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Math 130, Inverse FunctionsDerivative of the Natural Log, Study notes of Calculus

We use the natural log function to undo the exponential function x = ln 3. ... not have an inverse is because a horizontal line meets the graph twice.

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Inverse Functions
21.1Introduction
Often in the course of solving an equation we use one function to undo (the effect
of) another function. We do this without really thinking about it. For example,
suppose we want to solve
x3=8.
We use the cube root function to undo the cubing function
x=2.
Implicitly what we’ve said is that
x3=83
x33
8x=2.
The function g(x) = 3
x‘undid’ the cubing function f(x) = x3. This sounds Here’s another example: Solve
ex=3.
We use the natural log function to undo
the exponential function
x=ln 3.
Implicitly what we’ve said is that
ex=3ln(ex) = ln 3 x=ln 3.
trivial, but it is not always possible to do (or rather ‘undo’) this. Consider
x2=4
x2=4
x=±2
The square root function does not undo the squaring function. We can’t undo the
squaring function unless we know more about x. . . if we knew xwere positive,
we’d be able to say x=10. So why can we undo some functions and not others?
When we try to ’undo’ a function y=f(x), we are trying to reverse the function
process. Given a particular y-value, we want to know the original corresponding
input x-value. As the squaring function shows, this is problematic when a single
y-value comes from more than one x-value. Comparing the graphs of f(x) = x3
and f(x) = x2shows what the problem is geometrically.
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f(x) = x3
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............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. .............
4
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f(x) = x2
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............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. .............
In the first case, for the y-value 8, there is only one corresponding x-value because
the horizontal line y=8 meets the graph just once. In the second case, for the
y-value 4, the the horizontal line y=4 meets the graph the graph twice: at x=2
and x=2. So we can’t say which value xwas.
pf3
pf4
pf5

Partial preview of the text

Download Math 130, Inverse FunctionsDerivative of the Natural Log and more Study notes Calculus in PDF only on Docsity!

Inverse Functions

21. 1 Introduction

Often in the course of solving an equation we use one function to undo (the effect

of) another function. We do this without really thinking about it. For example,

suppose we want to solve

x

We use the cube root function to undo the cubing function

x = 2.

Implicitly what we’ve said is that

x

x

8 ⇒ x = 2.

The function g(x) =

x ‘undid’ the cubing function f (x) = x

. This sounds Here’s another example: Solve

e

x

We use the natural log function to undo

the exponential function

x = ln 3.

Implicitly what we’ve said is that

e

x

= 3 ⇒ ln(e

x

) = ln 3 ⇒ x = ln 3.

trivial, but it is not always possible to do (or rather ‘undo’) this. Consider

x

x

x = ± 2

The square root function does not undo the squaring function. We can’t undo the

squaring function unless we know more about x... if we knew x were positive,

we’d be able to say x = 10. So why can we undo some functions and not others?

When we try to ’undo’ a function y = f (x), we are trying to reverse the function

process. Given a particular y-value, we want to know the original corresponding

input x-value. As the squaring function shows, this is problematic when a single

y-value comes from more than one x-value. Comparing the graphs of f (x) = x

and f (x) = x

shows what the problem is geometrically.

f (x) = x

f (x) = x

In the first case, for the y-value 8, there is only one corresponding x-value because

the horizontal line y = 8 meets the graph just once. In the second case, for the

y-value 4, the the horizontal line y = 4 meets the graph the graph twice: at x = − 2

and x = 2. So we can’t say which value x was.

Key Idea: One-to-one Functions

DEFINITION 21. 1. 1. A function f is one-to-one if whenever x

1

6 = x

2

, then f (x

1

) 6 = f (x

2

).

This means that f never has the same output twice.

You should memorize this definition.

EXAMPLE 21. 1. 2. The function on the left is NOT one-to-one, because f (x 1

) = f (x

2

) even

though x 1

6 = x

2

. Similarly f (x

3

) = f (x

4

) even though x

3

6 = x

4

.

x

x

x

x

x

x

The function on the right is one-to-one; it never has the same output value twice.

EXAMPLE 21. 1. 3. Show that the function f (x) =

x

2

  • 9 is not one-to-one.

Solution. Using the definition of one-to-one, we need to find two different values

x

6 = x

so that f (x

) = f (x

). We know that the squaring function produces

the same output for inputs a and −a. So if we use x

= 1 and x

= −1, then

f (x

) =

1

  • 9 =

10 and f (x

) =

(− 1 )

  • 9 =

  1. We get the same output

twice, so f is not one-to-one.

YOU TRY IT 21. 1. Show that the function f (x) = x

2

  • cos x is not one-to-one.

Look back at the graphs in Example 21. 1. 2. The reason that one function does

not have an inverse is because a horizontal line meets the graph twice. The hori-

zontal line represents a particular y-value. If it meets the graph twice, there must

be two different x-values that have the same y-value. So we have the following

theorem.

THEOREM 21. 1. 4 (Horizontal Line Test, HLT). A function is one-to-one if and only if no hori-

zontal line meets the graph more than once.

Fails HLT

Passes HLT

YOU TRY IT 21. 2. Draw a function that passes the HLT and one that fails it.

Fails HLT

Passes HLT

21. 2 Inverse Functions

Let’s make the notion of two functions of undoing each other precise.

