EXAMPLE 6 Lecture 6: Inverse Functions - Part...

Lecture 6: Inverse Functions - Part 1

Note:

Section 1.5 in Stewart is quite heavy. So the topic of inverses and logarithmic functions willsplit over two lectures. In this set of notes, only one-to-one functions and the properties of log-arithms will be covered. Graphs of logarithmic functions and inverse trigonometric functionswill be covered in Lecture 7.

Objective:

• Present the concept of a one-to-one function and demonstrate how to compute itsinverse.

• Discuss how the domain and range of a function and its inverse are related.

• Present the logarithmic function and its properties.

Introduction

Recall, that given the graph of a curve, a relation qualifies as a function if it passes the verticalline test.

SECTION 1.1 Four Ways to Represent a Function 13

following domain convention: the domain of the function is the set of all inputs for which the formula makes sense and gives a real-number output.

EXAMPLE 6 Find the domain of each function.

(a) f sxd − sx 1 2 (b) tsxd −1

x 2 2 x

SOLUTION(a) Because the square root of a negative number is not defined (as a real number), the domain of f consists of all values of x such that x 1 2 > 0. This is equivalent to x > 22, so the domain is the interval f22, `d.(b) Since

tsxd −1

x 2 2 x−

1

xsx 2 1d

and division by 0 is not allowed, we see that tsxd is not defined when x − 0 or x − 1. So the domain of t is

hx | x ± 0, x ± 1j

which could also be written in interval notation as

s2`, 0d ø s0, 1d ø s1, `d� ■

■ Which Rules Define Functions?Not every equation defines a function. The equation y − x 2 defines y as a function of x because the equation determines exactly one value of y for each value of x. However, the equation y 2 − x does not define y as a function of x because some input values x corre-spond to more than one output y; for instance, for the input x − 4 the equation gives the outputs y − 2 and y − �2.

Similarly, not every table defines a function. Table 3 defined C as a function of w — each package weight w corresponds to exactly one mailing cost. On the other hand, Table 4 does not define y as a function of x because some input values x in the table correspond to more than one output y; for instance, the input x − 5 gives the outputs y − 7 and y − 8.

Table 4

x 2 4 5 5 6

y 3 6 7 8 9

What about curves drawn in the xy-plane? Which curves are graphs of functions? The following test gives an answer.

The Vertical Line Test A curve in the xy-plane is the graph of a function of x if and only if no vertical line intersects the curve more than once.

The reason for the truth of the Vertical Line Test can be seen in Figure 13. If each vertical line x − a intersects a curve only once, at sa, bd, then exactly one function value is defined by f sad − b. But if a line x − a intersects the curve twice, at sa, bd and sa, cd, then the curve can’t represent a function because a function can’t assign two different values to a.

a

x=a

(a, b)

0

a

(a, c)

(a, b)

x=a

0 x

y

x

y

(a) This curve represents a function.

(b) This curve doesn’t represent a function.

FIGURE 13

Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

SECTION 1.1 Four Ways to Represent a Function 13

following domain convention: the domain of the function is the set of all inputs for which the formula makes sense and gives a real-number output.

EXAMPLE 6 Find the domain of each function.

(a) f sxd − sx 1 2 (b) tsxd −1

x 2 2 x

SOLUTION(a) Because the square root of a negative number is not defined (as a real number), the domain of f consists of all values of x such that x 1 2 > 0. This is equivalent to x > 22, so the domain is the interval f22, `d.(b) Since

tsxd −1

x 2 2 x−

1

xsx 2 1d

and division by 0 is not allowed, we see that tsxd is not defined when x − 0 or x − 1. So the domain of t is

hx | x ± 0, x ± 1j

which could also be written in interval notation as

s2`, 0d ø s0, 1d ø s1, `d� ■

■ Which Rules Define Functions?Not every equation defines a function. The equation y − x 2 defines y as a function of x because the equation determines exactly one value of y for each value of x. However, the equation y 2 − x does not define y as a function of x because some input values x corre-spond to more than one output y; for instance, for the input x − 4 the equation gives the outputs y − 2 and y − �2.

Similarly, not every table defines a function. Table 3 defined C as a function of w — each package weight w corresponds to exactly one mailing cost. On the other hand, Table 4 does not define y as a function of x because some input values x in the table correspond to more than one output y; for instance, the input x − 5 gives the outputs y − 7 and y − 8.

