g x( ) = 3x +1 and f u( ) = u .

xg :

g x( ) = 3x +13x +1, our u

g x( )

f :

f u( ) = u

f g

3x +1

f g x( )

There are infinitely many possible answers.

For example, we could let g x( ) = 3x and

f u( ) = u +1 .

xg :

g x( ) = 3x3x, our u

g x( )

f :

f u( ) = u +1

f g

3x +1

f g x( )

§


PART F: APPLICATIONS

Example 9 (Conical Water Tank)

Water is flowing into a tank shaped as a right circular cone with base radius

5 meters and height 10 meters. Express the volume of the “water cone” in

the tank as a function of the base radius of the water cone.

§ Solution

• Define variables.

• Draw a diagram.

• Indicate known information.

Let r = the base radius of the water cone, in feet.

Let h = the height of the water cone, in feet.

Let V = the volume of the water cone, in cubic feet.

• Express the key formula.

We need the formula for the volume of a right circular cone of height

(or altitude) h and base radius r.

V =

1

3r 2h

As it stands, we are expressing V as a function of two variables,

r and h. However, r and h are not independent here, because the

shape of the tank forces a relationship between them.


• Express any other relationships between variables.

The height of the tank is twice its base radius.

This proportion must hold for the water cone, as well.

h = 2r

This may be thought of as a constraint equation.

It expresses h as a function of r.

•• This may be shown using similar triangles, for which corresponding

side lengths are in equal proportion.

h

10=

r

5

h = 2r

• Combine the key formula with any constraint equations.

We will make the substitution h = 2r( ) into

V =

1

3r 2h .

V =1

3r 2 2r( )

V =2

3r3

We are now expressing V as a function of one variable, r.


• Find the domain.

We may rewrite V as f r( ) , or even

V r( ) .

r, the base radius of the water cone, can be anything from

0 meters (a degenerate case corresponding to no water at all) to

5 meters, the base radius of the tank (corresponding to a full tank).

If V = f r( ) , then

Dom f( ) = 0, 5 . Units (m) aren’t written here.

Answer: V =

2

3r3 on 0, 5 .

• Note: We can also express V as a function of h on 0,10 .

We solve the constraint equation for r and express r as a function of h.

h = 2r

r =h

2

We then make the substitution

r =h

2 into V =

1

3r 2h .

V =1

3

h

2

2

h

V =1

3

h2

4h

V =h3

12

• To find the volume of the water cone at a particular instant, we only need

its radius or its height (the water level) at that instant. Then, we can use the

appropriate formula, V =2

3r3 , or

V =

h3

12.

•• If h = 3 m , say, then V =

h3

12=

3( )3

12=

27

12=

9

4m3 7.07 m3

.

• TIP 1: Observe that, if r or h is given in meters, the unit of volume we

obtain is (correctly) cubic meters. This dimensional analysis can be a useful

check. §


Example 10 (Conical Water Tank Problem with a Time Variable;

Revisiting Example 9)

Let’s say the tank from Example 9 is empty at time t = 0 , at which time

water starts flowing in. The radius of the water cone increases at the rate of

3m

min until the tank is full.

• In calculus, we call this rate

dr

dt, the derivative of the radius with respect to

time. (Does this mean that the water is flowing in at an increasing rate or a

decreasing rate as it fills the tank?)

Express the volume of the water cone as a function of t, the time elapsed,

until the tank is full.

§ Solution

• In Example 9, we developed as our new key formula:

V = f r( ) =

2

3r3

• We express r as a function of t.

If r starts at 0 m and increases at 3

m

min for t min, we can let:

r = g t( ) = 3t

• We can now express V as a composite function of t.

V = f g t( )( )

V =2

3r3 r = 3t( )

V =2

33t( )

3

V =2

327t3( )

V = 18 t3


• Find the domain of the composite function f g .

•• As in Example 9, r, the base radius of the water cone, can be

anything from 0 meters to 5 meters, the base radius of the tank.

•• Find the corresponding values of t.

0 r 5

0 3t 5

0 t5

3

Dom f g( ) = 0,5

3. (t is measured in minutes.)

Let’s review the composition of functions here.

tg :

g t( ) = 3t3t, our r

g t( )

f :

f r( ) =2

3r3

f g

18 t3

f g t( )

A tree diagram illustrates the relationships between variables.

• In calculus, the Chain Rule for differentiating (taking the derivative of)

composite functions states:

dV

dt=

dV

dr

dr

dt

• In multivariable calculus, the tree gets bushy! §

(Section 1.7: Symmetry Revisited) 1.7.1

SECTION 1.7: SYMMETRY REVISITED

LEARNING OBJECTIVES

• Recognize short cuts for identifying even functions and odd functions.

• Be able to justify those short cuts using proofs.

• Use equations to identify symmetries in their graphs.

PART A: DISCUSSION

• In Section 1.3, we defined even functions and odd functions, whose graphs in the

xy-plane were symmetric about the y-axis and the origin, respectively.

• We will now systematize our study of these functions and their graphical

symmetries. We will discuss useful short cuts for identifying even functions and

odd functions, and we will show how to prove them rigorously. We will also

extend related symmetries to graphs of general equations in x and y.

• A zero function f , where f x( ) = 0 (on any domain), will be called a trivial function.

If its domain is symmetric about 0, then it is both even and odd.

• Symmetry will prove helpful when graphing functions and equations and finding areas, surface

areas, and volumes using definite integrals in calculus. In multivariable calculus, it will be a

huge time-saver when finding volumes, masses, and centers of mass of three-dimensional solids.

PART B: LINEAR COMBINATIONS

We introduced linear combinations in Section 1.6, Part C.

For example, 2 f + 5g is a linear combination of f and g.

Linear combinations of even functions are even.

Linear combinations of odd functions are odd.

• In particular, sums, differences, and constant multiples of even functions

are even. The same goes for odd functions. Informally, E ± E E ,

O ± O O , cE E , and cO O .

• The domain of a linear combination of functions is the overlap

(intersection) of the domains of those functions. We assume this is

nonempty.

• Domains of even functions and odd functions must be symmetric about 0.


Example 1 (A Proof: The Sum of Odd Functions is Odd)

Prove that the sum of two odd functions is also odd.

§ Solution

• Let f and g be odd functions. Let h = f + g .

• Let D = Dom f( ) Dom g( ) . Then,

D = Dom h( ) .

• Because f and g are odd on D,

x D , f x( ) = f x( ) , and g x( ) = g x( ) .

• Show that h is odd. That is, x D , h x( ) = h x( ) .

x D ,

h x( ) = f + g( ) x( )= f x( ) + g x( )

by definition of f + g; see Section 1.6( )= f x( ) g x( )

because f and g are odd( )= f x( ) + g x( )= h x( )

Q.E.D. §


Example 2 (Demonstrating Example 1)

Let f x( ) = x and

g x( ) = x3 . We will use tables and graphs for f and g to

develop a table and a graph for f + g .

• The domain is for f , g, and f + g .

x

f x( )

x g x( )

x3

f x( ) + g x( )

x + x3

3

3

27

30

2

2

8

10

1 1 1 2

0 0 0

0

1 1 1

2

2 2 8

10

3 3 27

30

• f , g, and f + g are all odd functions. This explains why opposite

x-values always yield opposite function values for all three functions.

f x( ) = x yields the blue graph below;

g x( ) = x3 yields the red graph.

f x( ) + g x( ) = x + x3 yields the brown graph; we add y-coordinates.

• All three graphs are symmetric about the origin. §


PART C: POLYNOMIAL FUNCTIONS

Short Cuts for Determining whether a Polynomial Function is

Even, Odd, or Neither

Let f x( ) be a nontrivial (or “nonzero”) polynomial written in descending

powers of x, though the terms could be reordered.

• Delete any terms with zero coefficients.

• Nonzero constant terms have degree 0, which is an even degree.

•• We could rewrite the constant term 3, for example, as 3x0.

•• We define 00 to be 1 here.

If every term of f x( ) has even degree, then f is even.

If every term of f x( ) has odd degree, then f is odd.

• WARNING 1: If f x( ) has a nonzero constant term, then

f can’t be odd.

If f x( ) has a term of even degree and a term of odd degree, then

f is neither even nor odd.

Example 3 (An Even Polynomial Function)

Let f x( ) = 3x4 7x2

+1. Prove that f is even.

§ Solution 1 (Short Cut)

f x( ) is polynomial and only has terms of even degree (4, 2, and 0).

Therefore, f is even.

WARNING 2: 1 is a (nonzero) constant term, so it has degree 0.

• 1 has even degree; the fact that 1 is an odd integer is a different idea.

