Welcome to Precalculus

• Although there will not be a lecture over Appendix A.1, it will be assumed that you are familiar with the material covered there. Make sure you read and understand Appendix A.1.

• Maple software will often be used in class to plot graphs. You will not be responsible for learning Maple unless you want to. Maple is available on computers in math labs on campus.

Pythagoras' Theorem

• For a right triangle with hypotenuse of length c and sides of lengths a and b, it is true that a2 + b2 = c2. The converse is also true. That is, if a2 + b2 = c2 for some labeling of the sides of a triangle, then the triangle is a right triangle.

a

b

c

The Distance Formula• The distance d between the points (x1, y1) and (x2, y2) in the

plane is

• Example. Show that the triangle with vertices (–2,3), (3,–2), and (6, 1) is a right triangle.

• If we can verify that the sum of the squares of the sides equals the square of the hypotenuse, then this is a right triangle, and it follows since

.yyxxd 212

212

.68 )31()26(d

18 )21()36(d

50)32()23(d

223

222

221

.ddd 23

22

21

The Midpoint Formula

• The coordinates (xm, ym) of the midpoint M of the line segment

with endpoints (x1, y1) and (x2, y2) are given by

• Problem. Find the coordinates of the midpoint of the line segment with endpoints (2,1) and (2,4). Solution. The midpoint is (2, 5/2).

.2

yyy ,

2

xxx 21

m21

m

Translating points in the plane• A square is shown in the figure. Find the coordinates of the

vertices of the square shifted 2 units to the right and 1 unit downward.

Original vertex (x, y). Translated vertex (x+2, y–1).

(1,2) (2,2)

(1,3) (2,3)

(3,1)

Original Square

Translated Square

(4,1)

(4,2)(3,2)

x

y

Sketching a graph

• Sketch the graph of –x2 + y + x =0.

• First, we solve for y: y = x2 – x.

• Next, we make a table of values.

• Next, we plot the points and connect them with a smooth line.

x –2 –1 0 1 2 3

y = x2–x 6 2 0 0 2 6

(x, y) (–2, 6) (–1,2) (0, 0) (1, 0) (2,2) (3, 6)

Intercepts of a graph A point at which a graph has 0 as the y-coordinate is called an

x-intercept. Likewise, a point at which a graph has 0 as the x-coordinate is called a y-intercept. Sometimes we will refer to the nonzero coordinate as the x-intercept, y-intercept, resp.

To find x-intercepts, set y to zero and solve for x. To find y-intercepts, set x to zero and solve for y.

• Example. The graph of y = (x+2)2–1 is shown below.From the graph, we see that the x-intercepts are (–1,0) and (–3,0), and the y-intercept is (0,3).

Algebraic Tests for Symmetry

• The graph of an equation is symmetric with respect to the (i) x-axis if replacing y with –y results in an equivalent equation.

(ii) y-axis if replacing x with –x results in an equivalent equation.

(iii) origin if replacing x with –x and y with –y results in an equivalent equation.

• Example. Test the equation xy = 5 for symmetry. Since (x)(–y) = 5 is not equivalent, it is not symmetric wrt x-axis.

Since (–x)(y) = 5 is not equivalent, it is not symmetric wrt y-axis.

Since (–x)(–y) = 5 is equivalent, it is symmetric wrt the origin.

Circles• A point (x, y) is on a circle with center (h, k) if and only if its

distance from the center is equal to the radius r. By the distance formula,

• If we square both sides of the latter equation, we get the standard form of the equation of a circle with center (h, k) and radius r.

r.kyhx 22

.rkyhx 222

If h = k = 0 and r = 1, we get the equation of the unit circle which will be important later when we study trigonometric functions.

CircleUnit

The slope-intercept form of a linear equation

• The slope-intercept form is

where m is the slope and b is the y-intercept.

• The y-intercept, b, tells us where the line crosses the y-axis. If the slope, m, is positive, the line climbs from left to right. If the slope, m, is negative, the line falls from left to right.

• The larger the value of m (either positive or negative), the steeper the graph of the linear equation y = mx +b.