Table 4

x 2 4 5 5 6

y 3 6 7 8 9

What about curves drawn in the xy-plane? Which curves are graphs of functions? The following test gives an answer.

The Vertical Line Test A curve in the xy-plane is the graph of a function of x if and only if no vertical line intersects the curve more than once.

The reason for the truth of the Vertical Line Test can be seen in Figure 13. If each vertical line x − a intersects a curve only once, at sa, bd, then exactly one function value is defined by f sad − b. But if a line x − a intersects the curve twice, at sa, bd and sa, cd, then the curve can’t represent a function because a function can’t assign two different values to a.

a

x=a

(a, b)

0

a

(a, c)

(a, b)

x=a

0 x

y

x

y

(a) This curve represents a function.

(b) This curve doesn’t represent a function.

FIGURE 13



Function Not a function

Functions can further subdivided into two categories: one-to-one or many-to-one. If it is one-to-one, then the function has an inverse. This set of notes covers: what one-to-one functions are;how to compute their inverse; and investigates how the domain and range of these functionsare related. We will also examine the class of functions which are inverses of the exponentialfunction - these are the logarithmic functions.

1

September 8, 2020 201-NYA-05: Calculus I Sect. 00006 and 00015

One-to-One Functions

A function is considered to be one-to-one if no two elements from the domain are mapped tothe same element in the range. In other words, the function never takes the same value twice.

Here are some examples of one-to-one functions.

x

y

x

y

x

y

f(x) = ex f(x) = −x3 f(x) = ln x

Observe that in each case, no two values of x produce the same value of y. Here is the formaldefinition of a one-to-one function.

Definition 1

A function f is called one-to-one if it never assumes the same value twice. Mathematically,this can be expressed as

f(x1) 6= f(x2) whenever x1 6= x2

Example 1

A function is given by the table of values below. Determine if it is one-to-one or not.

x 1 2 3 4 5 6f(x) 1.5 2.0 3.6 5.3 2.8 2.0

Solution 1

The above table of values qualifies as a one-to-one function. Every value of x is mapped toa different value of f(x).

Example 2

A function is given by the table of values below. Determine if it is one-to-one or not.

x 1 2 3 4 5 6f(x) 1.2 0.2 1.2 4.3 5.8 6.0

2


Solution 2

The above table of values does not qualify as a one-to-one function. x = 1 and x = 3 ismapped to the value, 1.2

Example 3

For each of the following functions, state if it is one-to-one or not. Justify your answerwith a sentence or two, or with some calculations.

a. f(x) = x2

b. f(x) = x+ 1

c. f(x) =(12

)xd. f(x) = sin(x)

Solution 3

a. f(x) = x2

This is a parabola opening upwards. Moreover, it is an even function ∵ f(x) = f(−x).If x = −2 and x = 2 then f(−2) = 4 = f(2). So f(x) = x2 is not one-to-one.

b. f(x) = x+ 1

This is a line that has a slope of 1. Since the slope is positive, this function is strictlyincreasing on R. Thus, for every x1 6= x2, the function will produce a distinct valueof y (ie. f(x1) 6= f(x2)). So, f is one-to-one.

c. f(x) =(12

)x.

This is an exponential function. Since 0 < a < 1, the function is strictly decreasingon R. Because of this, whenever x1 6= x2 ⇒ f(x1) 6= f(x2). So f is one-to-one.

d. f(x) = sin(x)

This is a trigonometric function. It exhibits periodic behaviour (repeats itself aftersome interval). If x = 0 and x = π then f(0) = 0 = f(π). Since f(x1) = f(x2) eventhough x1 6= x2, we can conclude that f is not one-to-one.

The Horizontal Line Test

In addition to using a table of values or the technical definition to ascertain if a function isindeed one-to-one or not, there is the horizontal line test. It works in a similar fashion to thevertical line test.

3


The Horizontal Line Test

A curve in the xy−plane is the graph of a one-to-one function if and only if no horizontalline intersects the curve more than once.

x

y

One-to-one

x

y

Not one-to-one

Example 4

For each of the following questions, use the horizontal line test to determine if the functionis one-to-one.

a.x

y

b.x

y

c.x

y

d.x

y

Solution 4

a. One-to-one.

b. Not one-to-one.

c. Not one-to-one.

a. One-to-one.