• The graph of y = 3x4 7x2 is symmetric about the y-axis, and so is

the graph of y = 3x4 7x2+1. (How do the graphs differ? Review

Section 1.4.) §


§ Solution 2 (Rigorous)

Dom f( ) = . x ,

f x( ) = 3 x( )4

7 x( )2

+ 1

= 3x4 7x2+ 1

= f x( )

Q.E.D. §

Example 4 (An Odd Polynomial Function)

Let f x( ) = 4x5

+ x . Prove that f is odd.


f x( ) is polynomial and only has terms of odd degree (5 and 1).

Therefore, f is odd. §


Dom f( ) = . x ,

f x( ) = 4 x( )5

+ x( )= 4x5 x

= 4x5+ x( )

= f x( )

Q.E.D. §


Example 5 (A Polynomial Function that is Neither Even nor Odd)

Let g t( ) = t3

+ t +1. Is g even, odd, or neither? Justify your answer.


g t( ) is polynomial and has terms of odd degree (3 and 1) and a term

of even degree (0). Therefore, g is neither even nor odd. §


Dom g( ) = . Evaluate g t( ) and compare it to g t( ) and g t( ) .

g t( ) = t( )3

+ t( ) +1

= t3 t +1

• g is not even, because g t( ) is not equivalent to

g t( ) on .

• g is not odd, because g t( ) is not equivalent to g t( ) on .

• A counterexample suffices to justify both statements.

For instance, g 1( ) = 1, and

g 1( ) = 3.

Then, g 1( ) g 1( ) , and

g 1( ) g 1( ) .

WARNING 3: If it had been true that g 1( ) = g 1( ) or

g 1( ) = g 1( ) , that would not have been enough to prove that

g was even or odd. Examples cannot prove that a function g is

even or odd, unless Dom g( ) is a finite set.

g is neither even nor odd. §

Let f be a (nontrivial) even function, and let g be a (nontrivial) odd function.

Then, f + g and f g must be neither even nor odd.

(Experiment with graphs and numbers. See the Exercises.)

Example 6 (Revisiting Example 5)

Let g t( ) = t3

+ t +1; this is not g in the box above. Since g t( ) = 1( ) + t3+ t( ) ,

g can take the form even( ) + odd( ) , and thus g is neither even nor odd. §


PART D: PRODUCT AND QUOTIENT RULES

Product and Quotient Rules for Even Functions and Odd Functions

The rules for multiplying and dividing (nontrivial) even functions and

odd functions are similar to the sign rules for multiplying and dividing

positive numbers and negative numbers, respectively.

In the table below, we denote even functions by P (for “Positive”) and

odd functions by N (for “Negative”). The resulting statements in the

“Sign Patterns” column correspond to the sign rules for multiplying and

dividing real numbers.

Even Function and

Odd Function Patterns Sign Patterns

even( ) even( )= even( )

P( ) P( ) = P( )

even( ) odd( )= odd( )

P( ) N( ) = N( )

odd( ) odd( )= even( )

N( ) N( ) = P( )

even( )even( )

= even( )

P( )P( )

= P( )

even( )odd( )

= odd( )

P( )N( )

= N( )

odd( )even( )

= odd( )

N( )P( )

= N( )

odd( )odd( )

= even( )

N( )N( )

= P( )

Special Cases: Reciprocals of Even Functions and Odd Functions

The reciprocal of a (nontrivial) even function is even.

The reciprocal of a (nontrivial) odd function is odd.

Patterns:

1

even( )= even( ) , and

1

odd( )= odd( ) .

• Remember that f x( ) = 1 defines an even function.


Example 7 (Proving a Quotient Rule)

Prove: even( )odd( )

= odd( ) . That is, prove that, if a (nontrivial) even function is

divided by a (nontrivial) odd function, the resulting function is odd.

§ Solution

• Let f be an even function, and let g be an odd function. Let h =f

g.

• Let D = Dom f( ) Dom g( ) \ x g x( ) = 0{ } . Then, D = Dom h( ) .

• Because f is even on D,

x D , f x( ) = f x( ) .

• Because g is odd on D,

x D , g x( ) = g x( ) .

• Show that h is odd. That is, show that:

x D , h x( ) = h x( ) .

x D ,

h x( ) =f

gx( )

=f x( )g x( )

by definition of f

g; see Section 1.6

=f x( )g x( )

because f is even, and g is odd( )

=f x( )g x( )

= h x( )

Q.E.D. §


Example 8 (Applying a Product Rule)

Let f x( ) = 7x6 x2

+ 4( ) x3 5x( ) . Is f even, odd, or neither?

§ Solution

f x( ) = even( ) odd( )= odd( )

Sign Rule: P( ) N( ) = N( ) .

f is an odd function. §

Example 9 (Applying Product and Quotient Rules)

In Chapter 4, we will see that the sine function (abbreviated “sin” when

applied to an argument) is odd, and the cosine function (abbreviated “cos”)

is even. Let q( ) =cos

sin. Is q even, odd, or neither?

§ Solution

WARNING 4: The “ ” sign in front of the fraction is inconsequential in

our analysis; multiplication by the constant 1 corresponds to multiplication

by an even function. When considering corresponding sign rules, the

notations (P) and (N) may be preferable to +( ) and ( ) due to this

potentially confusing matter.

Rewrite the given formula as:

q( ) =( ) cos( )

sin( ).

q( ) =

odd( ) even( )odd( )

= even( )

Sign Rule: N( ) P( )

N( )= P( ) .

q is an even function. §

TIP 1: When in doubt, use the definitions of even function and odd function.


PART E: SYMMETRY AND EQUATIONS

Consider an equation in x and y and its graph in the usual xy-plane.

1) The graph is symmetric about the y-axis

Replacing all occurrences of x with x( ) yields an equivalent equation.

Special Case: y = f x( ) , where f is an even function.

The equations below are then equivalent:

y = f x( )y = f x( )

2) The graph is symmetric about the x-axis

Replacing all occurrences of y with

y( ) yields an equivalent equation.

3) The graph is symmetric about the origin

Replacing x with x( ) and y with y( ) yields an equivalent equation.

Special Case: y = f x( ) , where f is an odd function.

The equations below are then equivalent:

y( ) = f x( )y = f x( ) because f is odd( )y = f x( )

4) The graph is symmetric about the line y = x

The equation is symmetric in x and y, meaning that

switching x and y throughout the equation yields an equivalent equation.

• In multivariable calculus, symmetry in x and y allows us to trim down

repetitive work such as finding partial derivatives of a function with respect

to x and y.


Example 10 (Identifying Graphical Symmetries from an Equation)

The graph of x2+ y2

= 9 exhibits all four aforementioned symmetries.

1) Symmetry about the y-axis

x( )2

+ y2= 9

x2+ y2

= 9

x, y( ) is on the graph

x, y( ) is on the graph.

2) Symmetry about the x-axis

x2+ y( )

2

= 9

x2+ y2

= 9



3) Symmetry about the origin

x( )2

+ y( )2

= 9

x2+ y2

= 9



4) Symmetry about the line y = x

y2+ x2

= 9

x2+ y2

= 9


y, x( ) is on the graph.

§

(Section 1.8: x = f y( ) ) 1.8.1

SECTION 1.8 : x = f y( )

LEARNING OBJECTIVES

• Know how to graph equations of the form x = f y( ) .

• Compare these graphs with graphs of equations of the form y = f x( ) .

• Recognize when a curve or an equation describes x as a function of y, and

apply the Horizontal Line Test (HLT) for this purpose.

• Know basic graphs in this new context.

• Adapt rules and techniques for function behavior, symmetry, and transformations

to this new context.

PART A: DISCUSSION

• Sometimes, x is treated as a function of y, and we graph equations of the form

x = f y( ) in the xy-plane. These graphs must pass the Horizontal Line Test (HLT).

• The ordered pairs we associate with f are now of the form output, input( ) .

• x and y switch roles. y, not x, is our independent variable, and x, not y, is our

dependent variable. Function values correspond to x values.

• Function evaluations and point-plotting are modified accordingly. The order of

the coordinates of ordered pairs must be kept in mind.

• We will investigate similarities and differences between graphs of equations of

the form x = f y( ) and those of the familiar form y = f x( ) . We will adapt tools

for drawing the new graphs.

• For example, we will consider intervals on which f is increasing or decreasing,

as well as intercepts. We will also manipulate equations to help us graph them.

• We will see how our rules and techniques for symmetry (from Sections 1.3 and

1.7) and transformations (from Section 1.4) need to be modified.

• We will develop a gallery of basic graphs similar to the one in Section 1.3.

• In Section 1.9, we will see that the graphs of x = f y( ) and y = f x( ) are

reflections about the line y = x .

(Section 1.8: x = f y( ) ) 1.8.2

PART B : GRAPHING x = f y( )

Graphing x = f y( )

As a set of ordered pairs, the graph of x = f y( ) is given by:

f y( ) , y( ) y Dom f( ){ } .

• The graph of x = f y( ) is typically different from the graph of y = f x( )

in the usual xy-plane.

• When we do function evaluations and point-plotting, we treat y as the

independent (“input”) variable and x as the dependent (“output”) variable.

