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 function
f 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 every
vertical 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 a
greater 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 a
greater 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 the
left 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 it
lies below a straight line segment connecting two points corresponding to an arbitrary pair of
x-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 it
lies above a straight line segment connecting two points corresponding to an arbitrary pair of
x-values from this interval.
A point on the graph of a function y = f (x) is called an inflection point if, at this point, the
graph changes the character of its concavity.
1.3. LINEAR FUNCTIONS
1.3
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 distinct
x-values, x1 and x2 , and corresponding y-values, y1 = f (x1 ) and y2 = f (x2 ), the average rate
of change
y2 − y1
=m
x2 − x1
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.
Then
y = y1 + m(x − x1 )
Indeed, for an arbitrary x distinct from x1 , let y be a corresponding value. Then
y − y1
=m
x − x1
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 the
y-intercept b.
Special Cases
Direct Proportionality
In 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 varies
directly with x.
Constant
In 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 inputs
x1 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
rise
=
x2 − x1
run
is called the average rate of change (ARC) of y relative to x as x changes from x1 to x2 .
An Observation
Note that the ARC is independent of the order in which the inputs and the corresponding
outputs are taken. Indeed:
y2 − y1
y1 − y2
=
x2 − x1
x1 − x2
Thus, the ARC of y relative to x as x changes from x1 to x2 is absolutely the same as the
ARC of y relative to x as x changes from x2 to x1 .
Units
Unit of ARC =
y-Unit
x-Unit
Geometric Interpretation
Consider 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
= ARC
x2 − x1
(see figure).
Mechanical Interpretation
Let x(t) be the coordinate of a material point moving along the x-axis, in meters (m), and t
be 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 change
is t2 − t1 seconds, the corresponding coordinate change (displacement) is x2 − x1 , the ARC of x
relative to t is nothing but the average velocity, i.e., the average displacement per unit of time
of 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 the
coordinate of the material point, in meters, and t is time, in seconds. Find the average
velocity of the material point over the time interval 1 ≤ t ≤ 3.
We have:
average velocity =
−2 · 33 − (−2)13
−54 + 2
=
= −26 m/s
3−1
2
2. Consider the table
Perceived Temperature T (◦ F) 7 4 -2
Wind Speed w (mph)
16 20 30
Then
4−7 ◦
3
F/mph = − ◦ F/mph.
20 − 16
4
7 − (−2) ◦
9 ◦
• For w1 = 30 mph and w2 = 16 mph: ARC =
F/mph = −
F/mph.
16 − 30
14
• For w1 = 16 mph and w2 = 20 mph: ARC =
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. Then
run = 2 − (−1) = 3, rise = −2 − 4 = −6, and
ARC =
−6
= −2
3
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 of
y relative to x as x changes from x2 to x1 (inverse order of the inputs), are you getting
equal answers? Why?
3. If the x is measured in days and y in kilograms, what are the units for ARC of y relative
to x?
4. What is the geometric interpretation of ARC?
5. What is the mechanical interpretation of ARC?
8
CHAPTER 2. RATES OF CHANGE
2.2
Instantaneous Rate of Change
Let y = f (x) be a function with numerical inputs and outputs. Let a be an input such that
the function f is defined for all x’s sufficiently close to a from the left and right, i.e., for all
sufficiently 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 but
distinct from it (h 6= 0) and, for evry such value, consider the ARC of y as x changes from a
to a + h:
f (a + h) − f (a)
f (a + h) − f (a)
ARC =
=
a+h−a
h
Note that to every h corresponds its own ARC.
If the ARC approaches a finite value
as the increment h approaches 0, being not equal to 0,
we call the value
the 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 of
the secant line through P and Q.
As h changes, the point Q changes its position on the graph. Note that the point P remains
in the fixed position. The secant line P Q 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 P Q
approaches a finite value m as h → 0, i.e., the secant P Q approaches a non-vertical line trough
P with the slope m (see figure).
This line,
being the limit position of the secant lines P Q 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.
The IRC of y = f (x) relative to x at a exists if and only if there exists a
Conclusion:
non-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 at
that 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 + h
is the average velocity of the material point between the moments a and a + h, in this case, in
meters per second (m/s).