4


Inverses

If a function is one-to-one, then it has an inverse. Loosely speaking, the inverse reverses themapping that is produced when the function associates one element from the domain to therange. So, if f(x) is a one-to-one function, then given some value of y, it is possible to map itback to the value of x which generated it.

Definition 2

Let f be a one-to-one function with domain A and range B. Then its inverse function,f−1 has domain B and range A and is defined by

f−1(y) = x ⇔ f(x) = y

for any y ∈ B

Remarks

• The definition simply says that if f maps x to y, then f−1 maps y back into x

• The domain of f = the range of f−1

• The domain of f−1 = the range of f

Graphically, f(x) and f−1(x) are reflections of each other along the line y = x.

SECTION 1.5 Inverse Functions and Logarithms 57

EXAMPLE 4 Find the inverse function of f sxd − x 3 1 2.

SOLUTION According to (5) we first write

y − x 3 1 2

Then we solve this equation for x:

x 3 − y 2 2

x − s3 y 2 2

Finally, we interchange x and y:

y − s3 x 2 2

Therefore the inverse function is f 21sxd − s3 x 2 2 . ■

The principle of interchanging x and y to find the inverse function also gives us the method for obtaining the graph of f 21 from the graph of f. Since f sad − b if and only if f 21sbd − a, the point sa, bd is on the graph of f if and only if the point sb, ad is on the graph of f 21. But we get the point sb, ad from sa, bd by reflecting about the line y − x. (See Figure 8.)

0

y

x

(b, a)

(a, b)

y=x

0

y

x

f –!

y=x f

FIGURE 8 FIGURE 9

Therefore, as illustrated by Figure 9:

The graph of f 21 is obtained by reflecting the graph of f about the line y − x.

EXAMPLE 5 Sketch the graphs of f sxd − s21 2 x and its inverse function using the same coordinate axes.

SOLUTION First we sketch the curve y − s21 2 x (the top half of the parabola y 2 − 21 2 x, or x − 2y 2 2 1) and then we reflect about the line y − x to get the graph of f 21. (See Figure 10.) As a check on our graph, notice that the expression for f 21 is f 21sxd − 2x 2 2 1, x > 0. So the graph of f 21 is the right half of the parabola y − 2x 2 2 1 and this seems reasonable from Figure 10. ■

■ Logarithmic FunctionsIf b . 0 and b ± 1, the exponential function f sxd − bx is either increasing or decreasing and so it is one-to-one by the Horizontal Line Test. It therefore has an inverse function f 21, which is called the logarithmic function with base b and is denoted by logb. If we use the formulation of an inverse function given by (3),

f 21sxd − y &? f syd − x

In Example 4, notice how f 21 reverses the effect of f . The function f is the rule “Cube, then add 2”; f 21 is the rule “Subtract 2, then take the cube root.”

0

y=xy=ƒ

(0, _1)

y=f –!(x)

(_1, 0)

y

x

FIGURE 10



Example 5

64 CHAPTER 1 Functions and Models

We know that the lines x − 6�y2 are vertical asymptotes of the graph of tan. Since the graph of tan21 is obtained by reflecting the graph of the restricted tangent function about the line y − x, it follows that the lines y − �y2 and y − 2�y2 are horizontal asymptotes of the graph of tan21.

The remaining inverse trigonometric functions are not used as frequently and are sum-marized here.

12 y − csc21x (| x | > 1) &? csc y − x and y [ s0, �y2g ø s�, 3�y2g

y − sec21x (| x | > 1) &? sec y − x and y [ f0, �y2d ø f�, 3�y2d

y − cot21x sx [ Rd &? cot y − x and y [ s0, �d

The choice of intervals for y in the definitions of csc21 and sec21 is not universally agreed upon. For instance, some authors use y [ f0, �y2d ø s�y2, �g in the definition of sec21. [You can see from the graph of the secant function in Figure 26 that both this choice and the one in (12) will work.]

0

y

x_1

2ππ

FIGURE 26 y − sec x

1.5 Exercises

1. (a) What is a one-to-one function? (b) How can you tell from the graph of a function whether it

is one-to-one?