• WARNING 1: We still order coordinates of points as

x, y( ) , but they now

take the form output, input( ) .

PART C: SQUARING FUNCTION and EVEN FUNCTIONS

Let f y( ) = y2 . f here is the same squaring function on we have always known.

• Its rule could also be given as: f x( ) = x2 .

•

Dom f( ) = , and

Range f( ) = 0, ) , as before.

• A table for x = f y( ) is below.

Input

y

Output

x

f y( )

y2

Point

x, y( )

Input

y

Output

x

f y( )

y2

Point

x, y( )

0 0 0, 0( ) 0 0

0, 0( )

1 1 1,1( )

1 1

1, 1( )

2 4 4, 2( )

2 4

4, 2( )

3 9 9, 3( )

3 9

9, 3( )

(Section 1.8: x = f y( ) ) 1.8.3

• The graph of x = y2 below is a parabola opening to the right.

WARNING 2: The graph never goes to the left of the y-axis, because

squares of real numbers are never negative.

• Look at the table. f is even, so each pair of opposite y values yields

a common function value f y( ) , or x. Graphically, this means that every

point x, y( ) on the graph has a “mirror image partner” x, y( ) that is also

on the graph, and the graph is symmetric about the x-axis (WARNING 3).

A function f is even f y( ) = f y( ) , y Dom f( )

The graph of x = f y( ) is

symmetric about the x -axis.

• f is decreasing on the y-interval

, 0( , so x values decrease there, and

the graph moves to the left as we move up the graph on that interval.

• f is increasing on the y-interval

0, ) , so x values increase there, and

the graph moves to the right as we move up the graph on that interval.

TIP 1: We could try to solve the equation x = f y( ) for y in terms of x.

Here, x = y2 y = ± x . The graph of y = x is the top half of

the parabola, while the graph of y = x is the bottom half.

(Section 1.8: x = f y( ) ) 1.8.4

PART D: THE HORIZONTAL LINE TEST (HLT)

In Section 1.2, we said that an equation in x and y describes y as a function of x

its graph passes the Vertical Line Test (VLT) in the xy-plane. We now have:

The Horizontal Line Test (HLT)

A curve in a coordinate plane passes the Horizontal Line Test (HLT)

There is no horizontal line that intersects the curve more than once.

An equation in x and y describes x as a function of y

Its graph in the xy-plane passes the HLT.

• Then, there is no input y that yields more than one output x.

• Then, we can write x = f y( ) , where f is a function.

• A “curve” could be a straight line.

Observe that the graph of x = y2 from Part C passes the HLT.

PART E: ESTIMATING DOMAIN AND RANGE FROM A GRAPH

The domain of f is the set of all y-coordinates of points on the graph of

x = f y( ) . (Think of projecting the graph onto the y-axis.)

The range of f is the set of all x-coordinates of points on the graph of

x = f y( ) . (Think of projecting the graph onto the x-axis.)

WARNING 4:

Domain

Think: yf

Range

Think: x

• If f y( ) = y2 , then, once again,

Dom f( ) = , and

Range f( ) =

0, ) .

The graph in Part D demonstrates this.

(Section 1.8: x = f y( ) ) 1.8.5

PART F: CUBING FUNCTION and ODD FUNCTIONS

Let f y( ) = y3 . Graph

x = f y( ) .

• If we solve x = y3 for y, we obtain the equivalent equation y = x3

.

We saw the graph of y = x3

in Section 1.3, and it is the graph that we want here.

• Observe that f is increasing on , meaning that the graph moves to the right as

we move up the graph.

WARNING 5: When graphing x = f y( ) , trace the graph from bottom-to-

top (in the direction of increasing y), not top-to-bottom. A common error is

to reflect the graph about the x-axis, which can happen if you rotate your

head clockwise and draw the shape of the graph of y = x3 .

• f is an odd function, and we have graphical symmetry about the origin,

just as before.

A function f is odd f y( ) = f y( ) , y Dom f( )

The graph of x = f y( ) is

symmetric about the origin.

(Section 1.8: x = f y( ) ) 1.8.6

PART G: TRANSLATIONS (“SHIFTS”) and INTERCEPTS

WARNING 6: This time, horizontal shifts are more intuitive than vertical shifts.

Let G be the graph of x = f y( ) .

Let c be a positive real number.

Horizontal Translations (“Shifts”)

The graph of x = f y( )+ c is G shifted right by c units.

• We are increasing the x-coordinates.

The graph of x = f y( ) c is G shifted left by c units.

Vertical Translations (“Shifts”)

The graph of x = f y c( ) is G shifted up by c units.

The graph of x = f y + c( ) is G shifted down by c units.

Example 1 (Translations)

Let f y( ) = y2 . G is the central, purple graph of

x = f y( ) below.

§

(Section 1.8: x = f y( ) ) 1.8.7

The “coordinate shift method” from Section 1.4, Part F still applies:

Example 2 (Coordinate Shifting and Intercepts)

We will translate the parabola on the left so that its turning point (its vertex)

is moved from 0, 0( ) to

4,1( ) ; that is, it is moved 4 units to the left and

1 unit up.

Graph of x = y2 Graph of x + 4 = y 1( )2

, or

x = y 1( )

2

4

Look at the graph on the right. As usual,

• To find the x-intercept, substitute y = 0 into an equation, and solve for x.

• To find the y-intercepts, substitute x = 0 into an equation, and solve for y.

(Finding these intercepts algebraically will be an Exercise.) §

When finding intercepts in the x = f y( ) setting, there are some differences.

• There can be at most one x-intercept, corresponding to f 0( ) , if it exists.

• There can be any number of y-intercepts (possibly none), or infinitely

many, corresponding to the real zeros of f .

(Section 1.8: x = f y( ) ) 1.8.8

PART H: REFLECTIONS

Let G be the graph of x = f y( ) .

The graph of x = f y( ) is G reflected about the y-axis. (WARNING 7)

The graph of x = f y( ) is G reflected about the x-axis.

The graph of x = f y( ) is G reflected about the origin.

Example 3 (A Reflection about the y-axis)

The graph of x = y2 is in blue below. The graph of x = y2 is in red.

§

Example 4 (Reflections)

Let f y( ) = y . G is the upper right, purple graph of

x = f y( ) below.

Observe that x = y y = x2 x 0( ) , so G is the right half of an

upward-opening parabola.

§

(Section 1.8: x = f y( ) ) 1.8.9

PART I: NONRIGID TRANSFORMATIONS; STRETCHING AND SQUEEZING

WARNING 8: Just as for translations (“shifts”) for the x = f y( ) case, the

horizontal transformations are now the more intuitive ones.

The graph of x = cf y( ) is:

a horizontally stretched version of G if c > 1

a horizontally squeezed version of G if 0 < c < 1

The graph of x = f cy( ) is:

a vertically squeezed version of G if c > 1

a vertically stretched version of G if 0 < c < 1

If c < 0 , then perform the corresponding reflection either before or after the

horizontal or vertical stretching or squeezing.

Example 5 (Stretching and Squeezing)

Let f y( ) = y2 . Consider x = cf y( ) on the left and x = f cy( ) on the right.

§

(Section 1.8: x = f y( ) ) 1.8.10

PART J: A GALLERY OF GRAPHS

• Most of the graphs below differ from those in the table in Section 1.3, Part P.

•• However, x = y3 is equivalent to y = x3

, so they share a common graph.

(Which other graphs are familiar from Section 1.3?)

• Corresponding domains and ranges are the same as those in Section 1.3.

•• Again, they can be inferred from the graphs.

•• The domain corresponds to y-coordinates this time, the range to x.

• Graphs of even functions are now symmetric about the x-axis.

• We associate “increasing” with “move right” as we move up the graphs.

Equation (Sample)

Graph

Even/Odd;

Symmetry Equation

(Sample)

Graph

Even/Odd;

Symmetry

x = c

Even;

x-axis

x =1

y2

Even;

x-axis

x = y

Odd;

origin

x = y

Neither

x = cy + d

c 0( )

Odd

d = 0 ;

then,

origin

x = y3

Odd;

origin

x = y2

Even;

x-axis x = y2/3

Even;

x-axis

x = y3

Odd;

origin x = y

Even;

x-axis

x=1

y

Odd;

origin

x = a2 y2

a > 0( )

Even;

x-axis

(Section 1.9: Inverses of One-to-One Functions) 1.9.1

SECTION 1.9: INVERSES OF ONE-TO-ONE FUNCTIONS

LEARNING OBJECTIVES

• Understand the purposes and properties of inverse functions.

• Recognize when a function is one-to-one and invertible, and apply the

Horizontal Line Test (HLT) for this purpose.

• Know how to find the inverse of a one-to-one function numerically and

graphically, and inverse formulas conceptually and mechanically.

PART A: DISCUSSION

• In Section 1.8, we saw that the graphs of x = y3 and y = x3

were the same.

