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Chapter 1

Functions

1.1 What is a Function?

A function is a rule assigning to each element of a set exactly one value.

• The elements to which a function assigns values are called the inputs.

• The values assigned by a function are called the outputs.

Thus, a rule assigning outputs to certain inputs is a function if and only if

to every input it unabiguously assigns a unique output.

Four Ways to Define a Function

A function can be defined in the following four ways:

1. By a word discription.

2. By a table.

3. By a graph or a diagram.

4. By an equation (analytically)

Domain and Range of a Function

The domain of a function is the set of all its inputs.

The range of a function is the set of all its outputs.

Notations

Let f be a function and x be an input, then f(x) designates the output assigned by the functionf to the input x.

1

2 CHAPTER 1. FUNCTIONS

Graphs

The graph of a function f is the set of all points in the xy-plain with the coordinates (x, f(x)),where x is an input.

Vertical Line Test. A curve in the xy-plane is a graph of a function if and only if everyvertical line intersects it at most once.

1.2 Functions’ Behavior and Graphs’ Shapes

Increasing/Decreasing

A function y = f(x) is said to increase over an interval of its domain if, in this interval, to agreater input corresponds a greater output.

A function y = f(x) is said to decrease over an interval of its domain if, in this interval, to agreater input corresponds a less output.

Local Extrema

A function y = f(x) is said to have a local maximum at c if, for all x’s near c:

f(x) ≤ f(c)

A function y = f(x) is said to have a local minimum at c if, for all x’s near c:

f(x) ≥ f(c)

Note: Here, saying ”for all x’s near c”, we understand all x’s sufficiently close to c from theleft and from the right.

Local maxima and minima are called local extrema.

Concavity and Inflections

Let y = f(x) be a real-valued function defined on an interval.

The graph of a function y = f(x) is said to be a concave up over an interval of x-values if itlies below a straight line segment connecting two points corresponding to an arbitrary pair ofx-values from this interval.

The graph of a function y = f(x) is said to be a concave down over an interval of x-values if itlies above a straight line segment connecting two points corresponding to an arbitrary pair ofx-values from this interval.

A point on the graph of a function y = f(x) is called an inflection point if, at this point, thegraph changes the character of its concavity.

1.3. LINEAR FUNCTIONS 3

1.3 Linear Functions

Let y = f(x) be a functions whose inputs and outputs are numbers.

We say that y = f(x) is a linear function, or that y depends on x linearly if, for any two distinctx-values, x1 and x2, and corresponding y-values, y1 = f(x1) and y2 = f(x2), the average rateof change

y2 − y1

x2 − x1

= m

is the same, i.e., the rate of change of y relative to x is constant.

m has the following interpretation: y changes by m y-units as x changes by one x-unit.

Equation of a Linear Function

Suppose that we know that, for a linear function y = f(x),

• a value y1 corresponds to a certain value x1: y1 = f(x1),

• the rate of change is m.

Theny = y1 + m(x− x1)

Indeed, for an arbitrary x distinct from x1, let y be a corresponding value. Then

y − y1

x− x1

= m

Whence:

y − y1 = m(x− x1) and y = y1 + m(x− x1)

The latter can be simplified to the form:

y = mx + b

As is easily seen, b = y(0).

Conversely, any function of the form y = mx + b is linear whose rate of change is m (verify).

4 CHAPTER 1. FUNCTIONS

Graph of a Linear Function

The graph of a linear function y = mx + b is a nonvertical line with the slope m and they-intercept b.

Special Cases

Direct ProportionalityIn case when b = 0, we have:

y = mx

and call it a direct proportionality. We say that y is directly proportional to x or that y variesdirectly with x.

ConstantIn case when m = 0, we have a constant:

y = b

In particular, y = 0, determines the x-axis.

Chapter 2

Rates of Change

2.1 Average Rate of Change (ARC)

Let y = f(x) be a function whith numerical inputs and outputs. Consider two distinct inputsx1 and x2 and the corresponding outputs y1 = f(x1) and y2 = f(x2). Then:

• x2 − x1 is the change (absolute change) of x also called the run;

• y2 − y1 is the the corresponding change (absolute change) of y also called the rise.

The quotient

y2 − y1

x2 − x1

=rise

run

is called the average rate of change (ARC) of y relative to x as x changes from x1 to x2.

An ObservationNote that the ARC is independent of the order in which the inputs and the corresponding

outputs are taken. Indeed:y2 − y1

x2 − x1

=y1 − y2

x1 − x2

Thus, the ARC of y relative to x as x changes from x1 to x2 is absolutely the same as theARC of y relative to x as x changes from x2 to x1.

Units

Unit of ARC =y-Unit

x-Unit

Geometric InterpretationConsider two distinct points on the graph of the function y = f(x), P (x1, y1) and Q(x2, y2).Then the ARC y relative to x as x changes from x1 to x2 is the the slope of the line (secant)

through these points,

5

6 CHAPTER 2. RATES OF CHANGE

secant slope =y2 − y1

x2 − x1

= ARC

(see figure).

Mechanical Interpretation

Let x(t) be the coordinate of a material point moving along the x-axis, in meters (m), and tbe time, in seconds (s). Consider two distinct moments of time t1 and t2 (we can regard t1 < t2)and the corresponding positions of the material point x1 = x(t1), x2 = x(t2). The time changeis t2− t1 seconds, the corresponding coordinate change (displacement) is x2−x1, the ARC of xrelative to t is nothing but the average velocity, i.e., the average displacement per unit of timeof the material point between the moments t1 and t2, in meters per second (m/s).

Examples:

1. Let a material point move rectilinearly according to the law x = −2t3, where x thecoordinate of the material point, in meters, and t is time, in seconds. Find the averagevelocity of the material point over the time interval 1 ≤ t ≤ 3.