2. (a) Suppose f is a one-to-one function with domain A and range B. How is the inverse function f 21 defined? What is the domain of f 21 ? What is the range of f 21 ?

(b) If you are given a formula for f, how do you find a formula for f 21 ?

(c) If you are given the graph of f, how do you find the graph of f 21 ?

3–16 A function is given by a table of values, a graph, a formula, or a verbal description. Determine whether it is one-to-one.

3. x 1 2 3 4 5 6

f sxd 1.5 2.0 3.6 5.3 2.8 2.0

4. x 1 2 3 4 5 6

f sxd 1.0 1.9 2.8 3.5 3.1 2.9

5. 6. y

xx

y

y

xx

y

7. 8.

y

xx

y

y

xx

y

9. f sxd − 2x 2 3 10. f sxd − x 4 2 16

11. r std − t 3 1 4 12. tsxd − s3 x

13. tsxd − 1 2 sin x 14. f sxd − x 4 2 1, 0 ⩽ x ⩽ 10

15. f std is the height of a football t seconds after kickoff.

16. f std is your height at age t.

17. Assume that f is a one-to-one function. (a) If f s6d − 17, what is f 21s17d? (b) If f 21s3d − 2, what is f s2d?

18. If f sxd − x 5 1 x 3 1 x, find f 21s3d and f s f 21s2dd.

19. If tsxd − 3 1 x 1 e x, find t21s4d.

20. The graph of f is given. (a) Why is f one-to-one? (b) What are the domain and range of f 21 ? (c) What is the value of f 21s2d? (d) Estimate the value of f 21s0d.

y

x0 1

1

21. The formula C − 59 sF 2 32d, where F > 2459.67, expresses

the Celsius temperature C as a function of the Fahrenheit temperature F. Find a formula for the inverse function and interpret it. What is the domain of the inverse function?



Consider the graph of f(x) above.

a. What is the domain of f(x)?

b. What is the range of f(x)?

c. What is the domain of f−1(x)?

5


d. What is the range of f−1(x)?

Solution 5

a. Df(x) = [−3, 3]

b. Rf(x) = [−1, 3]

c. Df−1(x) = Rf(x) = [−1, 3]

d. Rf−1(x) = Df(x) = [−3, 3]

Example 6

64 CHAPTER 1 Functions and Models

We know that the lines x − 6�y2 are vertical asymptotes of the graph of tan. Since the graph of tan21 is obtained by reflecting the graph of the restricted tangent function about the line y − x, it follows that the lines y − �y2 and y − 2�y2 are horizontal asymptotes of the graph of tan21.

The remaining inverse trigonometric functions are not used as frequently and are sum-marized here.

12 y − csc21x (| x | > 1) &? csc y − x and y [ s0, �y2g ø s�, 3�y2g

y − sec21x (| x | > 1) &? sec y − x and y [ f0, �y2d ø f�, 3�y2d

y − cot21x sx [ Rd &? cot y − x and y [ s0, �d

The choice of intervals for y in the definitions of csc21 and sec21 is not universally agreed upon. For instance, some authors use y [ f0, �y2d ø s�y2, �g in the definition of sec21. [You can see from the graph of the secant function in Figure 26 that both this choice and the one in (12) will work.]

0

y

x_1

2ππ

FIGURE 26 y − sec x

1.5 Exercises

1. (a) What is a one-to-one function? (b) How can you tell from the graph of a function whether it

is one-to-one?

2. (a) Suppose f is a one-to-one function with domain A and range B. How is the inverse function f 21 defined? What is the domain of f 21 ? What is the range of f 21 ?

(b) If you are given a formula for f, how do you find a formula for f 21 ?

(c) If you are given the graph of f, how do you find the graph of f 21 ?

3–16 A function is given by a table of values, a graph, a formula, or a verbal description. Determine whether it is one-to-one.

3. x 1 2 3 4 5 6

f sxd 1.5 2.0 3.6 5.3 2.8 2.0

4. x 1 2 3 4 5 6

f sxd 1.0 1.9 2.8 3.5 3.1 2.9

5. 6. y

xx

y

y

xx

y

7. 8.

y

xx

y

y

xx

y

9. f sxd − 2x 2 3 10. f sxd − x 4 2 16

11. r std − t 3 1 4 12. tsxd − s3 x

13. tsxd − 1 2 sin x 14. f sxd − x 4 2 1, 0 ⩽ x ⩽ 10

15. f std is the height of a football t seconds after kickoff.

