The document discusses inverse functions, logarithmic functions, and their properties. It defines an inverse function f^-1(x) as satisfying f(f^-1(x)) = x. It also defines the logarithm log_a(x) as the inverse of the exponential function a^x. Key properties of inverse functions and logarithms are outlined, including: the derivative of an inverse function using the inverse function theorem; logarithm rules such as log_a(xy) = log_a(x) + log_a(y); and converting between logarithmic bases using ln(x)/ln(a). Examples of evaluating and graphing inverse functions and logarithms are provided.

1. Section 3.2
Inverse Functions and Logarithms
V63.0121, Calculus I
March 4/9/10, 2009
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.
Image credit: Roger Smith
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2. Outline
Inverse Functions
Derivatives of Inverse Functions
Logarithmic Functions
. . . . . .

3. What is an inverse function?
Definition
Let f be a function with domain D and range E. The inverse of f is
the function f−1 defined by:
f−1 (b) = a,
where a is chosen so that f(a) = b.
. . . . . .

4. What is an inverse function?
Definition
Let f be a function with domain D and range E. The inverse of f is
the function f−1 defined by:
f−1 (b) = a,
where a is chosen so that f(a) = b.
So
f−1 (f(x)) = x, f(f−1 (x)) = x
. . . . . .

5. What functions are invertible?
In order for f−1 to be a function, there must be only one a in D
corresponding to each b in E.
Such a function is called one-to-one
The graph of such a function passes the horizontal line test:
any horizontal line intersects the graph in exactly one point
if at all.
If f is continuous, then f−1 is continuous.
. . . . . .

6. Graphing an inverse function
The graph of f−1
interchanges the x and y f
.
coordinate of every
point on the graph of f
.
. . . . . .

7. Graphing an inverse function
The graph of f−1
interchanges the x and y f
.
coordinate of every
point on the graph of f
.−1
f
The result is that to get
the graph of f−1 , we .
need only reflect the
graph of f in the
diagonal line y = x.
. . . . . .

8. How to find the inverse function
1. Write y = f(x)
2. Solve for x in terms of y
3. To express f−1 as a function of x, interchange x and y
. . . . . .

9. How to find the inverse function
1. Write y = f(x)
2. Solve for x in terms of y
3. To express f−1 as a function of x, interchange x and y
Example
Find the inverse function of f(x) = x3 + 1.
. . . . . .

10. How to find the inverse function
1. Write y = f(x)
2. Solve for x in terms of y
3. To express f−1 as a function of x, interchange x and y
Example
Find the inverse function of f(x) = x3 + 1.
Answer √
y = x3 + 1 =⇒ x = y − 1, so
3
√
f−1 (x) = 3
x−1
. . . . . .

11. Outline
Inverse Functions
Derivatives of Inverse Functions
Logarithmic Functions
. . . . . .

12. derivative of square root
√ dy
Recall that if y = x, we can find by implicit differentiation:
dx
√
x =⇒ y2 = x
y=
dy
=⇒ 2y =1
dx
dy 1 1
=√
=⇒ =
dx 2y 2x
d2
y , and y is the inverse of the squaring function.
Notice 2y =
dy
. . . . . .

13. Theorem (The Inverse Function Theorem)
Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in an
open interval containing b = f(a), and
1
(f−1 )′ (b) = ′ −1
f (f (b))
. . . . . .

14. Theorem (The Inverse Function Theorem)
Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in an
open interval containing b = f(a), and
1
(f−1 )′ (b) = ′ −1
f (f (b))
“Proof”.
If y = f−1 (x), then
f(y) = x,
So by implicit differentiation
dy dy 1 1
f′ (y) = 1 =⇒ =′ = ′ −1
dx dx f (y) f (f (x))
. . . . . .