We have:

average velocity =−2 · 33 − (−2)13

3− 1=−54 + 2

2= −26 m/s

2. Consider the table

Perceived Temperature T (◦ F) 7 4 -2Wind Speed w (mph) 16 20 30

Then

• For w1 = 16 mph and w2 = 20 mph: ARC =4− 7

20− 16◦ F/mph = −3

4◦ F/mph.

• For w1 = 30 mph and w2 = 16 mph: ARC =7− (−2)

16− 30◦ F/mph = − 9

14◦ F/mph.

3. Let y = x2 − 3x. Find the ARC of y relative to x as x changes form −1 to 2.

In this case x1 = −1, x2 = 2, y1 = (−1)2 − 3(−1) = 4, and y2 = 22 − 3 · 2 = −2. Thenrun = 2− (−1) = 3, rise = −2− 4 = −6, and

ARC =−6

3= −2

In this case the ARC is unitless as well as x and y.

Questions:

2.1. AVERAGE RATE OF CHANGE (ARC) 7

1. What is the ARC of y (output) relative to x (input) as x changes from x1 to x2?

2. When you evaluate ARC of y relative to x as x changes from x1 to x2 and then ARC ofy relative to x as x changes from x2 to x1 (inverse order of the inputs), are you gettingequal answers? Why?

3. If the x is measured in days and y in kilograms, what are the units for ARC of y relativeto x?

4. What is the geometric interpretation of ARC?

5. What is the mechanical interpretation of ARC?


2.2 Instantaneous Rate of Change

Let y = f(x) be a function with numerical inputs and outputs. Let a be an input such thatthe function f is defined for all x’s sufficiently close to a from the left and right, i.e., for allsufficiently small positive or negative increments h, a + h is also in the domain of f .

For such a function, we can ask:

How fast y changes relative to x at a?or

What is the instantaneous rate of change of y relative to x at a?

One can assign to the increment h values (positive or negative) arbitrarily close to 0 butdistinct from it (h 6= 0) and, for evry such value, consider the ARC of y as x changes from ato a + h:

ARC =f(a + h)− f(a)

a + h− a=

f(a + h)− f(a)

h

Note that to every h corresponds its own ARC.

If the ARC approaches a finite valueas the increment h approaches 0, being not equal to 0,

we call the valuethe instantaneous rate of change of y relative to x at a (IRC).

ARC → IRC as h → 0

Units

Unit of IRC = Unit of ARC =y-Unit

x-Unit

2.2. INSTANTANEOUS RATE OF CHANGE 9

Geometric Interpretation

Consider the two points on the graph of the function y = f(x) corresponding to a and a+h,respectively, P (a, f(a)) and Q(a + h, f(a + h)).

We know that the ARC of y relative to x, as x changes from a, to a + h is the the slope ofthe secant line through P and Q.

As h changes, the point Q changes its position on the graph. Note that the point P remainsin the fixed position. The secant line PQ also changes as h changes.

Thus, the IRC of y relative to x at a exists and if and only if the slope of the secant PQapproaches a finite value m as h → 0, i.e., the secant PQ approaches a non-vertical line troughP with the slope m (see figure).

This line,being the limit position of the secant lines PQ as h → 0,

is called the tangent line to the graph at the point P (a, f(a)).

Note that there are many secant lines through P but there is only one tangent line.

Conclusion: The IRC of y = f(x) relative to x at a exists if and only if there exists anon-vertical tangent line to the graph of f at the point P (a, f(a)). Moreover:

IRC = tangent slope

We call the slope of the tangent line to the graph at a point P the slope of the graph atthat point. Thus,

IRC = graph slope

Mechanical Interpretation

Let x(t) be the coordinate of a material point moving along the x-axis, say, in meters (m),and t be time, in seconds (s).

We know that the ARC of the coordinate x relative to time t as t changes from a to a + his the average velocity of the material point between the moments a and a + h, in this case, inmeters per second (m/s).

The IRC of the coordinate x at the moment a is called the instantaneous velocity at themoment a.

2.2.1 More on Tangent Lines

Let y = f(x) be a function with numeric inputs and outputs and a be such an input that fis defined for all x’s sufficiently close to a to the left and right. In this case we say that f isdefined in a neighborhood of a.

Under the above assumptions, one can define the IRC of y relative to x at a and the tangentline to the graph of f at the point P (a, f(a)).


Questions/Answers

Q: What is the tangent line to the graph at the point P (a, f(a))?A: The tangent line to the graph at the point P (a, f(a)) is the limit position of the secant

through P and Q(a + h, f(a + h)), as h → 0. Its slope is the limit value of the slope of thesecant PQ as h → 0.

Q: Does a tangent line exists at any point on the graph?A: No, it need not. There may be such points on the graph at which a tangent does not

exist (see figures).

Q: Can there be more than one tangent line to the graph at a point?A: No. When a tangent line to the graph at a point exists, it is unique.

Q: How is the IRC of y relative to x at a related to the tangent line to the graph atP (a, f(a))?

A: The IRC of y relative to x at a exists if and only if there is a non-vertical tangent lineto the graph at P (a, f(a)), in which case:

IRC = tangent slope

Q: For the tangent line to the graph trough P (a, f(a)), is P the only point of intersectionof the line with the graph?

A: No. It may intersect the graph at other points, even infinitely many times (see figure),i.e., the tangent through P can also be a secant trough P (see figure).

Q: What is the slope of the graph at a point ?A: It is the slope of the tangent to the graph at that point, provided the tangent exists.

Thus, we can talk about the slope of the graph at a point only when there is a tangent to thegraph at that point.