The IRC of the coordinate x at the moment a is called the instantaneous velocity at the
moment 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 f
is defined for all x’s sufficiently close to a to the left and right. In this case we say that f is
defined in a neighborhood of a.
Under the above assumptions, one can define the IRC of y relative to x at a and the tangent
line to the graph of f at the point P (a, f (a)).
10
CHAPTER 2. RATES OF CHANGE
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 the
secant P Q 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 at
P (a, f (a))?
A: The IRC of y relative to x at a exists if and only if there is a non-vertical tangent line
to 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 intersection
of 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 the
graph 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 a
local 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 at
the 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 graph
passes from one side of the tangent to the other (see figure).
2.2. INSTANTANEOUS RATE OF CHANGE
2.2.2
11
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 of
f 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 h
we observe to which finite limit value
the 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:
f (a + h) − f (a)
h→0
h→0
h
f (a + h) − f (a)
graph slope = lim secant slope = lim
h→0
h→0
h
IRC = lim ARC = lim
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 trough
P (1, 1) and Q(1 + h, (1 + h)2 ) is
f (1 + h) − f (1)
(1 + h)2 − 1
=
.
h
h
Hence,
(1 + h)2 − 1
1 + 2h + h2 − 1
=
= 2 + h → 2 as h → 0.
h
h
Thus, the slope of the graph at (1, 1) is 2.
12
CHAPTER 2. RATES OF CHANGE
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 and
1 + h is
(1 + h)1+h − 1
f (1 + h) − f (1)
=
.
h
h
Unfortunately, the limit of this expression as h → 0 is not as easily found as in the
preceding example.
Using a calculator, we obtain the following table:
h
f (1 + h) − f (1)
h
0.1
−0.01
0.001
−0.0001
0.00001
0.000001
1.105342
0.9900497
1.001001
0.9999
1.00001
1.000001
Therefore,
(1 + h)1+h − 1
= 1.
h→0
h
lim
2.3. DERIVATIVE
2.3
13
Derivative
Let y = f (x) be a function with numerical inputs and outputs and a be such an input that f
is defined in its neighborhood, i.e., for all x sufficiently close to a from the left and right.
If the limit
f (a + h) − f (a)
h→0
h
lim
exists and is finite,
it is called the derivative of the function f (of y) at a.
Notations:
f 0 (a)
df (a)
dx
”f prime at a”, y 0 (a)
”dee f -dee x at a”,
dy(a)
dx
”y prime at a”,
”dee y-dee x at a”,
df
dx
when x = a
”dee f -dee x when x = a”,
dy
dx
when x = a
”dee y-dee x when x = a”
Conclusions:
• The derivative of y at a, y 0 (a), is the same as the IRC of y relative to x at a or the slope
of the graph of f at P (a, f (a)).
• y 0 (a) is the measure of how fast y changes relative to x at a.
• The unit of y 0 (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 line
at P (a, f (a)).
Example:
Let T (p) be the number of tickets from Atlanta to Boston that a certain airline sells in one
week when the price of a ticket is p dollars. Interpret the following:
14
CHAPTER 2. RATES OF CHANGE
(a) T (350) = 1757.
(b) T 0 (350) = −15.
(c)
dT
dp
= 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 15
tickets per dollar.
(c) When the ticket price is $280, the number of sold tickets is increasing at the rate of 22
tickets per dollar.
2.3. DERIVATIVE
2.3.1
15
Derivative as a Function
We can now consider finding the derivative of a function y = f (x) for any x where it exists:
f (x + h) − f (x)
h→0
h
y 0 (x) = lim
As x varies, y 0 (x) does. Thus, we can look at the derivative y 0 (x) as a function of x.
Note that the notation y 0 (x) is usually abriviated to y 0 .