Each equation can be solved for the “other variable” to obtain the other equation.

We call the corresponding cubing and cube root functions a pair of inverse

functions. The graphs of y = x3 and y = x3

are reflections about the line y = x .

• The term inverse is used in many different contexts. We discussed logical

inverses in Section 0.2. We will review additive and multiplicative inverses in

this section. Inverse functions are inverses with respect to composition of

functions (see Section 1.6). (Typically, f1 1 / f .) Inverse properties are based

on the idea that a pair of inverse functions “undo” each other.

• Inverse functions are developed by switching inputs with outputs and thus

domains with ranges. In order for a function f to be invertible (to have an inverse

function f1 ), f must be one-to-one so that we obtain a function after the switch.

Sometimes, the implied domain must be restricted for a function to be invertible.

• In Section 1.8, we used the Horizontal Line Test (HLT) to see if an equation or

its graph represented x as a function of y. Now, we will use the HLT to see if a

function is one-to-one and invertible.

• We use inverses to solve equations uniquely. For example, when solving

x +1= 3, we use subtraction by 1 to invert addition by 1.

• In Chapter 3, we will see that exponential and logarithmic functions form pairs of inverse

functions; in Chapter 4, we will discuss trigonometric and inverse trigonometric functions. We

will solve related equations using inverse properties. In Section 8.4, we will define inverse

matrices as multiplicative inverses, although they will be related to inverse functions

(“transformations”) in linear algebra.


PART B: INVERSE FUNCTIONS

We will begin by reviewing additive and multiplicative inverses.

• 0 is the additive identity of , because, if we add 0 to any real number, the

result is identical to that number. Similarly, 1 is the multiplicative identity.

• The additive inverse of, say, the real number 3 is 3. That is, 3 is the inverse

of 3 with respect to addition. This is because, if the two numbers are added, the

result is the additive identity, 0.

• The multiplicative inverse (or reciprocal) of 3 is 1

3. That is,

1

3 is the inverse of

3 with respect to multiplication. This is because, if the two numbers are

multiplied, the result is the multiplicative identity, 1.

A pair of inverse functions have the property that, when they are composed in

either order, the result is the identity function, which outputs its input. They are

inverses with respect to function composition. (See Footnote 1.)

If f has an inverse function (i.e., f is invertible), then this inverse is unique, and

it is denoted by f1 (called f inverse).

• Also, the inverse of f1 is f .

WARNING 1: f1 , the inverse function of f , is not typically the multiplicative

inverse of f . That is, typically,

f 1 1

f. The “ 1” is just a superscript; it is not

an exponent. However,

f x( )1

is interpreted as

1

f x( ). (See Footnote 2.)

Inverse Properties (can be taken as the Definition of an Inverse Function)

f has a unique inverse function f1

f 1 f x( )( ) = x ,

x Dom f( ) , and

f f 1 x( )( ) = x ,

x Dom f 1( ) .

That is, f and f1 “undo” each other.


Example 1 (Temperature Conversion).

Let f be the function that converts temperature measures from the Celsius

scale to the Fahrenheit scale.

f1 then exists. It is the function that converts from Fahrenheit to Celsius.

• For example, f 0( ) = 32 and

f 1 32( ) = 0 , because 0 degrees

Celsius corresponds to 32 degrees Fahrenheit. That is, 0 C = 32 F .

Both measures give the freezing point of water at sea level.

• Also, f 100( ) = 212 and f 1 212( ) = 100 , because 100 C = 212 F .

Both measures give the boiling point of water at sea level.

A partial table for f and f1 is below.

f f1

Input

x

Output f x( )

Input

x

Output

f 1 x( ) 0 32 32 0

100 212 212 100

A partial arrow diagram for f and f1 is below.

The Inverse Properties are demonstrated below.

f 1 f 0( )( ) = f 1 32( ) = 0

f 1 f 100( )( ) = f 1 212( ) = 100

f f 1 32( )( ) = f 0( ) = 32 f f 1 212( )( ) = f 100( ) = 212

§


Example 1 demonstrates that, in going from a function to its inverse

(if it exists), inputs and outputs switch roles. This is a key theme.

“Input-Output” Properties of Inverse Functions

(can also be taken as the Definition of an Inverse Function)

If f has an inverse function f1 ,

then f a( ) = b f 1 b( ) = a .

• That is, a, b( ) f b, a( ) f 1 .

Domain and Range of Inverse Functions

If f has an inverse function f1 , then:

Dom f( ) = Range f 1( ) , and

Dom f 1( ) = Range f( ) .

That is, the domain of one function is the range of the other.

Example 2 (Inverse Functions: Verification, Domain, and Range)

a) Let f x( ) = x3 on . Find f1 , and verify that it is the inverse of f .

b) Let g x( ) = x3 on

0, 2 . Find g

1 .

c) Let h x( ) = x3 on

2,1,{ } . Find h

1.

§ Solution

The inverse of each cubing function above is a cube root function.

a) f 1 x( ) = x

3 on , which is Range f( ) . Let

r x( ) = x

3 on .

We will verify that r is the inverse of f by verifying the Inverse

Properties for f and r.

r f( ) x( ) = r f x( )( ) = r x3( ) = x33

= x , x , and

f r( ) x( ) = f r x( )( ) = f x

3( ) = x3( )

3

= x , x .

b) g 1 x( ) = x

3 on

0, 8 , which is

Range g( ) .

c) h 1 x( ) = x

3 on 8,1, 3{ } , which is

Range h( ) .

The reader can investigate further in the Exercises. §


PART C: ONE-TO-ONE FUNCTIONS and

THE HORIZONTAL LINE TEST (HLT)

We now discuss which functions are invertible.

From Section 1.1,

A relation f is a function Each input to f in its domain yields

exactly one output in its range.

The graph of y = f x( ) in the xy-plane

passes the Vertical Line Test (VLT).

A function cannot allow an input in its domain to yield two or more

different outputs. The following do not represent functions:

Input-Output Machine Arrow Diagram Graph and Equation

Fails VLT

y = ± 9 x2

Now,

A function f is one-to-one Each output from f in its range is yielded by

exactly one input in its domain.

Equivalently, f a( ) = f c( ) a = c .

That is, identical outputs imply identical inputs. (See Footnote 3.)

A one-to-one function cannot allow two or more different inputs (x values)

in its domain to yield the same output (y value).


Example 3 (A Function that is Not One-to-One: Unrestricted Squaring Function)

Let f x( ) = x2 on . Then, f is a function, but it is not one-to-one on , as

demonstrated by the figures below:

Input-Output Machine Partial Arrow Diagram Graph

Fails HLT (Section 1.8)

§

• Also, f a( ) = f c( ) a2

= c2 a = ± c , which is not equivalent to

a = c , if c 0 .

If we solve the equation f x( ) = 9 , or x

2= 9 , we obtain two solutions for x,

namely 3 and 3. These are the two answers to the question, “Whose square

is 9?”

• When we switch inputs with outputs, the expression f 1 9( ) is

not well-defined here, because there are two possible values it could

take on: 3 and 3.

As a consequence, f is not invertible, because it has no inverse function. §

However, a squaring function can be one-to-one and invertible on a restricted

domain, as we will see in Example 4.

• We will restrict domains when we define inverse trigonometric functions in Section

4.10.


Example 4 (A One-to-One Function: Squaring Function on a Restricted Domain;

Modifying Example 3)

Let g x( ) = x2 on the restricted domain

0, ) .

The graph of y = g x( ) below passes the Vertical Line Test (VLT) and also

the Horizontal Line Test (HLT).

Consequently, g is a one-to-one correspondence between Dom g( ) ,

the set of input x values, and Range g( ) , the set of output y values.

(Think of matched pairs.) Also, g a( ) = g c( ) a2

= c2 a = c ,

since only nonnegative inputs are allowed.

• If we solve the equation g x( ) = 9 , or

x2

= 9 x 0( ) , we obtain a

unique solution for x, namely 3. It is the unique input that yields 9.

On the graph of g above, the only point with y-coordinate 9 has

x-coordinate 3.

• More generally, g x( ) = b has a unique solution for x, the unique

input that yields b, whenever b is in Range g( ) , which is 0, ) .

We can define a unique inverse function g1 .

• Let Dom g 1( ) = Range g( ) , which is 0, ) .

• Define g 1 b( ) to be the unique solution to

g x( ) = b ,

for every b in Dom g 1( ) . For instance, g 1 9( ) = 3. (In Example 1,

since f 0( ) = 32 , we reverse the arrow and define f 1 32( ) to be 0.) §


In summary …

A function f is invertible

f x( ) = b has a unique solution, given by

x = f 1 b( ) ,

b Range f( )

f is one-to-one

The graph of y = f x( ) in the xy-plane passes the Horizontal Line Test (HLT).

• The one-to-one property is essential for a function to be invertible,

because we need the inverse to be a function after inputs are switched with

outputs.