• If (x0, y0) and (x1, y1) are two distinct points on the graph of a line, then

b,mx y

.xx

yy

in x change

yin changem

01

01

Slope as a rate of change

• The slope of a nonvertical line is the number of units the line rises (or falls) vertically for each unit of horizontal change from left to right.

x

y

1 unit

m units, m>0

x

y1 unit

m units, m<0

Finding a linear equation from a table

• Suppose the value of a Batman comic book is increasing as follows:

• If we let year 2000 correspond to t = 0, the table becomes:

• Since is constant,

we let m = and the equation is:

Year 2000 2002 2004 2006

Value ($) 65 90 115 140

t (year) 0 2 4 6

v ($) 65 90 115 140

in t) (changein v) (change

,)02()6590(

65.12.5t v

Usage of duplicating paper at Lee High School

• At present there are 400 packages of duplicating paper available. Each week 12 packages are used. A table is shown next which gives the number of packages left, L, versus the number of weeks from now, w.

• The relation between L and w is linear. Such a linear relation can be given a formula of the type:

where m is the slope or rate of change and b is the vertical intercept. Can you tell what the values of m and b are in this situation? What is the significance of the horizontal intercept?

w 1 2 3 4 5 6 7L 388 376 364 352 340 328 316

b,mw L

Other equations of lines

• The point-slope form is

where m is the slope and (x0, y0) is a point on the line.

• The general form is

where A, B, and C are constants.

),– xm(x y–y 00

0, CBy Ax

Useful facts about equations of lines• For any constant k:

The graph of the equation y = k is a horizontal line and its slope is zero.

The graph of the equation x = k is a vertical line and its slope is undefined.

• Let L1 and L2 be two lines having slopes m1 and m2, respectively. Then:

These lines are parallel if and only if m1 = m2.

These lines are perpendicular if and only if

.m

1– m

21

Use of Maple to plot perpendicular lines

> plot({2*x,(–1/2)*x+5}, x=0..4,3..5, color=black, scaling=constrained);

• Example. Use the point-slope form of the line to derive the equation which converts temperature in degrees Celsius, C, to degrees Fahrenheit, F. We are given that the slope is 9/5 and that (20,68) is a point on the line. That is, C0 = 20 and F0 = 68. Using the point-slope form, we have:

• The equation may be rewritten in slope-intercept form as:

or ),C– C(5

9 F– F 00

.)20– C(5

9 68– F

.32 C5

9 F

Use of Maple to graph temperature conversion

> plot((9/5)*C+32,C=0..40,color=black,labels=["C","F"]);

Common formulas for Area A, Perimeter P, and Circumference C

• For a rectangle: A = lw and P = 2l + 2w

• For a circle: A = r2 and C = 2r

l

w

r

Introduction to functions

• When two quantities are related to each other by some rule of correspondence, we call the correspondence a relation.

• A function f from a set A to a set B is a relation that assigns to each element x from A exactly one element y from the set B. The set A is the domain (or set of inputs) of f, and the set B contains the range (or set of outputs).

• Example. A certain child’s height (in inches) on his birthday is a function of his age (in years):

Age 1 2 3 4 5Height 24 30 37 41 48

Understanding the function definition

• This represents a function:

• This does not represent a function from A to B. (Why?)

Domain = {1, 2} = A

Range = {1, 5}

is subset of B

1

2

1

3

5

1

3

5

1

2

A B

Testing for functions represented algebraically

• Which of these equations represents y as a function of x?

• Solution.a. If x is known, y = x2 is determined

uniquely, so this is a function.

b. If x is known, there are two values for y, namely, so this is not a function.

xy

xy 2

2

b.

a.

,x

Four ways to represent a function• Each batch of 5 dozen sugar cookies requires 2 and one-

half cups of flour (verbally).

• We may also use a table (numerically).

• Graphically:

• Algebraically: F = 2.5B

No. of Batches (B) 1 2 3No. of Cups Flour (F) 2.5 5.0 7.5

1 2 3 B

F

7.5

5.0

2.5

• To indicate that a quantity y is a function of a quantity x, we abbreviate to:

y equals f of x and, using function notation, to:

• Here, y is the dependent (or output) variable and x is the independent (or input) variable.