Example: For y = 3x2 − 2, y 0 = 6x. Indeed:
3(x + h)2 − 2 − (3x2 − 2)
f (x + h) − f (x)
= lim
h→0
h→0
h
h
3x2 + 6xh + 3h2 − 3x2
3(x2 + 2xh + h2 ) − 2 − 3x2 + 2
= lim
= lim
h→0
h→0
h
h
2
6xh + 3h
= lim
= lim (6x + 3h) = 6x.
h→0
h→0
h
y 0 (x) = lim
2.3.2
Table Derivatives
By table derivatives, we shall understand those basic ones knoweledge of which is necessary and
is taken for granted in the future like 2 × 2 = 4 (compare to the table of multiplication ). Our
table of derivatives, for the time being, will be as follows:
Function
Derivative
y=b
y0 = 0
y = mx + b
y0 = m
y = xn
y 0 = nxn−1 (Power Rule)
f (x + h) − f (x)
b−b
= lim
= lim 0 = 0.
h→0
h→0
h→0
h
h
1. y 0 (x) = lim
f (x + h) − f (x)
m(x + h) + b − (mx + b)
= lim
h→0
h→0
h
h
2. y 0 (x) = lim
16
CHAPTER 2. RATES OF CHANGE
mh
mx + mh + b − mx − b
= lim
= lim m = m.
h→0 h
h→0
h→0
h
= lim
3. Let us illustrate the Power Rule for n = 2:
(x + h)2 − x2
x2 + 2xh + h2 − x2
f (x + h) − f (x)
y 0 (x) = lim
= lim
= lim
h→0
h→0
h→0
h
h
h
2
2xh + h
= lim
= lim (2x + h) = 2x.
h→0
h→0
h
Note that the Power Rule is valid for any real exponent n, e.g.:
1. y = x (n = 1),
y 0 = x0 = 1.
2. y = x3 (n = 3),
y 0 = 3x2 .
3. y = x−7 (n = −7),
4. y =
5. y =
y 0 = (−7)x−8 .
1
= x−2 (n = −2),
x2
√
x = x1/2 (n = 1/2),
y 0 = (−2)x−3 .
1
y 0 = 12 x−1/2 = √ .
2 x
1
1
6. y = √
= 2/3 = x−2/3 (n = −2/3),
3
2
x
x
7. y = xπ (n = π),
2.3.3
y 0 = (− 23 )x−5/3 .
y 0 = πxπ−1 .
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 derivatives
at x, then
0
1. f (x) ± g(x) = f 0 (x) ± g 0 (x)
(Sum/Difference Rule).
0
2. cf (x) = cf 0 (x)
(Constant Factor Rule).
Note that the Sum/Difference Rule is valid for any finite number of terms, e.g.:
0
f (x) + g(x) + h(x) = f 0 (x) + g 0 (x) + h0 (x).
Example:
Find the derivative of the function
y = 7x2 −
√
12
+ 3 x.
2
x
2.3. DERIVATIVE
Solution: First, let us rewrite all the terms in power form:
Using the Rules of Differentiation, we have:
17
y = 7x2 − 12x−2 + 3x1/2 .
y 0 = Sum/Difference Rule = (7x2 )0 − (12x−2 )0 + (3x1/2 )0
= Constant Factor Rule = 7(x2 )0 − 12(x−2 )0 + 3(x1/2 )0 = 7(2x) − 12(−2x−3 )
1
24
3
+ 3( x−1/2 ) = 14x + 3 + √ .
2
x
2 x
18
CHAPTER 2. RATES OF CHANGE
Product Rule. Let functions f and g both have derivatives at x, then
0
f (x)g(x) = f 0 (x)g(x) + f (x)g 0 (x)
(Product Rule).
Example:
Find the derivative of the function
y=
1 √
( 3 x − 7x + 2.75).
4.5x +
x
2
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 0 = Product Rule
= (4.5x2 + x−1 )0 (x1/3 − 7x + 2.75) + (4.5x2 + x−1 )(x1/3 − 7x + 2.75)0
= Sum/Difference and Constant Factor Rules
−2
= (9x − x )(x
1/3
1 −2/3
− 7x + 2.75) + (4.5x + x ) x
−7 .
3
2
−1
Note that the Product Rule is valid for any finite number of factors, e.g.:
0
f (x)g(x)h(x) = f 0 (x)g(x)h(x) + f (x)g 0 (x)h(x) + f (x)g(x)h0 (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 function
defined 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,
q
√
√
√
x = 4 x,
f ◦ f (x) = f (f (x)) = f ( 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 at
g(x). Then the composition f ◦ g has a derivative at x:
0
f (g(x)) = f 0 (g(x))g 0 (x)
(Chain Rule)
Examples:
1. y = (1 − 2x)100 ,
y 0 = Chain Rule = 100(1 − 2x)99 (1 − 2x)0
= 100(1 − 2x)99 (−2) = −200(1 − 2x)99 .
