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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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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
.
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
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
√
10 and f (x
) =
√
(− 1 )
√
twice, so f is not one-to-one.
YOU TRY IT 21. 1. Show that the function f (x) = x
2
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.
YOU TRY IT 21. 2. Draw a function that passes the HLT and one that fails it.
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.
Write y = f (x)
Solve for x in terms of y. This amounts to finding x = f
(y).
(x).
EXAMPLE 21. 2. 4. Assume that y = f (x) = 5 x
3
find f
− 1
(x), we follow the three steps above.
3
3
3
⇒
y − 7
5
= x
3
⇒
√
y − 7
5
= x
− 1
(x) = y =
√
x − 7
5
Notice that we could also find the inverse of y = f (x) = 5 x
3
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
. 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
, we can draw
the graph of its inverse.
f (x) = e
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
= 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
inverse functions interchange the roles of x and y, this means that the domains and
the ranges of the two functions are reversed.
= (0, ∞) or x > 0.
= (−∞, ∞) or all x.
Again using basic properties of inverses,
( f (x)) = x or ln(e
) = x for all x.
(x)) = x or e
= 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
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
= e
.
Simplify
e
= x. ( 21. 1 )
Take the derivative using the chain rule (implicit differentiation)
d
dx
(e
) =
d
dx
(x)
Simplify
e
dy
dx
= 1
Solve for
dy
dx
=
1
e
Use 21. 1 and substitute back for e
: e
= 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
∣
∣
∣
∣
]
= D t
[
ln e
t
√
t
2
∣
∣
∣
= D t
[
t + ln | cos t| −
1
2
ln |t
2
]
= 1 +
1
cos t
· (− sin t) −
2 t
2 (t
2
= 1 − tan t −
t
t
2
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
∣
∣
∣
∣
∣
]
(d) D t
[
ln(t
(e
t
)
)
]
(e) D x
[
ln
√
3 x
3
]
(f ) D s
[
ln ( 5
ln s
)
]
t