• In the previous cookie example, F is the dependent variable and B is the independent variable, and we can write

F = f(B) = 2.5B Note that we could use another letter instead of f. How about c for cookie? Then

F = c(B) = 2.5B

f(x) y

• Example of quantities which are related, but neither quantity is a function of the other.

• Both F and R are functions of t. However, F is not a function of R, and R is not a function of F (do you see why?). In other words, if we know which month is being discussed, we can determine the values of F and R uniquely. However, if we only know the value of F, then the value of R may not be determined uniquely. Similarly, if we only know the value of R, then F may not be uniquely determined.

t, month 1 2 3 4 5 6 7

R, no.rabbits 750 567 500 567 750 1000 1250

F, no.foxes 143 125 100 75 57 50 57

Evaluating a function

• Evaluating a function means figuring out the value of a function’s output from a particular value of the input.

• Example. Let the function g be defined by:

.x x g(x) 2

2ab

bbaa)ba)bab2(a

bbaa)ba(b)(ag(b) g(a) b)g(a

g(b). g(a) b)g(asimplify and Evaluate

2222

222

.633 g(3) Evaluate 2

Evaluating functions using a table

• Suppose that f is defined by the table:

• To find f(3), we look in the table and get f(3) = –4.

• Now define g(x) = f(x+1). We evaluate g(3) = f(4) = –3. The table for g appears below:

• Why is no value listed for g(4)? Why is g defined at x = –2 while f is not?

x –1 0 1 2 3 4

f(x) 2 1 –5 2 –4 – 3

x –2 –1 0 1 2 3 4

g(x) 2 1 –5 2 –4 – 3 --

• Given an input, we evaluate a function to find the output. Often the situation is reversed; we know the output value and we want to find the corresponding input value(s). If the function is given by a formula, the input values are solutions to an equation.

• Problem. Let A = f(r) be the area of a circle of radius r, where r is in cm. What is the radius of a circle whose area is 100 cm2 ? Solution. The output f(r) is an area. Solving the equation f(r) = 100 for r gives us the radius of a circle whose area is 100 cm2. Since the formula for the area of a circle is

we solve This yields and we take the positive

value.

,r f(r) 2 .100r f(r) 2 5.64, r

Finding input and output values using a table

• Suppose that f is defined by the table:

• As before, to find f(1), we look in the table and get f(1) = –5.

• Now suppose we want to solve f(x) = 2 for x.

There are two values for x which satisfy this condition,

namely, x = –1 and x = 2.

x –1 0 1 2 3 4

f(x) 2 1 –5 2 –4 – 3

Input and output from a graph showing an influenza epidemic

• We have I = f(w), where I is the number of individuals infected (in thousands) w weeks after the epidemic begins.

• Evaluate f(2) and explain its meaning.• Solve f(w) = 4.5 and explain the meaning of the solution.

Use the graph of f to the right to find or estimate u:

(a) f(4) = uIs u an input or an output?

(b) f(f(8)) = uIs u an input or an output?

(c) f(u) = 24 Is u an input or an output?

More on input and output from the graph of a function

Evaluating a difference quotient--important in calculus

• For a given function f, the ratio

is called a difference

quotient.

• Evaluate the difference quotient for f(x) = x2.

0h ,h

f(x)h)f(x

h2xh

h)h(2xh

hxh2

h

x)hxh2(x

h

xh)x(

h

f(x)h)f(x

2222

22

Evaluating a piecewise defined function

• Evaluate the function f at x = –2, x = 0, x = 3.

• Solution.

0x,x

0 x5,

0 x2x,

f(x)2

4)2f(

5f(0)

9f(3)

Suppose an employee is paid $5.00 per hour to work a standard 40 hour work week. If he works overtime, he is paid $7.50 per hour up to a maximum of 80 hours. The graph below shows his weekly pay as a function, f(t), of the time worked.

p2

p1

What are the values of p1 and p2?

Pay function in bracket form(80,500)

(40,200)

f(t) = 80t40 100,7.5t

40t,0 t 5

Domain of a function

• The domain of a function can be described explicitly:

Domain ={1, 2, 3, 4}

• The domain may be determined from context: Area of a circle as a function of the radius is g(r) = r2, and the domain is all real numbers r > 0.