YOU TRY IT 21. 3. Draw the graph of f

− 1

for the function f graphed below.

When we are given the function formula for f rather than the graph, we can

also find the formula for f

using the following three steps.

  1. Write y = f (x)

  2. Solve for x in terms of y. This amounts to finding x = f

(y).

  1. Interchange the variable names to get y = f

(x).

EXAMPLE 21. 2. 4. Assume that y = f (x) = 5 x

3

  • 7 is one-to-one so that it has an inverse. To

find f

− 1

(x), we follow the three steps above.

  1. Write y = 5 x

3

  • 7
  1. Solve for x: y = 5 x

3

  • 7 ⇒ y − 7 = 5 x

3

y − 7

5

= x

3

y − 7

5

= x

  1. Interchange the variables: f

− 1

(x) = y =

x − 7

5

Notice that we could also find the inverse of y = f (x) = 5 x

3

  • 7 by thinking about the order

of operations of f and reversing them: To carry out f , first cube x, then multiply by 5 , then

add 7. To undo this, first subtract 7 to get x − 7, then divide by 5 to get

x− 7

5

, and finally take

the cube root f

− 1

(x) = y =

x− 7

5

.

YOU TRY IT 21. 4. Assume that y = f (x) = 4 +

3

x

is one-to-one so that it has an inverse. Find

f

− 1

(x).

21. 3 The Inverse of the Exponential Function e

x

Now consider y = f (x) = e

x

. We have seen that this function is increasing and that

it appears to pass the HLT, so it has an inverse. Using the graph of e

x

, we can draw

the graph of its inverse.

f (x) = e

x

f

(x) = ln x

The inverse of the exponential function is the natural log function and is de-

noted by y = ln x. Using the definition of inverse:

f (a) = b ⇐⇒ f

(b) = a

e

a

= b ⇐⇒ ln b = a

This means that logs are exponents of e (note where a is in the last line). Since Because logs are exponents, logs have

the following very useful properties:

( 1 ) ln(xy) = ln x + ln y

( 2 ) ln

x

y

= ln x − ln y

( 3 ) ln(x

r

) = r ln x

inverse functions interchange the roles of x and y, this means that the domains and

the ranges of the two functions are reversed.

  • Domain of ln x = Range of e

x

= (0, ∞) or x > 0.

  • Range of ln x = Domain of e

x

= (−∞, ∞) or all x.

Again using basic properties of inverses,

  1. f

( f (x)) = x or ln(e

x

) = x for all x.

  1. f ( f

(x)) = x or e

ln x

= x for x > 0.

21. 4 The Derivative of y = ln x

As calculus students one of out first questions should be can we find the derivative

of ln x. So let y = ln x. Our goal is to find

dy

dx

or

dy

dx

(ln x). We will use the method

below a few more times to find derivatives of inverses of other functions. Assume

x > 0 so the the natural log is defined. Start with

y = ln x.

Take the inverse of both sides

e

y

= e

ln x

.

Simplify

e

y

= x. ( 21. 1 )

Take the derivative using the chain rule (implicit differentiation)

d

dx

(e

y

) =

d

dx

(x)

Simplify

e

y

dy

dx

= 1

Solve for

dy

dx

dy

dx

=

1

e

y

Use 21. 1 and substitute back for e

y

: e

y

= x

dy

dx

=

1

x

That is, we have shown

THEOREM 21. 4. 1. The natural log function is differentiable and

d

dx

(ln x) =

1

x

.

EXAMPLE 21. 4. 4. Here’s one that involves a number of log properties:

D t

[

ln

e

t

cos t

t

2

  • 1

]

= D t

[

ln e

t

  • ln | cos t| − ln

t

2

  • 1

= D t

[

t + ln | cos t| −

1

2

ln |t

2

  • 1 |

]

= 1 +

1

cos t

· (− sin t) −

2 t

2 (t

2

  • 1 )

= 1 − tan t −

t

t

2

  • 1

YOU TRY IT 21. 6. Try finding these derivatives. Use log rules to simplify the functions before

taking the derivative.

(a) D

x

[

ln | 6 x

3

sin x|

]

(b) D

x

[

6 x

3

ln | sin x|

]

(different)

(c) D

x

[

ln

x

4

− 1

x

2

  • 1

]

(d) D t

[

ln(t

(e

t

)

)

]

(e) D x

[

ln

3 x

3

  • x + 1

]

(f ) D s

[

ln ( 5

ln s

)

]

and log properties, then simplify. 3. 4. 21. Use Theorem 6. answer to you try it 21

D )a(

x

[

x 6 | ln

|x sin

]

D =

x

[

x 6 | ln

|x sin | ln + |

]

x

x ) 18b( x cot +

x 6 + |x sin | ln

x cot

D )c (

x

[

ln

x

x

]

D =

x

[

ln

x∣

ln − ∣

x|∣

]

x 4

x

x 2

x

D )d(

t

[

t( ln

e(

t

]

D =

t

[

e

t

t ln

]

e =

t

+ t ln

e

t

t

D )e (

x

[

ln

x 3

1 + x +

]

D =

x

[

ln ·

x 3

1 + x +

]

x 9

x 3

1 + x +

D ) f(

s

[

5 ( ln

s ln

]

D =

s

] =ln 5 s ln[

ln 5

s

In part (f), remember that ln 5 is just a constant.