16. f std is your height at age t.

17. Assume that f is a one-to-one function. (a) If f s6d − 17, what is f 21s17d? (b) If f 21s3d − 2, what is f s2d?

18. If f sxd − x 5 1 x 3 1 x, find f 21s3d and f s f 21s2dd.

19. If tsxd − 3 1 x 1 e x, find t21s4d.

20. The graph of f is given. (a) Why is f one-to-one? (b) What are the domain and range of f 21 ? (c) What is the value of f 21s2d? (d) Estimate the value of f 21s0d.

y

x0 1

1

21. The formula C − 59 sF 2 32d, where F > 2459.67, expresses

the Celsius temperature C as a function of the Fahrenheit temperature F. Find a formula for the inverse function and interpret it. What is the domain of the inverse function?



Consider the graph of f(x) above.

a. What is the value of f−1(2) and f(2)?

b. What is the value of f−1(−1) and f(−1)?

c. What is the value of f−1(3)?

Solution 6

a. f−1(2) = 0

f(2) ≈ 2.9

b. f−1(−1) = −3

f(−1) = 1

c. f−1(3) = 3

6


Computing The Inverse of A Function

Often times, we will be working with the algebraic representation of a function. As such, wewould also want to generate the formula for its inverse.

How To Compute The Inverse of A Function

Suppose that f(x) is one-to-one.

1. Write y = f(x)

2. Interchange the x’s with the y’s

3. Solve for y. Once y has been isolated, the resulting equation will be y = f−1(x);which is the algebraic representation/formula for the inverse.

One way to check that the expression for the inverse is correct is to take the compositionbetween f(x) and f−1(x).

Theorem 1

If f(x) is a one-to-one function with domain A and range B, and f−1(x) is its inverse, then

(f−1 ◦ f)(x) = f−1(f(x)) = x for every x ∈ A(f ◦ f−1)(x) = f(f−1(x)) = x for every x ∈ B

Remarks

• The first equation says that if we start with x, apply f , and then apply f−1, we arriveback at x.

• The second statement says the same thing, but with order of the composition inter-changed.

Example 7

Find the inverse for the function f(x) = 3√x+ 2

Solution 7

Let y = 3√x+ 2 ⇒ x = 3

√y + 2 ; interchange the variables

x3 = y + 2 ; isolate for y

x3 − 2 = y

x3 − 2 = f−1(x) ; this is the inverse

7


Example 8

Let f(x) =4x− 1

2x+ 3

a. Find the inverse of the function. You may assume that it is one-to-one.

b. What is the domain of f(x)? What is the range of f−1(x)?

c. What is the domain of f−1(x) and the range of f(x)?

Solution 8

a.

Let y =4x− 1

2x+ 3⇒ x =

4y − 1

2y + 3; swap the variables

x(2y + 3) = 4y − 1 ; isolate for y

2xy + 3x = 4y − 1

3x+ 1 = 4y − 2xy

3x+ 1 = y(4− 2x)

3x+ 1

4− 2x= y

3x+ 1

4− 2x= f−1(x) ; rename the function

b. Df(x) = Rf−1(x) =(−∞,−3

2

)∪(−3

2,∞)

c. Df−1(x) = Rf(x) = (−∞, 2) ∪ (2,∞)

Example 9

Let f(x) = 1 +√x+ 3

a. Find the inverse of the function. You may assume that it is one-to-one.

b. What is the domain of f(x)? What is the range of f−1(x)?

c. What is the domain of f−1(x) and the range of f(x)?

d. Sketch a graph of f(x) and f−1(x) on the same axis.

8


Solution 9

a. f(x) = 1 +√x+ 3

y = 1 +√x+ 3

x = 1 +√y + 3

x− 1 =√y + 3

(x− 1)2 = y + 3

(x− 1)2 − 3 = y

(x− 1)2 − 3 = f−1(x)

b. Df(x) = Rf−1(x) = [3,∞)

c. Df−1(x) = Rf(x) = [1,∞)

d. Below is a graph of f(x) and f−1(x)

(−3, 1)

(1,−3)

x

yf(x) = 1 +

√x+ 3

f−1(x) = (x− 1)2 − 3

Example 10

Show that f(x) = −x5 − 3 and g(x) = 5√−x− 3 are inverses of each other.