• The domain may be implied. That is, it is all real numbers for which the expression involved is defined. For example:

x 1 2 3 4

f(x) 5 7 –2 3

). [2, 2 xall isDomain ,2xs(x)

). ,1()1 ,( 1 x all isDomain ,1x

1r(x)

or

or

The graph of a function• The graph of a function f is the collection of ordered pairs

(x, f(x)) such that x is in the domain of f.

• The graph of f can be visualized by plotting it in a coordinate plane.

• Example. f(x) = x3–x has the graph shown below.

Finding the domain and range for a function whose graph is given

Domain

Range

Domain = [1.5, 5.5)

Range = [1,3]

Vertical Line Test for functions

• A set of points in a coordinate plane is the graph of y as a function of x if and only if no vertical line intersects the graph at more than one point.

• In which of the graphs below could y be a function of x?

• Clearly, the one on the right fails the vertical line test, so this graph does not represent y as a function of x.

x x

y y

Zeros of a function

• The zeros of a function f of x are the x-values for which f(x) = 0.

• In terms of the graph of f, the zeros are the x-coordinates of the points where the graph crosses the x-axis.

• To find the zeros of f, set f(x) = 0 and solve for x.

• Example. Find the zeros of f(x) = x2 – x – 2.

.1 xand 2 xare zeros theso

01)x)(2x()2xx( 2

Increasing, decreasing, and constant functions• A function f is increasing on an interval if for any x1 and x2

in the interval,

• A function f is decreasing on an interval if for any x1 and x2

in the interval,

• A function f is constant on an interval if for any x1 and x2 in the interval,

).f(x)f(x implies xx 2121

).f(x)f(x implies xx 2121

).f(x)f(x 21

Decreasing

on (– ,0) Constant

on (0,2)

Increasing

on (2,)

Relative minimum and relative maximum

• A function value f(a) is called a relative minimum of f if there exists an interval (x1, x2) that contains a such that

• A function value f(a) is called a relative maximum of f if there exists an interval (x1, x2) that contains a such that

).f(x)f(a implies xxx 21

).f(x)f(a implies xxx 21

x

y

Relative maximum

Relative minima

Average rate of change• For a function f, the average rate of change between two

points is the slope of the line through the two points. This line is called the secant line and its slope is denoted

))f(x,x( and ))f(x,x( 2211

.msec

))f(x,x( 11

))f(x,x( 22 Secant line

1x 2x

)f(x)f(x 12

12 xx

12

12sec xx

)f(x)f(xm change of rate Average

)f(xy

Tests for even and odd functions

• A function is called an even function if, for all values of x in the domain of f,

The graph of an even function is symmetric across the y-axis. Examples of even functions are power functions with even exponents, such as y = x2, y = x4, y = x6, ...

• A function is called an odd function if, for all values of x in the domain of f,

The graph of an odd function is symmetric about the origin. Examples of odd functions are power functions with odd exponents, such as y = x1, y = x3, y = x5, ...

f(x). x)f(

f(x). x)f(

• Problem. Is the function f(x) = x3+x even, odd, or neither?

Solution. Since –2 = f(–1) is not equal to f(1) = 2, it follows that f is not even.

Since f(–x) = = –f(x), it follows that f is odd.

xxx)(x)( 33

y = x3+x

Note the symmetry about the origin.

• Problem. Is the function f(x) = |x| even, odd, or neither?

Solution. Since f(–x) = |x| = f(x), it follows that f is even.

Since 1 = f(–1) is not equal to –f(1) = –1, it follows that f is not odd.

• Question. Is it possible for a function to be both even and odd?

y = |x|

Note the symmetry about the y-axis.

A library of parent functions

• Next, eight functions which are the most commonly used functions in algebra will be presented.

• These functions will be called parent functions because they can be used to create many other functions using transformations which will be introduced later.

• Memorize the names and characteristics of these functions.

The constant function

• Domain is all real numbers

• Range is {c}

• The function is even for all c.