On Local Extrema, Concavity, and Inflections

• At a point of local extremum (min or max), the tangent to the graph, provided it exists,is necessarily horizontal (has zero slope) [see figure].

Note however, that the graph may have a horizontal tangent at a point without having alocal extremum there, e.g., y = x3 at (0, 0).

• Near a point where the graph is concave up/down, provided the tangent to the graph atthe point exists, the graph lies above/below the tangent (see figure).

• At a point of inflection, provided the tangent to the graph at the point exists, the graphpasses from one side of the tangent to the other (see figure).

2.2. INSTANTANEOUS RATE OF CHANGE 11

2.2.2 Finding IRC/Slope Numerically

Given a function y = f(x) and an input value a, such that f is defined in a neighborhood of a,how do we find the IRC of y relative to x at a or, which is the same, the slope of the graph off at the point P (a, f(a))?

The procedure follows directly from the definition of IRC and is as follows:

Giving a increasingly small nonzero run hwe observe to which finite limit valuethe ARC of y between a and a + h,

or the slope of the secant trough P (a, f(a)) and Q(a + h, f(a + h)),

f(a + h)− f(a)

h

approaches.This very value is the IRC of y relative to x at a,or slope of the graph of f at the point P (a, f(a)).

Using limit notation, we can express the above as follows:

IRC = limh→0

ARC = limh→0

f(a + h)− f(a)

h

graph slope = limh→0

secant slope = limh→0

f(a + h)− f(a)

h

Read: The limit of . . . as h approaches 0.

Examples:

1. Find the slope of the graph of y = x2 at (1, 1).

In this case, f(x) = x2, a = 1, and f(1) = 1. For a run h, the slope of the secant troughP (1, 1) and Q(1 + h, (1 + h)2) is

f(1 + h)− f(1)

h=

(1 + h)2 − 1

h.

Hence,(1 + h)2 − 1

h=

1 + 2h + h2 − 1

h= 2 + h → 2 as h → 0.

Thus, the slope of the graph at (1, 1) is 2.


2. Find the IRC of y = xx at 1.

We have: f(x) = xx, a = 1, and f(1) = 1. For a run h, the ARC of y between 1 and1 + h is

f(1 + h)− f(1)

h=

(1 + h)1+h − 1

h.

Unfortunately, the limit of this expression as h → 0 is not as easily found as in thepreceding example.

Using a calculator, we obtain the following table:

hf(1 + h)− f(1)

h

0.1 1.105342−0.01 0.99004970.001 1.001001−0.0001 0.99990.00001 1.000010.000001 1.000001

Therefore,

limh→0

(1 + h)1+h − 1

h= 1.

2.3. DERIVATIVE 13

2.3 Derivative

Let y = f(x) be a function with numerical inputs and outputs and a be such an input that fis defined in its neighborhood, i.e., for all x sufficiently close to a from the left and right.

If the limit

limh→0

f(a + h)− f(a)

h

exists and is finite,it is called the derivative of the function f (of y) at a.

Notations:

f ′(a) ”f prime at a”, y′(a) ”y prime at a”,

df(a)

dx”dee f -dee x at a”,

dy(a)

dx”dee y-dee x at a”,

df

dxwhen x = a ”dee f -dee x when x = a”,

dy

dxwhen x = a ”dee y-dee x when x = a”

Conclusions:

• The derivative of y at a, y′(a), is the same as the IRC of y relative to x at a or the slopeof the graph of f at P (a, f(a)).

• y′(a) is the measure of how fast y changes relative to x at a.

• The unit of y′(a) is y-unit/x-unit.

• The derivative of y at a exists if and only if the graph of f has a non-vertical tangent lineat P (a, f(a)).

Example:

Let T (p) be the number of tickets from Atlanta to Boston that a certain airline sells in oneweek when the price of a ticket is p dollars. Interpret the following:


(a) T (350) = 1757.

(b) T ′(350) = −15.

(c) dTdp

= 22 wnen p = 280.

Solution:

(a) When the ticket price is $350, 1,757 tickets are sold.

(b) When the ticket price is $350, the number of sold tickets is decreasing at the rate of 15tickets per dollar.

(c) When the ticket price is $280, the number of sold tickets is increasing at the rate of 22tickets per dollar.

2.3. DERIVATIVE 15

2.3.1 Derivative as a Function

We can now consider finding the derivative of a function y = f(x) for any x where it exists:

y′(x) = limh→0

f(x + h)− f(x)

h

As x varies, y′(x) does. Thus, we can look at the derivative y′(x) as a function of x.Note that the notation y′(x) is usually abriviated to y′.

Example: For y = 3x2 − 2, y′ = 6x. Indeed:

y′(x) = limh→0

f(x + h)− f(x)

h= lim

h→0

3(x + h)2 − 2− (3x2 − 2)

h

= limh→0

3(x2 + 2xh + h2)− 2− 3x2 + 2

h= lim

h→0

3x2 + 6xh + 3h2 − 3x2

h

= limh→0

6xh + 3h2

h= lim

h→0(6x + 3h) = 6x.

2.3.2 Table Derivatives

By table derivatives, we shall understand those basic ones knoweledge of which is necessary andis taken for granted in the future like 2× 2 = 4 (compare to the table of multiplication ). Ourtable of derivatives, for the time being, will be as follows:

Function Derivative

y = b y′ = 0

y = mx + b y′ = m

y = xn y′ = nxn−1 (Power Rule)

1. y′(x) = limh→0

f(x + h)− f(x)

h= lim

h→0

b− b

h= lim

h→00 = 0.

2. y′(x) = limh→0

f(x + h)− f(x)

h= lim

h→0

m(x + h) + b− (mx + b)

h


= limh→0

mx + mh + b−mx− b

h= lim

h→0

mh

h= lim

h→0m = m.