Solution 10

To show that f(x) = −x5 − 3 and g(x) = 5√−x− 3 are inverses of each other, we need to

show that (f ◦ g)(x) = x and (g ◦ f)(x) = x

(f ◦ g)(x) = f(g(x)) = f( 5√−x− 3) = −( 5

√−x− 3)5 − 3 = −(−x− 3)− 3 = x+ �3− �3 = x

(g ◦ f)(x) = g(f(x)) = g(−x5 − 3) = 5√−(−x5 − 3)− 3 =

5√x5 + �3− �3 = �5

√x�5 = x

So f(x) and g(x) are inverses of each other.

9


Logarithmic Functions

Exponential functions are one-to-one, which implies that they have an inverse - these are thelogarithmic functions. What this means is that any exponential function can be expressed as alogarithmic function, and every logarithmic function can be written as an exponential function.

Using the definition of an inverse function, we have that

ay = x ⇔ loga x = y

Essentially a logarithm, answers the question of how many powers of a are needed in orderto generate x. Logarithms are useful because they allows us to work with very large numberswhile manipulating numbers of a much more manageable size.

Definition 3

A logarithmic function with base a is a function having form f(x) = loga x wherex > 0, a ∈ R>0, and a 6= 1.

Convention: The base, a, can be any positive real number. But some bases occur more oftenthen others and come with their own shorthand.

• When a = 10, log10 x is abbreviated to log x

• When a = e = 2.712 . . . , loge x is abbreviated to lnx

Example 11

Calculate the exact value of each expression

a. log2 32

b. log5 125

c. log 100

d. log 1100

e. log313

Solution 11

a. log2 32 = 5

b. log5 125 = 3

c. log 100 = 2

d. log 1100

= −2

10


e. log313

= −1

Example 12

Calculate the exact value of each expression

a. log4 2

b. log34√

3

c. ln e

d. ln 1e2

Solution 12

a. log4 2 = 12

b. log34√

3 = 14

c. ln e = 1

d. ln 1e2

= −2

Properties of Logarithms

Similar to exponential functions, logarithms have properties that allow you to simplify loga-rithms when their arguments are products, quotient, or values with exponents.

Theorem 2

If x and y are positive numbers then,

1. loga(xy) = loga x+ loga y

2. loga

(xy

)= loga x− loga y

3. loga(xr) = r loga x ; where r ∈ R

4. If loga x = loga y ⇒ x = y

5. loga 1 = 0

6. loga a = 1

7. loga ax = x ; can be generalized to loga a

f(x) = f(x)

8. aloga x = x ; can be generalized to aloga f(x) = f(x)

11


Example 13

Rewrite each expression as a single logarithm

a. 2 loga x+ 13

loga y

b. 12

ln(x− 2)− ln y + 3 ln z

c. 5 log(x+ y)− 2 log y − 8 log x

d. 2 log x− 2 log(x2 + 1) + 2 log(x− 1)

Solution 13

a. 2 loga x+ 13

loga y = loga x2 + loga y

1/3 = loga(x2y1/3)

b. 12

ln(x− 2)− ln y + ln z3 = ln(x− 2)1/2 − ln y + ln z3 = ln

[(x− 2)1/2z3

y

]

c. 5 log(x+ y)− 2 log y − 8 log x = log(x+ y)5 − log y2 − log x8 = log

[(x+ y)5

y2x8

]d. 2 log x− 2 log(x2 + 1) + 2 log(x− 1) = log x2 − log(x2 + 1)2 + log(x− 1)2

= log

[x2(x− 1)2

(x2 + 1)2

]= log

[x(x− 1)

(x2 + 1)

]2

Example 14

Use the properties of logarithms to expand the expression.

a. ln

[x− 2

x+ 3

]b. log [ 4

√x(x+ 3)5]

c. ln

[a2√b

cd5

]

d. log 5

√(x+ 2)3(y − 8)10

z5(x+ y)