• The function is odd when c = 0.

cf(x)

The identity function

• Domain is all real numbers

• Range is all real numbers

• The function is odd

xf(x)

The absolute value function

• Domain is all real numbers

• Range is all nonnegative real numbers

• The function is even

xf(x)

The squaring function

• Domain is all real numbers

• Range is all nonnegative real numbers

• The function is even

2xf(x)

The cubic function

• Domain is all real numbers

• Range is all real numbers

• The function is odd

3xf(x)

The square root function

• Domain is all nonnegative real numbers

• Range is all nonnegative real numbers

• The function is neither even nor odd

xf(x)

The reciprocal function

• Domain is all real numbers except 0

• Range is all real numbers except 0

• The function is odd

x

1f(x)

The greatest integer function

• Domain is all real numbers

• Range is all integers

• The function is neither even nor odd

xf(x)

Name the functions

(i) (ii)

(iii) (iv)

Name the functions

(v) (vi)

(vii) (viii)

Graphing a piece-wise defined function

• Sketch the graph of

0x2,x

0x1,x f(x)

Transformation of functions--vertical and horizontal shifts

1. Vertical shift c units upward: h(x) = f(x) + c

2. Vertical shift c units downward: h(x) = f(x) – c

3. Horizontal shift c units to the right: h(x) = f(x – c)

4. Horizontal shift c units to the left: h(x) = f(x + c)

Let c be a positive real number. Vertical and horizontal shifts in the graph of y = f(x) are represented as follows.

2xf(x)

2x(x)h 21 2xf(x)

23 )2x((x)h

Example showing that shifts can be combined

2xf(x)

1)2x((x)h 2

The squaring function is shifted right by 2 units and upward by 1 unit.

Transformation of functions--reflectionsReflections in the coordinate axes of the graph of y = f(x) are represented as follows.

1. Reflection in the x-axis: h(x) = –f(x)

2. Reflection in the y-axis: h(x) = f(–x)

xf(x)

x(x)h1

xf(x) x(x)h 2

Example showing that reflections and shifts can be combined

xf(x)

2xh(x)

Note that domain of h(x) is: .2x

Transformation of functions--nonrigid transformations

• Horizontal shifts, vertical shifts, and reflections are rigid transformations because the basic shape of the graph is unchanged. Only the position of the graph is transformed.

• Nonrigid transformations, which are introduced next, cause a change in the shape of the original graph.

Transformation of functions--vertical stretch and vertical shrink

2xf(x)

21 x5(x)h

221

2 x(x)h

Vertical nonrigid transformations of the graph of y = f(x) are represented as follows.

1. Vertical stretch h(x) = cf(x), c > 1

2. Vertical shrink h(x) = cf(x), 0 < c < 1

Transformation of fcts--horizontal stretch and horizontal shrink

Horizontal nonrigid transformations of the graph of y = f(x) are represented as follows.

1. Horizontal shrink h(x) = f(cx), c > 1

2. Horizontal stretch h(x) = f(cx), 0 < c < 1

See the next slide for examples.

• Let f(x) = 4–x2, h1(x) = 4 – (2x)2, and h2(x) = 4 – (0.5x)2. Then h1 is a horizontal shrink of f and h2 is a horizontal stretch of f.

2x4f(x) 21 2x)(4x)(h 2

2 0.5x)(4x)(h

Arithmetic combination of functions--sum function

Let f and g be two functions with overlapping domains. Then, for all x common to both domains, the sum of f and g is defined as follows. g(x)f(x)g)(x)(f

Example. f(x) = x2 + 2

g(x) = x2 + 1

(f+g)(x) = 2x2+3

1xg(x) 2

2xf(x) 2

3x2g)(x)(f 2

Arithmetic combination of functions--difference function

Let f and g be two functions with overlapping domains. Then, for all x common to both domains, the difference of f and g is defined as follows. g(x)f(x)g)(x)(f

Example. f(x) = x2 + 2 g(x) = x2 + 1 (f – g)(x) = 1

2xf(x) 2

1xg(x) 2

1g)(x)(f

Arithmetic combination of functions--product function

Let f and g be two functions with overlapping domains. Then, for all x common to both domains, the product of f and g is defined as follows. f(x)g(x)(fg)(x)

Example. f(x) = x2 + 2 g(x) = x2 + 1 (fg)(x) = (x2 + 2)(x2 + 1) = x4 + 3x2 + 2 2xf(x) 2

1xg(x) 2

2x3x(fg)(x) 24

Arithmetic combination of functions--quotient function

Let f and g be two functions with overlapping domains. Then, for all x common to both domains, the quotient of f and g is defined as follows.