3. Let us illustrate the Power Rule for n = 2:

y′(x) = limh→0

f(x + h)− f(x)

h= lim

h→0

(x + h)2 − x2

h= lim

h→0

x2 + 2xh + h2 − x2

h

= limh→0

2xh + h2

h= lim

h→0(2x + h) = 2x.

Note that the Power Rule is valid for any real exponent n, e.g.:

1. y = x (n = 1), y′ = x0 = 1.

2. y = x3 (n = 3), y′ = 3x2.

3. y = x−7 (n = −7), y′ = (−7)x−8.

4. y =1

x2= x−2 (n = −2), y′ = (−2)x−3.

5. y =√

x = x1/2 (n = 1/2), y′ = 12x−1/2 =

1

2√

x.

6. y =1

3√

x2=

1

x2/3= x−2/3 (n = −2/3), y′ = (−2

3)x−5/3.

7. y = xπ (n = π), y′ = πxπ−1.

2.3.3 Rules of Differentiation

By differentiation we understand finding the derivative of a function.

Sum/Difference and Constant Factor Rules. Let functions f and g both have derivativesat x, then

1.[f(x)± g(x)

]′= f ′(x)± g′(x) (Sum/Difference Rule).

2.[cf(x)

]′= cf ′(x) (Constant Factor Rule).

Note that the Sum/Difference Rule is valid for any finite number of terms, e.g.:[f(x) + g(x) + h(x)

]′= f ′(x) + g′(x) + h′(x).

Example:

Find the derivative of the function y = 7x2 − 12

x2+ 3

√x.

2.3. DERIVATIVE 17

Solution: First, let us rewrite all the terms in power form: y = 7x2 − 12x−2 + 3x1/2.Using the Rules of Differentiation, we have:

y′ = Sum/Difference Rule = (7x2)′ − (12x−2)′ + (3x1/2)′

= Constant Factor Rule = 7(x2)′ − 12(x−2)′ + 3(x1/2)′ = 7(2x)− 12(−2x−3)

+ 3(1

2x−1/2) = 14x +

24

x3+

3

2√

x.


Product Rule. Let functions f and g both have derivatives at x, then

[f(x)g(x)

]′= f ′(x)g(x) + f(x)g′(x) (Product Rule).

Example:

Find the derivative of the function y =

(4.5x2 +

1

x

)( 3√

x− 7x + 2.75).

Solution: Let us first rewrite all the terms in power form: y = (4.5x2+x−1)(x1/3−7x+2.75).Using the Rules of Differentiation, we have:

y′ = Product Rule

= (4.5x2 + x−1)′(x1/3 − 7x + 2.75) + (4.5x2 + x−1)(x1/3 − 7x + 2.75)′

= Sum/Difference and Constant Factor Rules

= (9x− x−2)(x1/3 − 7x + 2.75) + (4.5x2 + x−1)

(1

3x−2/3 − 7

).

Note that the Product Rule is valid for any finite number of factors, e.g.:[f(x)g(x)h(x)

]′= f ′(x)g(x)h(x) + f(x)g′(x)h(x) + f(x)g(x)h′(x).

2.3. DERIVATIVE 19

Compositions and Chain Rule

For two functions y = g(x) and z = f(y), one can consider composition, i.e., the functiondefined as follows:

f ◦ g(x) := f(g(x)).

Read: The composition of f and g.

The composition woks as the diagram shows:

x 7−→ y = g(x) 7−→ z = f(y) = f(g(x))

Example: For f(x) =√

x and g(x) = x− 1,

f ◦ g(x) = f(g(x)) = f(x− 1) =√

x− 1,

g ◦ f(x) = g(f(x)) = g(√

x) =√

x− 1,

f ◦ f(x) = f(f(x)) = f(√

x) =

√√x = 4

√x,

g ◦ g(x) = g(g(x)) = g(x− 1) = (x− 1)− 1 = x− 2.

Chain Rule. Let a function g have a derivative at x and a function f have a derivative atg(x). Then the composition f ◦ g has a derivative at x:

[f(g(x))

]′= f ′(g(x))g′(x) (Chain Rule)

Examples:

1. y = (1− 2x)100, y′ = Chain Rule = 100(1− 2x)99(1− 2x)′

= 100(1− 2x)99(−2) = −200(1− 2x)99.

2. y =27

3x2 − 5= 27(3x2 − 5)−1, y′ = Constant Factor Rule

= 27[(3x2 − 5)−1

]′= Chain Rule = 27

[−(3x2 − 5)−2(3x2 − 5)′

]= 27

[−(3x2 − 5)−2(6x)

]= −162x(3x2 − 5)−2 =

−162x

(3x2 − 5)2.

3. y =5√

x7 − 4x = (x7 − 4x)1/5, y′ = Chain Rule


=1

5(x7 − 4x)−4/5(x7 − 4x)′ =

1

5(x7 − 4x)−4/5(7x6 − 4)

=7x6 − 4

5 5√

(x7 − 4x)4.

4. y = (5x3 − 31)2(1− 3x)11, y′ = Product Rule

=[(5x3 − 31)2

]′(1− 3x)11 + (5x3 − 31)2

[(1− 3x)11

]′= Chain Rule = 2(5x3 − 31)(5x3 − 31)′(1− 3x)11

+ (5x3 − 31)211(1− 3x)10(1− 3x)′ = 2(5x3 − 31)(15x2)(1− 3x)11

+ (5x3 − 31)211(1− 3x)10(−3) = 30x2(5x3 − 31)(1− 3x)11 − 33(5x3 − 31)2(1− 3x)10.