• (See Footnote 3; we assume the “onto” property.)

PART D: GRAPHING INVERSE FUNCTIONS

By the “Input-Output” Properties, a, b( ) f b, a( ) f 1 .

Graphical Properties of Inverse Functions

If f is invertible, then:

The point a,b( ) lies on the graph of f

The point b,a( ) lies on the graph of f1 .

To obtain the graph of f1 , reflect the graph of f

about the line y = x .

Example 5 (Restricted Squaring Function; Revisiting Example 4)

Again, let g x( ) = x2 on

0, ) .

Partial tables for g and g1 can be constructed as follows:

g g1

x g x( ) Point

x g 1 x( ) Point

0 0 0, 0( ) 0 0 0, 0( )

1 1 1,1( ) 1 1

1,1( )

2 4 2, 4( ) 4 2

4, 2( )

3 9 3, 9( ) 9 3

9, 3( )


The graph of g, or y = g x( ) , is in blue below. It passes the HLT, meaning

that g is one-to-one.

Therefore, when we reflect the graph about the line y = x (drawn as a

dashed line, though it is not an asymptote), we obtain the red graph of

y = g 1 x( ) below, and it passes the VLT. It should look familiar …..

• The graph of y = g 1 x( ) is the graph of y = x , because

g 1 x( ) = x . It

makes sense that the corresponding square root function undoes what the

(restricted) squaring function does.

• It is also the graph of x = g y( ) , or x = y2 , with y restricted to 0, ) .

This is consistent with the aforementioned Graphical Properties of Inverse

Functions.

The graph of x = g y( ) for a function g is the reflection of the graph

of y = g x( ) about the line y = x .

• This holds even if g is not one-to-one. §


PART E: FINDING FORMULAS FOR INVERSE FUNCTIONS

If f is a one-to-one function, and if its formula can be expressed algebraically,

then we should be able to find a formula for f1 .

Conceptual Approach to Finding the Inverse of a One-to-One Function f

To determine f1 , we need to invert (or “undo”) the steps applied by f

in reverse order (WARNING 2). (See the Exercises.)

• Also, if necessary, impose restrictions and ensure that


Dom f 1( ) = Range f( ) . (See Example 10.)

TIP 1: Similarly, when dressing, you put on your socks before your shoes,

but, when undressing, you remove your shoes before your socks.

Example 6 (Conceptual Approach to Finding an Inverse)

Let f x( ) =

x3

+ 5

7 on . Find

f 1 x( ) .

§ Solution

• What does f do to x?

1) Takes the cube root: “ input3 ”

2) Adds 5 to the result from 1): “ input +5”

3) Divides the result from 2) by 7: “

input

7”

• What should f1 do to its input? (The asterisk “*” denotes inverting.)

3*) Multiplies by 7: “ input 7 ”

2*) Subtracts 5 from the result from 3*): “ input 5”

1*) Cubes the result from 2*): “ input( )

3

”

• Therefore, f 1 x( ) = 7x 5( )

3

.

• Dom f( ) = Range f 1( ) = , and Dom f 1( ) = Range f( ) = , so no

restrictions are necessary. §


Mechanical Approach to Finding the Inverse of a One-to-One Function f

Given a formula for f in terms of x, we can attempt to find a formula for

f1 as follows:

Step 1: Replace f x( ) with y.

Step 2: Switch x and y.

(Remember the theme of switching inputs with outputs.)

Step 3: Solve for y, if possible.

Step 4: Replace y with f 1 x( ) .

Step 5: If necessary, impose restrictions and ensure that

Dom f( ) = Range f 1( ) , and Dom f 1( ) = Range f( ) .

• In Steps 1-3, we are essentially solving x = f y( ) for y.

Example 7 (Mechanical Approach to Finding an Inverse; Revisiting Example 6)

Again, let f x( ) =

x3

+ 5

7 on . Find

f 1 x( ) .

§ Solution

Step 1: Replace f x( ) with y.

y =

x3

+ 5

7

Step 2: Switch x and y.

(Remember the theme of switching inputs with outputs.)

x =

y3+ 5

7


Step 3: Solve for y.

7x = y3 + 5 Multiplied both sides by 7( )

7x 5 = y3 Subtracted 5 from both sides( )

7x 5( )3

= y Cubed both sides( )

WARNING 3 : Squaring both sides of an equation may or

may not yield an equivalent equation.

y = 7x 5( )3

• Compare these steps with Steps 3*, 2*, and 1* in Example 6.

Step 4: Replace y with f 1 x( ) , if we obtain a function.

f 1 x( ) = 7x 5( )

3

• f1 is a function, because f was one-to-one.

Step 5: If necessary, impose restrictions and ensure that


Dom f 1( ) = Range f( ) .

• Dom f( ) = Range f 1( ) = , and

Dom f 1( ) = Range f( ) = , so

no restrictions are necessary. §

Example 8 (Evaluating an Inverse Function; Revisiting Examples 6 and 7)

Again, let f x( ) =x

3+ 5

7 on . Evaluate f 1 1( ) .

§ Solution

f 1 x( ) = 7x 5( )3

on from Examples 6 and 7( )

f 1 1( ) = 7 1( ) 5( )3

= 8

Absent Examples 6 and 7, we could have also solved f x( ) = 1 for x. §


Example 9 (Checking an Inverse Function Formula; Revisiting Examples 6 and 7)

Again, let f x( ) =x

3+ 5

7 on . Let g x( ) = 7x 5( )

3

on .

Check that g = f 1 .

§ Solution

We will verify that the compositions g f and f g are identity functions.

Either check below is sufficient, because f is one-to-one, and

Range f( ) = Dom g( ) (each set is ). If in doubt, do both checks.

(See Footnote 1.)

Check that

g f( ) x( ) = x , x . Check that

f g( ) x( ) = x , x .

x , x ,

g f( ) x( ) = g f x( )( )

= gx

3+ 5

7

= 7x

3+ 5

75

3

= x3

+ 5 53

= x3

3

= x

f g( ) x( ) = f g x( )( )

= f 7x 5( )3( )

=7x 5( )

33 + 5

7

=7x 5+ 5

7

=7x

7

= x

Therefore, g = f 1 . §


Example 10 (Temperature Conversion; Revisiting Example 1)

Let f be the one-to-one function that converts from Celsius to Fahrenheit.

Then, f1 is the function that converts from Fahrenheit to Celsius.

The following is left for the reader in the Exercises:

• Show that f x( ) =9

5x + 32 by developing a linear model for f such

that f 0( ) = 32 and f 100( ) = 212 .

• Show that f 1 x( ) =5

9x 32( ) in three different ways:

•• Develop a linear model for f1 such that

f 1 32( ) = 0 and

f 1 212( ) = 100 .

•• Begin with f x( ) =

9

5x + 32 and apply the Conceptual

Approach used in Example 6.

•• Begin with f x( ) =

9

5x + 32 and apply the Mechanical

Approach used in Example 7.

We will now determine domains and ranges.

(See Step 5 in the Mechanical Approach.)

• Temperatures cannot go below absolute zero, which is 273.15 C .

We require: Dom f( ) = Range f 1( ) = x x 273.15{ } .

• Find the Fahrenheit equivalent of 273.15 C .

f x( ) =9

5x + 32

f 273.15( ) =9

5273.15( ) + 32

= 459.67 F( )

• We require: Dom f 1( ) = Range f( ) = x x 459.67{ } .


Observe that the red graph for f and the brown graph for f1 below are

reflections about the line y = x .

FOOTNOTES

1. Identity functions and compositions of inverse functions. There are technically different

identity functions on different domains.

(See Footnote 3 below and Section 1.1, Footnote 1.)

• Let f be an invertible function that maps from domain X to codomain Y ; i.e., f : X Y .

If f is invertible, then f is onto, meaning that the range of f is the codomain Y.

f1 maps from Y to X ; i.e., f

1 :Y X .

• Let I

X be the identity function on Dom f( ) , which is X .

I

X: X X .

• Let IY

be the identity function on Dom f 1( ) , which is Y . I

Y:Y Y .

• Then, f 1 f = I

X, and

f f 1

= IY

.

• If g is a function such that g f = I

X, then g is a left inverse of f ; f has a left inverse

f is one-to-one. For example, let X = 1, 2{ } ,

Y = 10, 20, 30{ } ,

f = 1,10( ), 2, 20( ){ } , and

g = 10,1( ), 20, 2( ), 30, 2( ){ } . Then, g is a left inverse of f .

• If h is a function such that f h = I

Y, then h is a right inverse of f ; f has a right inverse

f is onto. For example, let X = 1, 2, 3{ } ,

Y = 10, 20{ } ,

f = 1,10( ), 2, 20( ), 3, 20( ){ } ,

and h = 10,1( ), 20, 2( ){ } . Then, h is a right inverse of f , although h is not a left inverse of f .