0g(x) ,g(x)

f(x)x)(

g

f

Example. f(x) = x2 + 2 g(x) = x2 + 1

1x

2xx)(

g

f2

2

2xf(x) 2

1xg(x) 2

1x

2xx)(

g

f2

2

An example showing reduction in domain

• Let

• The quotient of f and g is

• The domain of f is [0, ) and the domain of g is [–1, 1].

• The intersection of these domains is [0, 1], but we must also exclude x = 1 since g(1) = 0.

• Therefore, the domain of is [0, 1).

.x1g(x) and xf(x) 2

.x1

xx)(

g

f2

g

f

Composition of functionsThe composition of the function f with the function g is

The domain of is the set of all x in the domain of g such that g(x) is in the domain of f.

.(x))f(gg)(x)(f g)(f

Example. f(x) = x2 – 1

g(x) = x +1

x2x

11)(x

1)f(xg)(x)(f

2

2

Note: In this example, the graph of is the same as the graph of f shifted left by 1 unit.

g)(f 1xf(x) 2 x2xg)(x)(f 2

Composition of Functions--Example

• Suppose we have two money machines, both of which increase any money inserted into them. Machine A doubles our money while Machine B adds five dollars. The money that comes out is described by a(x) = 2x for Machine A and b(x) = x + 5 for Machine B, where x is the number of dollars inserted. The machines can be hooked together so that the money coming out of one machine goes into the other. There are two ways of hooking up the machines which result in the formulas shown below. The first formula is while the second formula is

• Which of these two compositions would you prefer? Why?

A B$ $: 5.2x 5a(x)b(a(x))

B A$ $: 10.2x 5)2(x b(x)2a(b(x))

ab b.a

Finding the domain of a composite function--an example

1xg)(x)f(

1xg(x)

xf(x)

Domain of f : nonnegative real numbers

Domain of g : all real numbers

Numbers in domain of g such that g(x) = x – 1 is nonnegative :

Domain of [1,)

1x

:g)f(

Decomposition of a given function

If h is a given function, the problem of finding functions f and g such that h(x) = f(g(x)) is called decomposing. The function f is called the outer function and the function g is called the inner function.

Example. Find two different ways of decomposing

1.

2.

.1)2(xh(x)

1)2(xg(x) ,xf(x)

1xg(x) ,2xf(x)

Inverse functions

Let f and g be two functions such that

f(g(x)) = x for every x in the domain of g,

g(f(x)) = x for every x in the domain of f.

Under these conditions, the function g is the inverse function of the function f. The function g is denoted by f –1 (read "f-inverse"). Therefore,

f(f –1 (x)) = x and f –1(f (x)) = x.

The domain of f must be equal to the range of f –1, and the range of f must be equal to the domain of f –1.

Note: f –1 and are different.

Example. f(x) = 2x and f –1(x) = (1/2)x. Note that f –1 "undoes" what f "does".

f

1

Interchanging input and output: Inverse Functions• The roles of input and output are not necessarily fixed. In

an earlier example, we derived a function f which converts degrees Celsius, C, to degrees Fahrenheit, F. The formula for f was:

Suppose now that we know the value of F and we wish to compute the value of C. We can define a new function g such that C = g(F). For this function, F is the input and C is the output. Of course, g = f –1.

• Find a formula for the inverse function C = f –1(F). We solve the previous equation for C to obtain:

.32 C5

9f(C) F

.)32– (F9

5(F)f C 1

A function and its inverse "undo" each other.

• What happens if we convert from Celsius to Fahrenheit and then back to Celsius? Answer: We are back where we started. In terms of function composition,

• If, on the other hand, we convert from Fahrenheit to Celsius and then back to Fahrenheit, we have

C(f(C))f -1

F(F))f(f -1

A function and its inverse "undo" each other, continued.