Note that there is a Quotient Rule for differentiation a quotientf(x)

g(x), which is as follows:

[f(x)

g(x)

]′=

f ′(x)g(x)− f(x)g′(x)[g(x)

]2 (Quotient Rule).

However, applying this rule can be avoided by representing the quotient as a product

f(x)

g(x)= f(x)

[g(x)

]−1

and then applying the Product and Chain Rules.

Example: Let y =3x5 − 2

(x2 − 4x3)2. Then y = (3x5 − 2)(x2 − 4x3)−2 and

y′ = Product Rule

= (3x5 − 2)′(x2 − 4x3)−2 + (3x5 − 2)[(x2 − 4x3)−2

]′= Chain Rule = 15x4(x2 − 4x3)−2 + (3x5 − 2)(−2)(x2 − 4x3)−3(x2 − 4x3)′

= 15x4(x2 − 4x3)−2 + (3x5 − 2)(−2)(x2 − 4x3)−3(2x− 12x2).

= 15x4(x2 − 4x3)−2 − 4(3x5 − 2)(x2 − 4x3)−3(x− 6x2).

Note that the Chain Rule can be applied as many times as needed. Thus, for a composition ofthree functions f , g, and h, we have:

[f(g(h(x))

]′= f ′(g(h(x)))

[g(h(x))

]′= f ′(g(h(x)))g′(h(x))h′(x).

Chapter 3

Eponentials and Logarithms

3.1 Powers and Exponent Laws

Let a be a real number and n be a natural number. The nth power of a is defined as follows:

an := a · a · · · a︸︷︷︸n

In particular, a1 = a.

The numbers a and n are called the base and exponent of the power, respectively.

Such a natural definition has a number of implications.

Exponent Laws for Natural Exponents

aman = am+n

am

an= am−n (m > n)

(am

)n= amn

(ab)n = anbn

(a

b

)n

=an

bn

Powers with Zero Exponent

To preserve the Exponent Laws the only way to define a power with zero exponent is:

a0 := 1

21

22 CHAPTER 3. EPONENTIALS AND LOGARITHMS

Indeed, for a 6= 0: a0a = a0+1 = a. Whence: a0 = 1.Note that, by the definition, 00 = 1.

Powers with Negative Integer Exponents

Let a 6= 0 and n be a natural number. In order to preserve the Exponent Laws we cannotbut define a−n as follows:

a−n :=1

an

Indeed, a−nan = a−n+n = a0 = 1. Whence: a−n = 1/an.

Examples:

1. 5−1 =1

51=

1

5.

2. (−3)−3 =1

(−3)3= − 1

27.

3. 0−4 is undefined.

Powers with Fractional Exponents

Defining powers with fractional exponents is also based upon preserving the Exponent Laws.

Let a fractionm

n, where m is integer, n is natural, be in the lowest terms. Then:

amn := n

√am

Indeed,(a

mn

)n= a

mn

n = am. Whence: amn = n

√am.

Examples:

1. 51/2 =√

5.

2. (−3)−2/3 =1

3√

(−3)2=

13√

9.

3. (−7)3/4 is undefined.

Powers with Irrational ExponentsFor a positive base a one can consider a power with an irrational exponent x, ax, e.g., 3

√2.

The powers with irrational exponents are also defined in such a way that the Exponent Lawshold.

With powers so defined, the Exponent Laws are valid for arbitrary real exponents.

Exponent Laws

3.1. POWERS AND EXPONENT LAWS 23

axay = ax+y ax

ay= ax−y

(ax

)y= axy

(ab)x = axbx

(a

b

)x

=ax

bx

An expression of the form ax is called a power or an exponential, a and x beingcalled its base and exponent, respectively.

Note that the range of possible values for the base depends on the exponent, e.g.:

• an, where n is natural, is defined for any real a;

• a−n, where n is natural, is defined for any a 6= 0;

• a1/3 = 3√

a is defined for any real a;

• a−2/5 =1

5√

a2is defined for a 6= 0;

• a1/2 =√

a is defined for a ≥ 0;

• a−3/4 =1

4√

a3is defined for a > 0.

• aπ is defined for a > 0.

Examples:

1. Find the exact value of 10−1/3100001/3.

Applying the Exponent Laws, we have:

10−1/3100001/3 = 10−1/3(104

)1/3= 10−1/3104/3 = 10−1/3+4/3 = 101 = 10.

2. Simplify the expression and eliminate any negative and fractional exponents. Assumethat all letters designate positive numbers.

(y2z−1

8x6

)−1/3(x−1/3z−1/2

√2y

)2

Applying the Exponent Laws, we have:


(z−1

8x6y−2

)−1/3(x−1/3z−1/2

√2y

)2

=(8−1x−6y2z−1

)−1/3(2−1/2x−1/3y−1z−1/2

)2

=(8−1

)−1/3(x−6

)−1/3(y2

)−1/3(z−1

)−1/3(2−1/2

)2(x−1/3

)2(y−1

)2(z−1/2

)2

= 81/3x2y−2/3z1/32−1x−2/3y−2z−1 = 21

2x2−2/3y−2/3−2z1/3−1

= x4/3y−8/3z−2/3 =3√

x4

3√

y8 3√

z2.

3.1. POWERS AND EXPONENT LAWS 25

Problems

Find the exact value of the expression:

1.

(−27

8

)2/3

Answer:9

4.

2. 32−0.2 Answer: 0.5.

3. 31/291/4 Answer: 3.

Simplify the expression and eliminate any negative and fractional exponents. Assume that allletters designate positive numbers.

4. (4b)1/2(8b2/5) Answer: 1610√

b9.

5. (5x4y−4/5)3(8y2)2/3 Answer: 500x12

15√

y16.