27
2. y = 2
= 27(3x2 − 5)−1 ,
y 0 = Constant Factor Rule
3x − 5
0
= 27 (3x2 − 5)−1 = Chain Rule = 27 −(3x2 − 5)−2 (3x2 − 5)0
−162x
= 27 −(3x2 − 5)−2 (6x) = −162x(3x2 − 5)−2 =
.
(3x2 − 5)2
3. y =
√
5
x7 − 4x = (x7 − 4x)1/5 ,
y 0 = Chain Rule
20
CHAPTER 2. RATES OF CHANGE
1
1
= (x7 − 4x)−4/5 (x7 − 4x)0 = (x7 − 4x)−4/5 (7x6 − 4)
5
5
7x6 − 4
= p
.
5 5 (x7 − 4x)4
4. y = (5x3 − 31)2 (1 − 3x)11 ,
y 0 = Product Rule
0
0
= (5x3 − 31)2 (1 − 3x)11 + (5x3 − 31)2 (1 − 3x)11
= Chain Rule = 2(5x3 − 31)(5x3 − 31)0 (1 − 3x)11
+ (5x3 − 31)2 11(1 − 3x)10 (1 − 3x)0 = 2(5x3 − 31)(15x2 )(1 − 3x)11
+ (5x3 − 31)2 11(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 quotient
f (x)
g(x)
0
=
f 0 (x)g(x) − f (x)g 0 (x)
2
g(x)
f (x)
, which is as follows:
g(x)
(Quotient Rule).
However, applying this rule can be avoided by representing the quotient as a product
−1
f (x)
= f (x) g(x)
g(x)
and then applying the Product and Chain Rules.
3x5 − 2
Example: Let y = 2
. Then y = (3x5 − 2)(x2 − 4x3 )−2 and
3
2
(x − 4x )
y 0 = Product Rule
0
= (3x5 − 2)0 (x2 − 4x3 )−2 + (3x5 − 2) (x2 − 4x3 )−2
= Chain Rule = 15x4 (x2 − 4x3 )−2 + (3x5 − 2)(−2)(x2 − 4x3 )−3 (x2 − 4x3 )0
= 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 of
three functions f , g, and h, we have:
0
0
f (g(h(x)) = f 0 (g(h(x))) g(h(x)) = f 0 (g(h(x)))g 0 (h(x))h0 (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{z· · · 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
am an = am+n
am
= am−n (m > n)
an
n
am = amn
(ab)n = an bn
n
a
an
= n
b
b
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: a0 a = 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 cannot
but define a−n as follows:
a−n :=
1
an
Indeed, a−n an = a−n+n = a0 = 1. Whence: a−n = 1/an .
Examples:
1
1
= .
1
5
5
1
1
2. (−3)−3 =
=− .
3
(−3)
27
−4
3. 0
is undefined.
1. 5−1 =
Powers with Fractional Exponents
Defining powers with fractional exponents is also based upon preserving the Exponent Laws.
m
Let a fraction , where m is integer, n is natural, be in the lowest terms. Then:
n
√
m
a n := n am
√
m
m
m n
Indeed, a n = a n n = am . Whence: a n = n am .
Examples:
√
1. 51/2 = 5.
1
= √
.
3
9
(−3)2
2. (−3)−2/3 = p
3
3. (−7)3/4
1
is undefined.
Powers with Irrational Exponents
√
For 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 Laws
hold.
With powers so defined, the Exponent Laws are valid for arbitrary real exponents.
Exponent Laws
3.1. POWERS AND EXPONENT LAWS
23
ax
= ax−y
y
a
x
ax
a
= x
b
b
ax ay = ax+y
(ab)x = ax bx
ax
y
= axy
An expression of the form ax is called a power or an exponential, a and x being
called 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;
1
• a−2/5 = √
is defined for a 6= 0;
5
a2
√
• a1/2 = a is defined for a ≥ 0;
1
• a−3/4 = √
4
a3
• aπ
is defined for a > 0.
is defined for a > 0.
Examples:
1. Find the exact value of
10−1/3 100001/3 .
Applying the Exponent Laws, we have:
10−1/3 100001/3 = 10−1/3 104
1/3
= 10−1/3 104/3 = 10−1/3+4/3 = 101 = 10.