• What happens if we convert from Celsius to Fahrenheit and then back to Celsius?

• If, on the other hand, we convert from Fahrenheit to Celsius and then back to Fahrenheit, we have

• In this example, the domain of f is equal to the range of f –1and the range of f is equal to the domain of f –1 since they are “all real numbers”.

C)3232C5

9(

9

532)f(C)(

9

5(f(C))f 1-

F32))32F(9

5(

5

932(F)f

5

9(F))f(f 1-1-

The graph of an inverse function

The graphs of a function f and its inverse function are related to each other in the following way. If the point (a, b) lies on the graph of f, then the point (b, a) must lie on the graph of f –1, and vice versa. This means that the graph of f –1 is a reflection of the graph of f in the line y = x.

If we have the graph of two functions, we can check that one is the inverse of the other by seeing if the graphs are the reflections of each other in the line y = x.

See the next slide for an example.

Verifying inverse functions graphically

• Let f(x) = x2 with domain [0, ). Then f –1(x) = The squaring function does not have an inverse unless its domain is restricted.

• We will plot the graphs of f and f –1 on the same coordinate system and show that they are reflections of each other in the line y = x.

2 x f(x)

x (x)f 1

.x

2) (4,

4) (2,3) (9,

9) (3,

xy

The Horizontal Line Test for inverse functions

• If there is a horizontal line which intersects a function’s graph in more than one point, then the function does not have an inverse. If every horizontal line intersects a function’s graph at most once, then the function has an inverse. The graph of the function f(x) = x2, which is shown below, fails the horizontal line test. Consider y = 4.

The line y = 4 intersects graph of f(x) = x2 twice.

One-to-One functions

If every horizontal line intersects a function’s graph at most once, then the function is one-to-one and the inverse function exists.

Example. f(x) = x3 + x +1. From the graph, this function seems to pass the horizontal line test and is therefore one-to-one. The inverse function is difficult to find algebraically but we can use a calculator to trace along the graph to find individual values of f –1.

1xxf(x) 3

4y

213.1)4(f 1

Finding an inverse function algebraically

1. Use the horizontal line test to decide whether f has an inverse function.

2. In the equation for f(x), replace f(x) by y.

3. Interchange the roles of x and y, and solve for y.

4. Replace y by f –1(x) in the new equation.

5. Check your work by showing that the domain of f is equal to the range of f –1, the range of f is equal to the

domain of f –1, and f(f –1 (x)) = x and f –1(f

(x)) = x.

An example for finding an inverse function

• Consider the rectangle shown (which is defined only when x > 0)

• Let y = perimeter of rectangle = f(x) = 2x +2 where the domain of f is (0, ). The graph of f is shown next.

xx

1

1

Finding the inverse function algebraically

Consider f(x) = 2x +2 with domain (0, ) and range (2, ).

1. f(x) = 2x + 2 write original function

2. y = 2x + 2

3.

solve for y1x y

22y x

21

interchange x and y

replace f(x) by y

4. 1x x)(f 211 replace y by f –1(x)

x12)x2( f(x))(f

x,2)1x(2(x))f(f 5.

211

211

see that inverse formula works

Note: Domain of f –1 is (2, ) and range of f –1 is (0, ) .

In terms of the rectangle, how do you interpret f –1(100) = 49?

Graphs of f and f –1

0 x 2,2x y

2 x ,1x y 21

False Inverse function• If we are given a function y = f(x), then the function is y = g(x)

where g(x) = f(x) is not the inverse function of f.

• Example. If is given, then is not the inverse of f. The graph of this false inverse is obtained by reflecting the graph of f in the x-axis.

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

y = x3y = x3

Inverse functions—a review

• If we are given a function y = f(x), then the inverse function is the function y = g(x) satisfying

f(g(x)) = x for every x in the domain of g,

g(f(x)) = x for every x in the domain of f. As we know, the inverse function may fail to exist.

• Example. If is given, then is the inverse of f. The graph of the inverse is obtained by reflecting the graph of f in the line y = x.

3 xg(x)

3 xy

3xf(x)

3xy xy