• If f is one-to-one and onto, then f has a unique inverse function that serves as both a

unique left inverse and a unique right inverse.


2. fn . Many instructors reluctantly use the f 1 notation to represent the inverse function of f .

• This is because n often represents an exponent in the notation fn , except when n = 1 .

For example, f 2 is often taken to mean ff ; that is, f 2 x( ) = f x( ) f x( ) .

In Chapters 4 and 5, we will accept that sin2 x = sin x( ) sin x( ) , which is the standard

interpretation.

• On the other hand (and this compounds the confusion), some sources use n to indicate the

number of applications of f in compositions of f with itself; the result is called an iterated

function. For example, they would let f 2= f f , and they would use the rule:

f 2 x( ) = f f x( )( ) . This is typically different from the rule

f 2 x( ) = f x( ) f x( ) .

However, our use of the notation f1 for “ f inverse” is more consistent with this second

interpretation, since f

1 f is an identity function, which could be construed as f0 in this

context.

3. One-to-one, onto, and bijective functions. A function f is invertible it is one-to-one

(or injective) and onto (or surjective); then, f is a bijective function, or a one-to-one

correspondence.

• A one-to-one (or “injective”) function has the property that, whenever f a( ) = f c( ) for

domain elements a and c, it must be true that a = c . That is, two outputs are equal the

inputs are equal. This definition will lead to the One-to-One Properties for exponential and

logarithmic functions in Chapter 3.

• An onto (or “surjective”) function has the property that its range is the entire codomain

(see Section 1.1, Footnote 1). This means that every element of the codomain is the image

(that is, function value or output) of some element of the domain.

• When we develop the inverse of a bijective function, we switch inputs with outputs.

•• The one-to-one property guarantees that the resulting function does not allow one

input to yield more than one output.

•• The onto property guarantees that the resulting function is defined for all elements of

the codomain of the original function, which is now the domain of the inverse function.

The onto property is often ignored in discussions of inverse functions. This is because we

typically force the codomain of the original function to be the range (or “image”) of the

original function in this context; this then guarantees the onto property. (Precalculus

sources typically avoid the term “codomain” in the first place.) If this is the case, then

“one-to-one function” and “one-to-one correspondence” are interchangeable, and we only

have to worry about the one-to-one property when it comes to invertibility.

(Section 1.10: Difference Quotients) 1.10.1

SECTION 1.10: DIFFERENCE QUOTIENTS

LEARNING OBJECTIVES

• Define average rate of change (and average velocity) algebraically and graphically.

• Be able to identify, construct, and evaluate (and simplify) various forms of

difference quotients.

PART A: DISCUSSION

• This section will revisit Section 0.14 on slopes of lines, Section 1.1 on function

evaluations, and Section 1.2 on graphs.

• A difference quotient is used to find the slope of a secant line to a graph or to find

an average rate of change (perhaps an average velocity). Difference quotients will

form the basis for derivatives, tangent lines, and instantaneous rates of change in

Section 1.11.

PART B: SECANT LINES and AVERAGE RATE OF CHANGE

The secant line to the graph of a function f on the interval

a, b , where a < b ,

is the line that passes through the points a, f a( )( ) and

b, f b( )( ) .

The average rate of change of f on a, b is equal to the slope of this secant line,

which is given by:

rise

run=

f b( ) f a( )b a

.

We call this a difference quotient, because it has the form:

difference of outputs

difference of inputs.

• (See Footnote 1 on assumptions about f .)


PART C: AVERAGE VELOCITY

The following development of average velocity will help explain the association

between slope and average rate of change.

Example 1 (Average Velocity)

A car is driven due north 100 miles during a two-hour trip.

What is the average velocity of the car?

• Let t = the time (in hours) elapsed since the beginning of the trip.

• Let y = s t( ) , where s is the position function for the car (in miles).

s gives the signed distance of the car from the starting position.

•• The position (s) values would be negative if the car were south of

the starting position.

• Let s 0( ) = 0 , meaning that y = 0 corresponds to the starting position.

Therefore, s 2( ) = 100 (miles).

The average velocity on the time-interval

a, b is the

average rate of change of position with respect to time. That is,

change in position

change in time=

s

t

where (uppercase delta) denotes “change in”

=

s b( ) s a( )b a

, a difference quotient

Here, the average velocity on

0, 2 is:

s 2( ) s 0( )2 0

=100 0

2

= 50 miles

houror

mi

hr or mph

TIP 1: The unit of velocity is the unit of slope given by:

unit of s

unit of t.


The average velocity is 50 mph on

0, 2 in the three scenarios below.

It is the slope of the orange secant line.

We will define instantaneous velocity (or simply velocity) in Section 1.11.

• Here, the velocity is constant (50 mph).

• Here, the velocity is increasing; the car is accelerating.

• Here, the car overshoots the destination and then backtracks.

WARNING 1: The car’s velocity is negative in value when it is

backtracking; this happens when the graph falls.

• In calculus, the Mean Value Theorem for Derivatives will imply that the car must be

going exactly 50 mph at some time value t in 0, 2( ) . The theorem applies in all three

scenarios above, because s is continuous on

0, 2 and is differentiable on 0, 2( ) ,

meaning that its graph makes no sharp turns and does not exhibit “infinite steepness” on

0, 2( ) . Differentiability will be discussed in Section 1.11. §


PART D: FORMS OF DIFFERENCE QUOTIENTS

The purposes of these forms will be discussed in Section 1.11 on derivatives.

Forms of Difference Quotients

Form 1: Fixed interval

f b( ) f a( )b a

a, b are constants

Form 2: Variable endpoint (x)

f x( ) f a( )x a

a is constant;

x is variable

Form 3: Variable run (h)

f a + h( ) f a( )h

a is constant;

h is variable

Form 4: Variable endpoint (x) and Variable run (h)

f x + h( ) f x( )h

x, h are variable

• The denominators can be negative. (See Footnote 2 in Section 1.11.)


PART E: EVALUATING DIFFERENCE QUOTIENTS

Example 2 (Form 1 of a Difference Quotient; Profit)

A company sells widgets. Assume that all widgets produced are sold.

Let P be the profit function for the company; P x( ) is the profit (in dollars)

if x widgets are produced and sold. Our model: P x( ) = x2

+ 200x 5000 .

Find the average rate of change of profit between 60 and 90 widgets.

WARNING 2: We will treat the domain of P as 0, ) , even though one

could argue that the domain should only consist of integers. Be aware of this

issue with applications such as these.

§ Solution

P 90( ) P 60( )90 60

=

90( )2

+ 200 90( ) 5000 60( )2

+ 200 60( ) 5000

30

WARNING 3: Grouping symbols are essential when

expanding P 60( ) here, since we are subtracting an

expression with more than one term.

=

8100 +18,000 5000 3600 +12,000 5000

30

=8100 +18,000 5000 + 3600 12,000 + 5000

30

TIP 2: Instead of completely evaluating P 90( ) and

P 60( ) , it may help to take advantage of cancellations

such as the one above.

=1500

30

= 50dollars

widget


• The practical interpretation is that, if the company increases its production

level from 60 widgets to 90 widgets, then its profit will increase by

50dollars

widget, on average.

• The total increase in profit is $1500 over the 30 additional widgets.

• The graph of y = P x( ) is below.

•• The slope of the orange secant line is also

50dollars

widget.

•• The fact that the slope is positive reflects the fact that the profit

function is increasing on

60, 90 . We will revisit this idea in

Section 1.11 on derivatives. §


Example 3 (Form 2 of a Difference Quotient; Revisiting Example 2 on Profit)

Again, P x( ) = x2+ 200x 5000 . Evaluate

P x( ) P 60( )x 60

. (Form 2)

§ Solution

P x( ) P 60( )x 60

=

x2+ 200x 5000 60( )

2

+ 200 60( ) 5000

x 60

=x2

+ 200x 5000 3600 +12,000 5000

x 60

=x2

+ 200x 5000 + 3600 12,000 + 5000

x 60

=x2

+ 200x 8400

x 60

=x2 200x + 8400( )

x 60

• The Factor Theorem in Chapter 2 will imply that,

if P is polynomial, then

x 60( ) is a factor of the

numerator, which is equivalent to P x( ) P 60( ) .

This is because 60 is a zero of the numerator:

P 60( ) P 60( ) = 0 .

=x 60( ) x 140( )

x 60( )1

, x 60( )

= x 140( ) , x 60( )= 140 x, x 60( )


• The unit for the difference quotient is still

dollars

widget, though we tend to omit

it when a variable is present in our final expression.

• If we substitute x = 90 , we obtain 50

dollars

widget, our answer to Example 2.