2. Simplify the expression and eliminate any negative and fractional exponents. Assume
that all letters designate positive numbers.
y 2 z −1
8x6
−1/3 Applying the Exponent Laws, we have:
x−1/3 z −1/2
√
2y
2
24
CHAPTER 3. EPONENTIALS AND LOGARITHMS
−1/3 −1/3 −1/2 2
x
z
z −1
−1 −6 2 −1 −1/3 −1/2 −1/3 −1 −1/2 2
√
=
x
y
z
2
x
y
z
8
8x6 y −2
2y
−1/3 −1 −1/3 −1/2 2 −1/3 2 −1 2 −1/2 2
−1/3
−1/3
= 8−1
x−6
y2
z
2
x
y
z
1
= 81/3 x2 y −2/3 z 1/3 2−1 x−2/3 y −2 z −1 = 2 x2−2/3 y −2/3−2 z 1/3−1
2
√
3
4
x
√
= x4/3 y −8/3 z −2/3 = p
.
3
3
y8 z2
3.1. POWERS AND EXPONENT LAWS
25
Problems
Find the exact value of the expression:
1.
27
−
8
2/3
Answer:
2. 32−0.2
9
.
4
Answer: 0.5.
3. 31/2 91/4
Answer: 3.
Simplify the expression and eliminate any negative and fractional exponents. Assume that all
letters designate positive numbers.
√
10
4. (4b)1/2 (8b2/5 )
Answer: 16 b9 .
5. (5x4 y −4/5 )3 (8y 2 )2/3
(ab)3/2
(a3 b−4 )2/3
2 −3 −3 −2 −1 1/2
ab
x b
7.
x−1 y 1/2
a3/2 y 1/3
6.
x12
p .
Answer: 500 15
y 16
√
6 25
b
Answer: √ .
a
√ p
b17 3 y 4
√
Answer: 4
.
a27 x4
Produce the graph of the function on the calculator, find its derivative, and answer the questions.
8. y =
√
3
x + 2.
Question: Why does the function increase on (−∞, ∞)?
9. y =
1
.
+5
2x3
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 happens
at 0?
26
CHAPTER 3. EPONENTIALS AND LOGARITHMS
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).
concave up
y → 0 as x → −∞
y → ∞ as x → ∞
0 < a < 1 decreasing concave up
y → ∞ as x → −∞
y → 0 as x → ∞
a>1
increasing
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:
x
1
= lim (1 + x)1/x
e = lim 1 +
x→0
x→∞
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 Decay
A 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 form
y = 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 Change
Consider an exponential function
y = Cax .
As the input changes by 1, from x to x + 1, the output changes from Cax to Cax+1 = Cax a,
i.e., by a factor of a.
The absolute change is Cax+1 − Cax . The relative change is
Cax+1 − Cax
Cax a − Cax
Cax (a − 1)
=
=
= a − 1.
Cax
Cax
Cax
The relative change, expressed in percent form, i.e., as (a − 1)100%, is called the percentage
change.
• 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. It
depends 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. Then
a = 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
28
CHAPTER 3. EPONENTIALS AND LOGARITHMS
Problems
Determine whether the exponential function grows or decays. Find the percentage change.
1. y = 4ex .
t
2
.
2. m(t) = 3.75
5
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 the
origional amount. Find the amount A(t) on John’s account as a function of time t, in
years since the investment was made.
7. Julie invested $2,500 at a compound interest rate of 10% per year applied annually to the
amount of the previous year. Find the amount A(t) on Julie’s account as a function of
time 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.
3.2. EXPONENTIAL FUNCTIONS
29
Simple Interest vs. Compound Interest
Let
• 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 into
n compounding periods, e.g., annually, semiannually, quaterly, or monthly. This can be done
according 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 principal only.
The interest is earned based on
the amount of the previous compounding period.
r
A(t) = P + P
nt = P + (P r)t
n
r
A(t) = P 1 +
n
Linear
Problem 1:
=P
r
1+
n
n # t
Exponential
"
Rate of Change = P r $ per year
"
nt
Percentage Change =
r
1+
n
n
#
− 1 100 % per year
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 investment
was made.
(b) Find the amount after 5 years and 2 months.