6.(ab)3/2

(a3b−4)2/3Answer:

6√

b25

√a

.

7.

(a2b−3

x−1y1/2

)−3(x−2b−1

a3/2y1/3

)1/2

Answer:

√b17 3

√y4

4√

a27x4.

Produce the graph of the function on the calculator, find its derivative, and answer the ques-tions.

8. y = 3√

x + 2.

Question: Why does the function increase on (−∞,∞)?

9. y =1

2x3 + 5.

Question: Why does the function decrease on each interval of its domain?

10. y = (4x2 − 7)2

Questions: Over which intervals does the function increase/decrease? What happensat 0?


3.2 Exponential Functions

Let a > 0 and a 6= 1. A function of the form

y = ax, −∞ < x < ∞,

is called an exponential function.There are two principally different cases: a > 1 and 0 < a < 1 (see figure).

a > 1 increasing concave up y → 0 as x → −∞ y →∞ as x →∞

0 < a < 1 decreasing concave up y →∞ as x → −∞ y → 0 as x →∞

Base e

An important special case of an exponential with a base a > 1 is the natural exponential

y = ex,

where e = 2.718281828459045 · · · ≈ 2.7 is an irrational number obtained as follows:

e = limx→∞

(1 +

1

x

)x

= limx→0

(1 + x)1/x

More generally, we shall understand by an exponential function any function of the form:

y = Cax

where C is a nonzero number, a > 0 and a 6= 1. Note that C = y(0) is the initial value.

Exponential Growth and DecayA function

y = Cax,

where C > 0 and a > 1, is said to grow exponentially (see fig.).For such a function: y →∞ as x →∞.

Any function of the formy = Cax,

where C > 0 and 0 < a < 1, is said to decay exponentially (see fig.).For such a function: y → 0 as x →∞.

3.2. EXPONENTIAL FUNCTIONS 27

Percentage ChangeConsider an exponential function

y = Cax.

As the input changes by 1, from x to x+1, the output changes from Cax to Cax+1 = Caxa,i.e., by a factor of a.

The absolute change is Cax+1 − Cax. The relative change is

Cax+1 − Cax

Cax=

Caxa− Cax

Cax=

Cax(a− 1)

Cax= a− 1.

The relative change, expressed in percent form, i.e., as (a − 1)100%, is called the percentagechange.

• In the case of exponential growth (C > 0 and a > 1), we say that y increases by (a−1)100%per x-unit.

• In the case of exponential decay (C > 0 and 0 < a < 1), we say that y decreases by(1− a)100% per x-unit.

Note that the percentage change depends neither on C nor on the input value x. Itdepends on the base a only.

Examples:

1. Let P (t) = 5(3.2)t be a model of a bacteria culture, where t is time, in hours. Thena = 3.2 > 1 and, hence, the percentage change is (3.2− 1)100% = 220% per hour. Thus,the culture increases by 220% per hour.

2. Let y = 27(0.68)x. Then a = 0.68 < 1, the percentage change is (0.68− 1)100% = −32%per x-unit, i.e., y decreases by 32% per x-unit.

Linear vs. Exponential

Linear y = mx + b Constant Change per x-unit

Exponential y = Cax Constant Percentage Change per x-unit


Problems

Determine whether the exponential function grows or decays. Find the percentage change.

1. y = 4ex.

2. m(t) = 3.75

(2

5

)t

.

3. P (x) = 17.23(1.1)x

4. P (x) = 10.36(0.9)2x

5. A culture contains initially 1500 bacteria and doubles every 30 minutes.

(a) Find the number of bacteria n(t) as a function of time t, in minutes.

(b) Find the percentage change of the culture and interpret it in the context.

6. John invested $2,500 at a simple interest rate of 10% per year applied annually to theorigional amount. Find the amount A(t) on John’s account as a function of time t, inyears since the investment was made.

7. Julie invested $2,500 at a compound interest rate of 10% per year applied annually to theamount of the previous year. Find the amount A(t) on Julie’s account as a function oftime t, in years since the investment was made. Does it exibit an exponential growth? If”yes”, find the percentage change and interpret it in the context.


Simple Interest vs. Compound InterestLet

• P be an initial investment called the principal;

• r be an interest rate, in % per year;

• t be time, in years since the initial investment was made;

• A(t) be the amount after t years;

Assume that the interest is earned n times a year (n ≥ 1), i.e., a year is subdivided inton compounding periods, e.g., annually, semiannually, quaterly, or monthly. This can be doneaccording to the two following schemes (r is used below in its decimal form):

Simple Interest Scheme Compound Interest Scheme

The interest is earned based on The interest is earned based onthe principal only. the amount of the previous compounding period.

A(t) = P +

(P

r

n

)nt = P + (Pr)t A(t) = P

(1 +

r

n

)nt

= P

[(1 +

r

n

)n]t

Linear Exponential

Rate of Change = Pr $ per year Percentage Change =

[(1 +

r

n

)n

− 1

]100 % per year

Problem 1: Suppose you invest $1,500 in an account that pays 7% per year, com-pounded quaterly.

(a) Find the amount A(t) on the account as a function of time t, in years since the investmentwas made.

(b) Find the amount after 5 years and 2 months.


Continuous Compounding

By letting n →∞, we partition a year into shorter and shorter compounding periods arrivingat the situation when compounding happens continuously.

We have: (1 +

r

n

)n

=

[(1 +

1

n/r

)n/r]r

→ er as n →∞.

Simple Interest Scheme Compound Interest Scheme

A(t) = P + (Pr)t A(t) = Pert = P[er

]t

Linear Exponential

Rate of Change = Pr $ per year Percentage Change =[er − 1

]100 % per year

Problem 2: Answer the questions of Problem 1 for the same amount invested at thesame interest rate, compounded continuously.