15. Outline
Inverse Functions
Derivatives of Inverse Functions
Logarithmic Functions
. . . . . .

16. Logarithms
Definition
The base a logarithm loga x is the inverse of the function ax
y = loga x ⇐⇒ x = ay
The natural logarithm ln x is the inverse of ex . So
y = ln x ⇐⇒ x = ey .
. . . . . .

17. Logarithms
Definition
The base a logarithm loga x is the inverse of the function ax
y = loga x ⇐⇒ x = ay
The natural logarithm ln x is the inverse of ex . So
y = ln x ⇐⇒ x = ey .
Facts
(i) loga (x · x′ ) = loga x + loga x′
. . . . . .

18. Logarithms
Definition
The base a logarithm loga x is the inverse of the function ax
y = loga x ⇐⇒ x = ay
The natural logarithm ln x is the inverse of ex . So
y = ln x ⇐⇒ x = ey .
Facts
(i) loga (x · x′ ) = loga x + loga x′
(x)
(ii) loga ′ = loga x − loga x′
x
. . . . . .

19. Logarithms
Definition
The base a logarithm loga x is the inverse of the function ax
y = loga x ⇐⇒ x = ay
The natural logarithm ln x is the inverse of ex . So
y = ln x ⇐⇒ x = ey .
Facts
(i) loga (x · x′ ) = loga x + loga x′
(x)
(ii) loga ′ = loga x − loga x′
x
(iii) loga (xr ) = r loga x
. . . . . .

20. Logarithms convert products to sums
Suppose y = loga x and y′ = loga x′
′
Then x = ay and x′ = ay
′ ′
So xx′ = ay ay = ay+y
Therefore
loga (xx′ ) = y + y′ = loga x + loga x′
. . . . . .

21. Example
Write as a single logarithm: 2 ln 4 − ln 3.
. . . . . .

22. Example
Write as a single logarithm: 2 ln 4 − ln 3.
Solution
42
2 ln 4 − ln 3 = ln 42 − ln 3 = ln
3
ln 42
not !
ln 3
. . . . . .

23. Example
Write as a single logarithm: 2 ln 4 − ln 3.
Solution
42
2 ln 4 − ln 3 = ln 42 − ln 3 = ln
3
ln 42
not !
ln 3
Example
3
Write as a single logarithm: ln + 4 ln 2
4
. . . . . .

24. Example
Write as a single logarithm: 2 ln 4 − ln 3.
Solution
42
2 ln 4 − ln 3 = ln 42 − ln 3 = ln
3
ln 42
not !
ln 3
Example
3
Write as a single logarithm: ln + 4 ln 2
4
Answer
ln 12
. . . . . .

25. “.
. lawn”
.
Image credit: Selva
.
. . . . . .

26. Graphs of logarithmic functions
y
.
. = 2x
y
y
. = log2 x
. . 0, 1)
(
..1, 0) . x
.
(
. . . . . .

27. Graphs of logarithmic functions
y
.
. = 3x= 2x
y. y
y
. = log2 x
y
. = log3 x
. . 0, 1)
(
..1, 0) . x
.
(
. . . . . .

28. Graphs of logarithmic functions
y
.
. = .10x 3x= 2x
y y=. y
y
. = log2 x
y
. = log3 x
. . 0, 1)
(
y
. = log10 x
..1, 0) . x
.
(
. . . . . .

29. Graphs of logarithmic functions
y
.
. = .10=3xx 2x
yxy
y y. = .e =
y
. = log2 x
y
. = ln x
y
. = log3 x
. . 0, 1)
(
y
. = log10 x
..1, 0) . x
.
(
. . . . . .

30. Change of base formula for exponentials
Fact
If a > 0 and a ̸= 1, then
ln x
loga x =
ln a
. . . . . .

31. Change of base formula for exponentials
Fact
If a > 0 and a ̸= 1, then
ln x
loga x =
ln a
Proof.
If y = loga x, then x = ay
So ln x = ln(ay ) = y ln a
Therefore
ln x
y = loga x =
ln a
. . . . . .