30
CHAPTER 3. EPONENTIALS AND LOGARITHMS
Continuous Compounding
By letting n → ∞, we partition a year into shorter and shorter compounding periods arriving
at the situation when compounding happens continuously.
We have:
r
1+
n
"
n
=
1
1+
n/r
n/r #r
→ er
as n → ∞.
Simple Interest Scheme
Compound Interest Scheme
A(t) = P + (P r)t
t
A(t) = P ert = P er
Linear
Exponential
Rate of Change = P r $ per year Percentage Change = er − 1 100 % per year
Problem 2:
Answer the questions of Problem 1 for the same amount invested at the
same interest rate, compounded continuously.
APR and APY
• Annual Percentage Rate (APR), also called the nominal interest rate, is a compound
interest applied a certain number of times per year, say, semiannually, quaterly, or continuously.
• The Annual Percentage Yield (APY), also called the effective interest rate, for a given
APR, is the percentage by which the amount, invested at the APR, is exponentially
increasing 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:
2t
t
0.06
A(t) = P 1 +
= P (1.03)2t = P (1.03)2 = P (1.0609)t .
2
Hence, APY = (1.0609 − 1)100 = 6.09% per year.
3.2. EXPONENTIAL FUNCTIONS
31
Because both the size of the interest rate and the number of compounding periods contribute
into the amount of interest earned per year, comparing APR’s may be puzzling. Comparing
APY’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 semiannually
or 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 of
8.5% per year, compounded quaterly. Find the future value of these investments 1 year after
the third deposit is made.
The present value of a sum of money is the amount that must be invested now, at a given
APR, 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 order
to meet her January tuition payment. How much should she invest on August 1 at an APR of
5.5% 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 3
years, assuming that the effective rate is 4.8% per year?
32
CHAPTER 3. EPONENTIALS AND LOGARITHMS
Logistic Functions
A population, the assumption of limited resources not being imposed, increases without boundaries 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 a
tendentcy to level off and can be modeled by a so called logistic function:
y=
L
+ C,
1 + Ae−Bx
where L, A, B, and C are positive numbers.
As t → ∞, e−Bx → 0. Therefore,
"
#
L
lim
+ C = L + C.
x→∞ 1 + Ae−Bx
The function increases approaching L + C as t increases. The number L + C is called the
limiting value of the logistic function as t → ∞ (see fig.).
A function of the form:
L
+ C,
1 + AeBx
whith positive L, A, B, and C, is also called logistic.
In this, as x → ∞, eBx → ∞. Therefore,
"
#
L
lim
+ C = C.
x→∞ 1 + AeBx
y=
The function decreases approaching C as x increases. C is the limiting value x → ∞ in this
case (see fig.).
Problems: Determine whether the logistic function increases or decreases. Identify the
initial and limiting values of the function.
1. y =
123
+5
1 + 2e1.64x
2. P (t) =
18
+ 24
1 + 5e−0.24t
3.3. LOGARITHMS
3.3
33
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.
5−1 = 0.2.
−3
1
= 33 = 27.
3. log1/3 27 = −3 since
3
√
4. log49 7 = 1/2 since 491/2 = 49 = 7.
2. log5 0.2 = −1 since
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.
1
64
√
4. log 10.
Answer: −3.
3. log4
Express in exponential/logarithmic form.
5. log5 x = 3
1/2
9. 81
=9
6. log x = 2
−3
10. 10
= 0.001
7. ln x = 1/2
x
11. 10 = 2
8. ln x = −1/3
12. ex = 7
34
CHAPTER 3. EPONENTIALS AND LOGARITHMS
Immediate Properties of Logarithms
The following properties of logarithms are immediate cosequences of the definition.
loga 1 = 0
loga a = 1
loga ax = x,
aloga x = 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.
√
log 9
1/2
1
3. ( 10)log 9 = 101/2
= 10 2 log 9 = 10log 9
= 91/2 = 3.
Problems: Evaluate the expression.
1
1. log3 √
27
Answer: −3/2
2. log8 4
3. 5− log5 18
4. 23 log2 6 .