§



Again, P x( ) = x2

+ 200x 5000 . Evaluate

P 60 + h( ) P 60( )h

. (Form 3)

§ Solution

P 60 + h( ) P 60( )h

=

60 + h( )2

+ 200 60 + h( ) 5000 60( )2

+ 200 60( ) 5000

h

=3600 +120h + h2( ) +12,000 + 200h 5000 3600 +12,000 5000

h

=3600 120h h2

+ 12,000 + 200h 5000 + 3600 12,000 + 5000

h

=80h h2

h

=h 80 h( )

h1

, h 0( )

= 80 h, h 0( )

• If we substitute h = 30 , which corresponds to x = 90 , we obtain

50 dollars

widget, our answer to Example 2. §



Again, P x( ) = x2

+ 200x 5000 . Evaluate

P x + h( ) P x( )h

. (Form 4)

§ Solution

P x + h( ) P x( )h

=

x + h( )2

+ 200 x + h( ) 5000 x2+ 200x 5000

h

=x2

+ 2xh + h2( ) + 200x + 200h 5000 x2+ 200x 5000

h

=x2 2xh h2

+ 200x + 200h 5000 x2+ 200x 5000

h

=x2 2xh h2

+ 200x + 200h 5000 + x2 200x + 5000

h

=2xh h2

+ 200h

h

=h 2x h + 200( )

h1

, h 0( )

= 2x h + 200, h 0( )

• If we substitute x = 60 and h = 30 (which then corresponds to x = 90 ),

we obtain 50

dollars

widget, our answer to Example 2. §


FOOTNOTES

1. Assumptions made about a function. When defining the average rate of change of a

function f on an interval

a, b , where a < b , sources typically do not state the assumptions

made about f . The formula

f b( ) f a( )b a

seems only to require the existence of f a( ) and

f b( ) , but we typically assume more than just that.

• Although the slope of the secant line on a, b can still be defined, we need more for

the existence of derivatives (i.e., the differentiability of f ) and the existence of

non-vertical tangent lines, as we will see in Section 1.11.

• We ordinarily assume that f is continuous on

a, b . Then, there are no holes, jumps,

or vertical asymptotes on a, b when f is graphed. (See Section 1.5.)

• We may also assume that f is differentiable on

a, b . Then, the graph of f makes no

sharp turns and does not exhibit “infinite steepness” (corresponding to vertical tangent

lines). However, this assumption may lead to circular reasoning, because the ideas of

secant lines and average rate of change are used to develop the ideas of derivatives,

tangent lines, and instantaneous rate of change, as we will see in Section 1.11.

Differentiability is defined in terms of the existence of derivatives.

• We may also need to assume that f , the derivative of f , is continuous on a, b .

Then, the average rate of change of f on

a, b is equal to the average value of f on

a, b . In calculus, we will assume that a function (say, g) is continuous on a, b and

then define the average value of g on a, b to be

g x( ) dxa

b

b a; the numerator is a

definite integral, which will be defined as a limit of sums in calculus (see Section 9.8).

Then, the average value of f on

a, b is given by:

f x( )dxa

b

b a, which is equal to

f b( ) f a( )b a

by the Fundamental Theorem of Calculus. The theorem assumes that the

integrand [function], f , is continuous on

a, b .

(Section 1.11: Limits and Derivatives in Calculus) 1.11.1

SECTION 1.11: LIMITS AND DERIVATIVES IN CALCULUS

LEARNING OBJECTIVES

• Be able to develop limit definitions of derivatives and use them to find derivatives.

• Understand the graphical interpretation of a derivative as the slope of a tangent line.

• Understand the practical interpretation of a derivative as an instantaneous rate of

change, possibly velocity.

• Be able to find equations of tangent lines to graphs.

• Relate derivatives to local linearization of, and marginal change in, a function.

• Use derivatives to determine where a function is increasing, decreasing, or constant.

PART A: DISCUSSION

• In Section 1.5, we discussed limits. In Section 1.10, we saw how difference

quotients represented slopes of secant lines and average rates of change.

• We will now define derivatives as limits of difference quotients. Derivatives

represent slopes of tangent lines (which model or approximate graphs locally) and

instantaneous rates of change.

• Difference quotients on “small” intervals might be used to approximate

derivatives. Conversely, derivatives might be used to approximate marginal change

in a function, an idea used in economics.

• In calculus, we will use derivatives to help us graph functions by finding where

they are increasing, decreasing, or constant. (See Section 1.2.)

• Limits, differentiation (the process of taking derivatives), and integration

(which reverses differentiation; see Section 9.8) are the three key topics you will

find in the first half of a calculus book, and they will be key themes throughout.


PART B: TANGENT LINES and DERIVATIVES

In Section 1.10, we looked at four forms of difference quotients.

Form 1 gave us the average rate of change of a function f on a fixed interval

a, b , where a < b . This is equal to the slope of the secant line on a, b .

Form 1: Fixed interval

f b( ) f a( )b a

a, b are constants

The other forms dealt with variable intervals. Each form, together with the idea of

limits from Section 1.5, leads to a definition of derivative.

Form 2: Variable endpoint (x)

f x( ) f a( )x a

a is constant;

x is variable

• If we let x approach a denoted by: x a( ) , the corresponding secant lines

approach the red tangent line in the following figures.

• This “limiting process” makes the tangent line a creature of calculus, not

just precalculus. (See Footnote 1 on etymology.)


Below, we let x approach a Below, we let x approach a

from the right x a+( ) . from the left x a( ) .

(See Footnote 2.)

• We define the slope of the tangent line to be the (two-sided) limit of the

difference quotient as x approaches a, if that limit exists.

• We denote this slope by f a( ) , read as “ f prime of (or at) a.”

f a( ) , the derivative of f at a, is the slope of the tangent line to the graph

of f at the point a, f a( )( ) , if that slope exists (as a real number).

\

• If f a( ) exists, we then say that f is differentiable at a.

Limit Definition of the Derivative at a Point (Based on Form 2)

f a( ) = lim

x a

f x( ) f a( )x a

• The term “point” here technically refers to the real number a, which

corresponds to a point on the real number line.


Form 3: Variable run (h)

f a + h( ) f a( )h

a is constant;

h is variable

• If we let the run h approach 0 denoted by: h 0( ) , the corresponding

secant lines approach the red tangent line below.

Below, we let h approach 0 Below, we let h approach 0

from the right h 0+( ) . from the left

h 0( ) .

(See Footnote 2.)

Limit Definition of the Derivative at a Point (Based on Form 3)

f a( ) = lim

h 0

f a + h( ) f a( )h


Form 4: Variable endpoint (x) and Variable run (h)

f x + h( ) f x( )h

x, h are variable

• Again, if we let h 0 , the corresponding secant lines approach the red

tangent line below.

• The figures are similar to the ones for Form 3, except that x replaces a.

Below, we let h approach 0 Below, we let h approach 0

from the right h 0+( ) . from the left

h 0( ) .

(See Footnote 2.)

• f x( ) provides a rule for f , the derivative function (or simply derivative)

of f . TIP 1: Think of f as a slope function.

Limit Definition of the Derivative Function (Based on Form 4)

f x( ) = lim

h 0

f x + h( ) f x( )h


PART C: INSTANTANEOUS RATE OF CHANGE and

INSTANTANEOUS VELOCITY

The instantaneous rate of change of f at a is equal to f a( ) , if it exists.

The following development of instantaneous velocity will help explain the

association between the slope of a tangent line and instantaneous rate of change.

Example 1 (Velocity)

A car is driven due north for two hours, beginning at noon.

How can we find the instantaneous velocity of the car at 1pm?

(If this is positive, this can be thought of as the speedometer reading

at 1pm.)

Re Example 1: Definitions

• Let t = the time (in hours) elapsed since noon.

• Let y = s t( ) , where s is the position function for the car (in miles).

In Section 1.10, we saw that:


a, b is the

average rate of change of position with respect to time. That is,

change in position

change in time=

s

t

=s b( ) s a( )

b a

We will now consider average velocities on variable time intervals of the

form

a, a + h , if h > 0 , or the form

a + h, a , if h < 0 ,

where h is a variable run. (We can let h = t .)


Using Form 3 of a difference quotient,


a, a + h , if h > 0 , or

a + h, a , if h < 0 , is given by:

s a + h( ) s a( )h

• This equals the slope of the secant line to the graph of s on the interval.

• (See Footnote 2 on the h < 0 case.)

Let’s assume there exists a non-vertical tangent line to the graph of s at the

point a, s a( )( ) .

• Then, as h 0 , the slopes of the secant lines will approach the

slope of this tangent line, which is s a( ) .

• Likewise, as h 0 , the average velocities will approach the

instantaneous velocity at a.

Below, we let h 0+. Below, we let h 0 .

The instantaneous velocity (or simply velocity) at a is given by:

s a( ) , or v a( ) = lim

h 0

s a + h( ) s a( )h


Using Form 4 of a difference quotient,

The velocity function v is defined by: v t( ) = lim

h 0

s t + h( ) s t( )h

Re Example 1: Numerical approaches; Approximating derivatives

Let’s say the position function s is defined by: s t( ) = t3 on

0, 2 .