12


Solution 14

a. ln

[x− 2

x+ 3

]= ln(x− 2)− ln(x+ 3)

b. log[ 4√x(x+ 3)5] = log[x1/4(x+ 3)5] = 1

4log x+ 5 log(x+ 3)

c. ln

[a2√b

cd5

]= ln(a2

√b)− ln(cd5) = 2 ln a+ 1

2ln b− (ln c+ 5 ln d)

d. log 5

√(x+ 2)3(y − 8)10

z5(x+ y)= log

(x+ 2)3/5(y − 8)2

z(x+ y)1/5

=3

5log(x+ 2) + 2 log(y − 8)− [log x+

1

5log(x+ y)]

Example 15

Solve each equation for x

a. e7−4x = 6

b. ln(3x− 10) = 2

c. ln(x2 − 1) = 3

d. 2x−5 = 3

Solution 15

a. e7−4x = 6

(7− 4x) ·��*1

ln e = ln 6 ⇒ x = 7−ln 64

Check: e7−4( 7−ln 64

) = e7−(7−ln 6) = e7−7+ln 6 = eln 6 = 6

∴ x = 7−ln 64

is a solution.

b. ln(3x− 10) = 2

3x− 10 = e2 ⇒ x = e2+103

Check: ln(

3(

e2+103

)+ 10− 10

)= ln(e2 + 10− 10) = ln(e2) = 2

∴ x = e2+103

is a solution.

13


c. ln(x2 − 1) = 3

x2 − 1 = e3

x2 = e3 + 1 ⇒ x = ±√e3 + 1

Check: ln[ (±√e3 + 1)2 − 1 ] = ln(e3 + 1− 1) = ln e3 = 3

∴ x = ±√e3 + 1 are solutions to the equation.

d. 2x−5 = 3

(x− 5) · ln 2 = ln 3 ⇒ x ln 2− 5 ln 2 = ln 3 ⇒ x = ln 3+5 ln 2ln 2

= ln 3ln 2

+ 5

Check: 2ln 3/ ln 2+5−5 = 2ln 3ln 2 ≈ 3

∴ x =ln 3

ln 2+ 5 is a solution.

Example 16

Solve each equation for x

a. x− xe5x+2 = 0

b. 10t2−t = 100

c. 5(x2 − 4) = (x2 − 4)e7−x

d. log x+ log(x− 3) = 1

Solution 16

a. x− xe5x+2 = 0

x(1− e5x+2) = 0

This equation goes to 0 when x = 0 and when 1− e5x+2 = 0

1− e5x+2 = 0

e5x+2 = 1

(5x+ 2)��*1

ln e =��*0

ln 1

5x+ 2 = 0 ⇒ x = −2

5

This equation has two solutions: x = 0 and x = −25

14


b. 10t2−2 = 100

10t2−t = 100

log 10t2−t = log 100

(t2 − t)��:1log 10 =��

��:2log 100

t2 − t = 2

t2 − t− 2 = 0 ⇒ (t− 2)(t+ 1) = 0 ⇒ t = 2, t = −1

This equation has two solutions: t = −1 and t = 2

c. 5(x2 − 4) = (x2 − 4)e7−x

5(x2 − 4) = (x2 − 4)e7−x

5(x2 − 4)− (x2 − 4)e7−x = 0

(x2 − 4)(5− e7−x) = 0

(x− 2)(x+ 2)(5− e7−x) = 0

The equation goes to zero when x = ±2 and when 5− e7−x = 0

5− e7−x = 0

e7−x = 5

(7− x)��*1

ln e = ln 5

7− x = ln 5 ⇒ x = 7− ln 5

The equation has three solutions x = −2, x = 2, and x = 7− ln 5

d. log x+ log(x− 3) = 1

log x+ log(x− 3) = 1

log x(x− 3) = 1

x(x− 3) = 101

x2 − 3x− 10 = 0

(x− 5)(x+ 2) = 0 ⇒ x = −2, x = 5

This equation only has one solution: x = 5

The argument for a logarithm needs to positive. When x = −2, the original expressionwill be undefined because log(−2) + log(−5). Therefore x = −2 is not a solution tothe equation.

15

EXAMPLE 6 Lecture 6: Inverse Functions - Part...

Documents

Transcript of EXAMPLE 6 Lecture 6: Inverse Functions - Part...