Answer: 1/18
3.3. LOGARITHMS
35
Laws of Logarithms
For any positive A and B and any real c:
loga (AB) = loga A + loga B
A
loga
= loga A − loga B
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.
x3 ez
ln √
y
Solution:
1
x3 ez
√
ln √ = ln x3 ez − ln y = ln x3 + ln ez − ln y 1/2 = 3 ln x + z − ln y
y
2
2. Evaluate the expression log2 112 − log2 7
Solution: log2 112 − log2 7 = log2
112
7
= log2 16 = 4.
Problems:
Simplify the expression to a form tha does not contain logarithms of products, quotients, and
powers. Assume that all the letters designate positive numbers.
√
4
ab3
1
3
Answer:
ln a + ln b + c
1. ln −c
e
4
4
r√
a+b
3
2. log
10c d4
Evaluate the expression.
1
log12 81 + 4 log12 2
2
1
2. log 5 + 3 log 2 − log 64
3
1.
Answer: 2
36
CHAPTER 3. EPONENTIALS AND LOGARITHMS
Changing Base in Logarithms
Changing the base from a to a new base b performed according to the Change of Base
Formula:
loga x =
logb x
logb a
Indeed,
logb a logb x
logb x
logb x
a logb a = a = blogb a = blogb a logb a = b logb a = blogb x = x.
Calculator computations require changing the base to e or 10, e.g.:
ln 5
log 5
log2 5 =
=
.
ln 2
log 2
Example:
Evaluate the expression log3 0.2 + log9 25.
Solution:
log3 25
1
= log3 0.2 + log3 25
log3 9
2
= log3 0.2 + log3 5 = log3 (0.2 · 5) = log3 1 = 0.
log3 0.2 + log9 25 = log3 0.2 +
= log3 0.2 + log3 251/2
Problems: Use a calculator to evaluate the expression, correct to four decimal places.
Change the base when needed.
1. log 7
3. log3 17
2. ln 10
4. log7 14
3.3. LOGARITHMS
37
Changing Base to e in Exponentials
An exponential function
y = Cax
can be rewritten in the form:
y = Cekx .
The procedure is as follows:
x
y = Cax = a = eln a = C eln a = 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 ,
2. y = (0.7)x = eln 0.7x ,
k = ln 2.2 > 0.
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 decimal
places.
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
38
CHAPTER 3. EPONENTIALS AND LOGARITHMS
3.4
Exponential and Logarithmic Equations
Solve the equation. Evaluate the logarithms using a calculator and round off your
answers up to two decimal places.
1. 3x = 12.
2. (3.171)x = 73.
3. 102x−1 = 2.
4. 4(1 + 75x ) = 9.
5.
10
= 2.
1 + e−x
6. ln x = −2.
7. log(1 − 2x) = 0.
8. log2 (3x − 4) = 5.
3.5
Applications of Logarithms
Exponential Growth
Consider 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 3000
bacteria. 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 an
input 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 .
Cax ad = 2Cax
ad = 2
Therefore,
d = loga 2
Note that the doubling period, d, depends neither on C nor on x and is as characteristic 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, compounded
quaterly.
(a) Find the amount A(t) on the account as a function of time t, in years since the investment
was made.
(b) When will the amount attain the level of $8,000?
(c) Find the doubling period for the investment.
Exponential Decay
Consider 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:
40
CHAPTER 3. EPONENTIALS AND LOGARITHMS
The decay of a radioactive substance is exponential:
m(t) = m0 at ,
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 substance
to decay to one-half of its initial mass.
We have:
m 0 ah =
ah =
m0
2
1
2
Thus,
h = loga (1/2)
As is seen, the half-life does not depend on the initial mass, it depends on the kind
of radioactive substance only.
Given a half-life, h, the radioactive decay can be represented as follows:
t/h
1
m(t) = m0
2
Radioactive Carbon Dating
The radioactive carbon dating method is used by archeologists to determine the age of an
ancient object. It is based on the fact that the carbon dioxide, CO2 , in the atmoshere and in
all living organisms contains radioactive carbon-14, 14 C, with a half-life of about 5730 years,
and nonradioactive carbon-12, 12 C, in a fixed proportion.
After an organism dies, it stops assimilating 14 C, which begins to decay exponentially. The
time elapsed since the death of the organism is determined by measuring the amount of 14 C
left in it.
Problem 3: The burial cloth of an Egyptian mummy is estimated to contain 59% of
carbon-14 it contained initially. How long ago was the mummy buried? (The half-life of
carbon-14 is 5730 years).