APR and APY

• Annual Percentage Rate (APR), also called the nominal interest rate, is a compoundinterest applied a certain number of times per year, say, semiannually, quaterly, or con-tinuously.

• The Annual Percentage Yield (APY), also called the effective interest rate, for a givenAPR, is the percentage by which the amount, invested at the APR, is exponentiallyincreasing per year, i.e., the percentage change of the amount.

Note that, being the percentage change, the APY does not depend on the principal,it depends on the APR only.

Example: Consider the APR of 6% per year, compounded semiannually.We have:

A(t) = P

(1 +

0.06

2

)2t

= P (1.03)2t = P[(1.03)2

]t= P (1.0609)t.

Hence, APY = (1.0609− 1)100 = 6.09% per year.


Because both the size of the interest rate and the number of compounding periods contributeinto the amount of interest earned per year, comparing APR’s may be puzzling. ComparingAPY’s is easy. The greater is the APY the faster the investment grows.

Problem 3: Which is a better deal: an APR of 5% per year compounded semiannuallyor an APR of 4.95% per year compounded continuously?

If you invested $1, 000, by how much would these two options differ after 10 years?

Future Value and Present Value

The future value of a sum of money, invested at a given APR, is its value at a later date.

Problem 4: Suppose that on September 1 of each year, you deposit $700 at an APR of8.5% per year, compounded quaterly. Find the future value of these investments 1 year afterthe third deposit is made.

The present value of a sum of money is the amount that must be invested now, at a givenAPR, to produce the desired sum at a later date.

Problem 5: A student wants to put a part of her summer earnings in a bank in orderto meet her January tuition payment. How much should she invest on August 1 at an APR of5.5% compounded daily if on January 1 she needs $3,600?

Problem 5: What is the present value of an investment that will be worth $5,000 in 3years, assuming that the effective rate is 4.8% per year?


Logistic FunctionsA population, the assumption of limited resources not being imposed, increases without bound-aries and can be modeled by an exponentially growing function

y = Cax,

where C > 0 and a > 1.

A population, under the assumption of limited environment (e.g., fish in a pond), has atendentcy to level off and can be modeled by a so called logistic function:

y =L

1 + Ae−Bx+ C,

where L, A, B, and C are positive numbers.As t →∞, e−Bx → 0. Therefore,

limx→∞

[L

1 + Ae−Bx+ C

]= L + C.

The function increases approaching L + C as t increases. The number L + C is called thelimiting value of the logistic function as t →∞ (see fig.).

A function of the form:

y =L

1 + AeBx+ C,

whith positive L, A, B, and C, is also called logistic.In this, as x →∞, eBx →∞. Therefore,

limx→∞

[L

1 + AeBx+ C

]= C.

The function decreases approaching C as x increases. C is the limiting value x → ∞ in thiscase (see fig.).

Problems: Determine whether the logistic function increases or decreases. Identify theinitial and limiting values of the function.

1. y =123

1 + 2e1.64x+ 5

2. P (t) =18

1 + 5e−0.24t+ 24

3.3. LOGARITHMS 33

3.3 Logarithms

Let a be a positive number, a 6= 1. For any x > 0, there is a unique real number y such that

x = ay,

which is called the logarithm of x with base a.

Notation: loga x.

Logarithms were introduced by John Napier (1550–1617).

Examples: Evaluate the expression. Explain your answer.

1. log2 8 = 3 since 23 = 8.

2. log5 0.2 = −1 since 5−1 = 0.2.

3. log1/3 27 = −3 since

(1

3

)−3

= 33 = 27.

4. log49 7 = 1/2 since 491/2 =√

49 = 7.

5. log4 0 and log3(−2) are undefined.

Thus,y = loga x is the same as x = ay

The former is said to be in logarithmic form the latter in the exponential form.

Common and Natural Logarithms

• Common Logarithm: a = 10. Notation: log.

• Natural Logarithm: a = e. Notation: ln.

Problems: Evaluate the expression. Explain your answer.

1. ln 1 Answer: 0.

2. log2 0.25.

3. log4

1

64Answer: −3.

4. log√

10.

Express in exponential/logarithmic form.

5. log5 x = 3 6. log x = 2 7. ln x = 1/2 8. ln x = −1/3

9. 811/2 = 9 10. 10−3 = 0.001 11. 10x = 2 12. ex = 7


Immediate Properties of Logarithms

The following properties of logarithms are immediate cosequences of the definition.

loga 1 = 0

loga a = 1

loga ax = x, −∞ < x < ∞

aloga x = x, x > 0

Examples: Evaluate the expression.

1. log9

√3

2. eln π

3. (√

10)log 9

Solutions:

1. log9

√3 = log9 31/2 = log9

(91/2

)1/2= log9 91/4 = 1/4.

2. eln π = π.

3. (√

10)log 9 =(101/2

)log 9= 10

12

log 9 =[10log 9

]1/2= 91/2 = 3.

Problems: Evaluate the expression.

1. log3

1√27

Answer: −3/2

2. log8 4

3. 5− log5 18 Answer: 1/18

4. 23 log2 6.

3.3. LOGARITHMS 35

Laws of Logarithms

For any positive A and B and any real c:

loga(AB) = loga A + loga B

loga

(A

B

)= loga A− loga B

loga Ac = c loga A

Examples:

1. Simplify the expression to a form tha does not contain logarithms of products, quotients,and powers. Assume that all the letters designate positive numbers.

lnx3ez

√y

Solution:

lnx3ez

√y

= ln[x3ez

]− ln

√y = ln x3 + ln ez − ln y1/2 = 3 ln x + z − 1

2ln y

2. Evaluate the expression log2 112− log2 7

Solution: log2 112− log2 7 = log21127

= log2 16 = 4.