We want to find v 1( ) , the instantaneous velocity of the car at t = 1 or a = 1.

We will first consider average velocities on intervals of the form 1,1+ h .

Here, we let h 0+.

Interval Value of h

(in hours) Average velocity,

s 1+ h( ) s 1( )h

1, 2 1

s 2( ) s 1( )1

= 7 mph

1,1.1 0.1

s 1.1( ) s 1( )0.1

= 3.31 mph

1,1.01 0.01

s 1.01( ) s 1( )0.01

= 3.0301 mph

1,1.001 0.001

s 1.001( ) s 1( )0.001

= 3.003001 mph

0+ 3 mph

• These average velocities approach 3 mph, which is v 1( ) .

The reader can verify that v 1( ) = 3 mph in the Exercises.

Part D will help.

• WARNING 1: Tables can sometimes be misleading. The table here

does not represent a rigorous evaluation of v 1( ) . Answers are not

always integer-valued.

• If h is “close” to 0, we may be able to use the corresponding average

velocity as an approximation for v 1( ) .


We could also consider this approach:

Interval Value of h Average velocity,

s 1+ h( ) s 1( )h

(rounded off to six significant digits)

1, 2 1 hour

7.00000 mph

1,11

60 1 minute 3.05028 mph

1,11

3600 1 second 3.00083 mph

0+ 3 mph

Here, we let h 0 .

Interval Value of h

(in hours) Average velocity,

s 1+ h( ) s 1( )h

0,1 1

s 0( ) s 1( )1

= 1 mph

0.9,1 0.1

s 0.9( ) s 1( )0.1

= 2.71 mph

0.99,1 0.01

s 0.99( ) s 1( )0.01

= 2.9701 mph

0.999,1 0.001

s 0.999( ) s 1( )0.001

= 2.997001 mph

0 3 mph

• Because of the way we normally look at slopes, we may prefer

to rewrite the difference quotient

s 0( ) s 1( )1

as

s 1( ) s 0( )1

,

and so forth. (See Footnote 2.) §


PART D: FINDING DERIVATIVES

Example 2 (Profit; Revisiting Examples 2-5 in Section 1.10)

A company sells widgets. Assume that all widgets produced are sold.

Let P be the profit function for the company; P x( ) is the profit (in dollars)

if x widgets are produced and sold. Our model: P x( ) = x2

+ 200x 5000 .

Find the instantaneous rate of change of profit at 60 widgets.

• We will ignore integer restrictions on x.

§ Solution

We want to find P 60( ) . We will present three solutions based on three

different forms of difference quotients we developed in Section 1.10.

Form 2

P 60( ) = limx 60

P x( ) P 60( )x 60

= limx 60

140 x( ) , x 60( ) by Ex.3 in Section 1.10( )

• As x approaches 60, we see that 140 x approaches

140 60( ) , or 80. We can substitute x = 60 directly; the

restriction

x 60( ) is irrelevant here.

= 140 60( )

= 80 dollars

widget

• This is the slope of the red tangent line below.


Form 3

P 60( ) = limh 0

P 60 + h( ) P 60( )h

= limh 0

80 h( ) , h 0( ) by Ex.4 in Section 1.10( )

= 80 0( )

= 80 dollars

widget

Form 4

First, find the formula for P x( ) , the rule for the derivative of P.

P x( ) = limh 0

P x + h( ) P x( )h

= limh 0

2x h + 200( ) , h 0( ) by Ex.5 in Section 1.10( )

= 2x 0( ) + 200

= 2x + 200

Now, evaluate P 60( ) .

P 60( ) = 2 60( ) + 200

= 80 dollars

widget

Form 3 Form 4

§


PART E: EQUATIONS OF TANGENT LINES

Example 3 (Profit; Revisiting Example 2)

Again, P x( ) = x2

+ 200x 5000 . Find Slope-Intercept and Point-Slope

Forms of the equation of the tangent line to the graph of y = P x( ) at the

point 60, P 60( )( ) . (Review Section 0.14 on these forms.)

§ Solution

• Find P 60( ) , the y-coordinate of the desired point.

P 60( ) = 60( )2

+ 200 60( ) 5000

= 3400 dollars( )

• Find P 60( ) , the slope of the desired tangent line.

In Part D, we showed (three times) that:

P 60( ) = 80 dollars

widget

• Find a Point-Slope Form of the equation of the tangent line.

y y1= m x x

1( )y 3400 = 80 x 60( )

• Find the Slope-Intercept Form of the equation of the tangent line.

y = 80x 1400

• Note: Some sources would use P instead of y as the dependent variable.

§


PART F: LINEARIZATION and MARGINAL CHANGE

• The tangent line to the graph of a function f at the point a, f a( )( ) represents

the best linear approximation to the function close to a. The tangent line model

linearizes the function locally around a.

• The principle of “local linearity” states that, if the tangent line exists, the graph

of f resembles the line if we “zoom in” on the point a, f a( )( ) .

Example 4 (Marginal Profit; Revisiting Examples 2 and 3)

If P is a profit function, the derivative P is referred to as a marginal profit

(MP) function. It approximates the change in profit if we increase

production by 1 unit (from 60 to 61 widgets, for instance). That is,

P 60( ) P 61( ) P 60( ) .

• This is based on the idea that the tangent line to the graph of P at

60, P 60( )( ) represents the best linear approximation to P “close to”

x = 60 (or a = 60 ).

• P 61( ) P 60( ) is the actual rise along the graph of P as x changes from 60

to 61. We will approximate this by the rise along the tangent line as x

changes from 60 to 61. This rise is given by P 60( ) , the slope of the tangent

line, because, if run = 1, slope =

rise

run= rise . (We saw this in Section 0.14.)

• (The blue graph of P below has been distorted for convenience. The true

graph is very close to the red tangent line.)

§


PART G: GRAPHING FUNCTIONS USING DERIVATIVES

Example 5 (Profit; Revisiting Example 2)

Again, P x( ) = x2

+ 200x 5000 . Graph P.

§ Solution

We will discuss how to graph quadratic functions such as P in Section 2.2.

For now, we can use the derivative function P to help us graph P.

• In Part D (Example 2, Form 4), we found that P x( ) = 2x + 200 .

Think of this as a “slope function” that gives us slopes of tangent lines.

• Find any point(s) on the graph where the tangent line is horizontal;

its slope is 0.

Solve:

P x( ) = 0

2x + 200 = 0

x = 100

There is a horizontal tangent line at the point 100, P 100( )( ) , or

100, 5000( ) .

WARNING 2: Evaluate P 100( ) , not

P 100( ) , to obtain the

y-coordinate of the point. What is P 100( ) ?

•• In calculus, we call 100 a critical number of P, because it is a

number in Dom P( ) where P is either 0 in value or is undefined.

We are interested in critical numbers, because those are the possible

x-coordinates of turning points on the graph of P.

WARNING 3: Some precalculus sources use the term critical

number differently; it is where P, itself, is either 0 in value or is

undefined. P values could change sign around critical numbers

and discontinuities.


• Find x-coordinates of points where the tangent lines slope upward;

their slopes are positive.

Solve:

P x( ) > 0

2x + 200 > 0

2x > 200

x < 100

Therefore, P is increasing on the x-interval 0,100( ) . In fact, we can

say that P is increasing on

0,100 by a continuity argument.

(See Section 1.2, Part H on when a function increases on an interval.)

• Find x-coordinates of points where the tangent lines slope downward;

their slopes are negative.

Solve:

P x( ) < 0

2x + 200 < 0

2x < 200

x > 100

Therefore, P is decreasing on the x-interval 100,( ) . In fact, we can

say that P is decreasing on 100, ) by a continuity argument.

Here is the graph of P; the horizontal tangent line is in red:

§


FOOTNOTES

1. Etymology. The word “secant” comes from the Latin “secare” (to cut). The word “tangent”

comes from the Latin “tangere” (to touch). A secant line to the graph of f must intersect it in

at least two points. A tangent line only need intersect the graph in one point; the tangent line

might “just touch” the graph at that point (though there could be other intersection points).

2. Difference quotients with negative denominators. Our forms of difference quotients allow

negative denominators, as well. They still represent slopes of secant lines.

Form 1 Form 2

• Form 1 b < a( )

slope =rise

run=

f a( ) f b( )a b

=f b( ) f a( )

b a( )=

f b( ) f a( )b a

• Form 2

x < a( )

slope =rise

run=

f a( ) f x( )a x

=f x( ) f a( )

x a( )=

f x( ) f a( )x a

Form 3 Form 4

• Form 3

h < 0( )

slope =rise

run=

f a( ) f a + h( )a a + h( )

=f a + h( ) f a( )

h=

f a + h( ) f a( )h

• Form 4

h < 0( )

slope =rise

run=

f x( ) f x + h( )x x + h( )

=f x + h( ) f x( )

h=

f x + h( ) f x( )h

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