Problems:

Simplify the expression to a form tha does not contain logarithms of products, quotients, andpowers. Assume that all the letters designate positive numbers.

1. ln4√

ab3

e−cAnswer:

1

4ln a +

3

4ln b + c

2. log3

√√a + b

10cd4

Evaluate the expression.

1.1

2log12 81 + 4 log12 2 Answer: 2

2. log 5 + 3 log 2− 1

3log 64


Changing Base in Logarithms

Changing the base from a to a new base b performed according to the Change of BaseFormula:

loga x =logb x

logb a

Indeed,

alogb x

logb a = a = blogb a =[blogb a

] logb x

logb a = blogb a logb x

logb a = blogb x = x.

Calculator computations require changing the base to e or 10, e.g.:

log2 5 =ln 5

ln 2=

log 5

log 2.

Example: Evaluate the expression log3 0.2 + log9 25.

Solution:

log3 0.2 + log9 25 = log3 0.2 +log3 25

log3 9= log3 0.2 +

1

2log3 25

= log3 0.2 + log3 251/2 = log3 0.2 + log3 5 = log3(0.2 · 5) = log3 1 = 0.

Problems: Use a calculator to evaluate the expression, correct to four decimal places.Change the base when needed.

1. log 7 2. ln 10

3. log3 17 4. log7 14

3.3. LOGARITHMS 37

Changing Base to e in ExponentialsAn exponential function

y = Cax

can be rewritten in the form:y = Cekx.

The procedure is as follows:

y = Cax = a = eln a = C[eln a

]x= Celn a·x = Cekx

where k = ln a.

The number k is called the relative rate of change. Note that, if a > 1, k = ln a > 0,otherwise, k < 0.

Examples:

1. y = 5(2.2)x = 5eln 2.2x, k = ln 2.2 > 0.

2. y = (0.7)x = eln 0.7x, k = ln 0.7 < 0.

Note that, for the exponential growth (C > 0, a > 1), k = ln a > 0; for the exponential decay(C > 0, 0 < a < 1), k = ln a < 0.

Problems: Change the base to e. Find the relative rate of change correct to two decimalplaces.

1. y = 39.2(1.29)x 2. m(t) = 62.4(0.78)t

3. h(t) = 1.02(0.62)t 4. y = 2.93(1.00)x


3.4 Exponential and Logarithmic Equations

Solve the equation. Evaluate the logarithms using a calculator and round off youranswers up to two decimal places.

1. 3x = 12.

2. (3.171)x = 73.

3. 102x−1 = 2.

4. 4(1 + 75x) = 9.

5.10

1 + e−x= 2.

6. ln x = −2.

7. log(1− 2x) = 0.

8. log2(3x− 4) = 5.

3.5 Applications of Logarithms

Exponential GrowthConsider an exponential growth y = Cax (C > 0, a > 1).

Switching to the base e, we have:

y = Cax = Cekx,

where k = ln a > 0 is the relative rate of change.

Problem 1: A culture, containing 2000 bacteria initially, one hour later amounts to 3000bacteria. Assuming that the culture increases with a constant percentage change per hour.

(a) What is the percentage change?

(b) Find the number of bacteria n(t) as a function of time t, in hours.

(c) When will the culture contain 7700 bacteria?

(d) Change the base to e and find the relative rate of change, correct to two decimal places.

3.5. APPLICATIONS OF LOGARITHMS 39

Doubling Period: For an exponentially growing function, the doubling period is such aninput change d that results in the doubling of the output.

To find the doubling period, we set the equation:

Ca(x+d) = 2Cax.

Solving the equation, we have:

Cax+d = 2Cax.

Caxad = 2Cax

ad = 2

Therefore,d = loga 2

Note that the doubling period, d, depends neither on C nor on x and is as charac-teristic a parameter for the exponential growth as the percentage change.

Given a doubling period, d, the exponential growth can be represented as follows:

y = C2x/d

Problem 2: A man invests $5,000 in an account that pays 8.5% per year, compoundedquaterly.

(a) Find the amount A(t) on the account as a function of time t, in years since the investmentwas made.

(b) When will the amount attain the level of $8,000?

(c) Find the doubling period for the investment.

Exponential DecayConsider an exponential decay y = Cax (C > 0, 0 < a < 1).

Switching to the base e, we have:

y = Cax = Cekx,

where k = ln a < 0 is the relative rate of change.

Half-Life:


The decay of a radioactive substance is exponential:

m(t) = m0at,

where 0 < a < 1, m0 is the initial mass of the substance, m(t) is the current mass.

The half-life, h, of a radioactive substance is the time it takes for a sample of the substanceto decay to one-half of its initial mass.

We have:

m0ah =

m0

2

ah =1

2

Thus,

h = loga(1/2)

As is seen, the half-life does not depend on the initial mass, it depends on the kindof radioactive substance only.

Given a half-life, h, the radioactive decay can be represented as follows:

m(t) = m0

(1

2

)t/h

Radioactive Carbon Dating

The radioactive carbon dating method is used by archeologists to determine the age of anancient object. It is based on the fact that the carbon dioxide, CO2, in the atmoshere and inall living organisms contains radioactive carbon-14, 14C, with a half-life of about 5730 years,and nonradioactive carbon-12, 12C, in a fixed proportion.

After an organism dies, it stops assimilating 14C, which begins to decay exponentially. Thetime elapsed since the death of the organism is determined by measuring the amount of 14Cleft in it.

Problem 3: The burial cloth of an Egyptian mummy is estimated to contain 59% ofcarbon-14 it contained initially. How long ago was the mummy buried? (The half-life ofcarbon-14 is 5730 years).

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