HWQ 1/12/15
Evaluate the definite integral:
No calculator please.
e

1
1  ln x 
2
x
7
3
dx
Section 5.3 20143
Derivatives of Inverse Functions
Objective: To find the derivative of the
inverse of a function.
2
Inverse Functions - Review

Inverse functions are basically functions that
“cancel” when we perform a composition.

Formally, we say that the function g (x) is the
inverse function of f (x) if:
f g ( x)  x for all x in thedomainof g and
g  f ( x)  x for all x in thedomainof f
Inverse Functions

The graph of the inverse of f (x),
f -1(x), is the reflection of f (x)
across the line y = x.

The inverse function exists (without any
constraints) if it satisfies one condition:
1.
A function has an inverse iff it is one-to-one
 For every x, there is only one y and for every y,
there is only one x.
 If f is strictly monotonic for its entire domain, then
it is one-to-one and therefore has an inverse
function.
Example 1 – Verifying Inverse Functions
Show that the functions are inverse functions of each other.
and
Solution:
Because the domains and ranges of both f and g consist of
all real numbers, you can conclude that both composite
functions exist for all x.
The compositions f (g(x)) and g(f(x)) are given by:
Example 1 – Solution
Because f(g(x)) = x and g(f(x)) = x, you can conclude that
f and g are inverse functions of each other
(see Figure 5.11).
Figure 5.11
cont'd
Inverse Functions
The idea of a reflection of the graph of f in the line y = x is
generalized in the following theorem.
Figure 5.12
Inverse Functions
Finding an Inverse Function
f  x   2x  3
Find f
f
1
1
 x
x 3
 x 
2
2
Derivative of an Inverse Function
Find f  6
f  6  3, 
 f  3  6
Find f   6 and  f 1   3
f   6 
1

3
 f  3  3
1
If (x,y) is a point on the graph of a
function, the derivative of the
function at x will be the reciprocal of
the derivative of its inverse at y.
Graphs of inverse functions have
reciprocal slopes at inverse
points.
1
Derivative of an Inverse Function
If (x,y) is a point on the graph of a function, the derivative
of the function at x will be the reciprocal of the derivative
of its inverse at y.
Graphs of inverse functions have reciprocal slopes at
inverse points.
The Derivative of the Inverse

Let’s try to get a better understanding of what
this formula is actually stating:
1
g ( x ) 
, f  g ( x )   0
f  g ( x ) 
y
f (x)
g (x)
Let this function
be f (x)
The inverse
function will be
reflected across the
line y = x
Let this function
be f -1 (x) = g (x)
x
The Derivative of the Inverse

Let’s try to get a better understanding of what
this formula is actually stating:
1
g ( x ) 
, f  g ( x )   0
f  g ( x ) 
f (x)
y
g (x)
(5, g (5))
x
5
Let’s say we wish
to find the slope of
the tangent line of
g (x) at x = 5
The coordinate
will be (5, g (5))
The slope of the
tangent line will
be some constant;
we’ll call it g ‘(x)
The Derivative of the Inverse

Let’s try to get a better understanding of what
this formula is actually stating:
1
g ( x ) 
, f  g ( x )   0
f  g ( x ) 
This coordinate
will be the
reverse or the
inverse (g (5), 5)
f (x)
y
g (x)
(g (5), 5)
(5, g (5))
x
5
Let’s say we wish
to find the slope of
the tangent line of
g (x) at x = 5
We know that if
there exists a point
on g (x), then there
is an inverse point
on f (x)
The Derivative of the Inverse

Let’s try to get a better understanding of what
this formula is actually stating:
1
g ( x ) 
, f  g ( x )   0
According to this
In other words, if we
f  g ( x ) 
formula, the
y
reciprocal of the
slope of the
tangent line of f(x)
at x = g(5) is the
(g (5), 5)
SAME as the
slope of the
tangent line of
g(x) at x = 5
f (x)
(5, g (5))
want to find the slope
of the tangent line of
g(x) for some x, all we
have to do is find the
g (x) reciprocal of the
derivative of the f(x)
when f(x) = (the given
x of the inverse)
x
5
The Derivative of the Inverse

Let’s try to get a better understanding of what
this formula is actually stating:
1
g ( x ) 
, f  g ( x )   0
f  g ( x ) 
f (x)
y
If we let the coordinate be
more generic such as (a, b),
then we could say:
g (x)
(b, a)
(a,b)
x
a
g ( a ) 
1
, f b   0
f b 
Graphs of inverse functions
have reciprocal slopes at
inverse points.
Examples

1. Find f 1 (8) if f ( x)  x 3
 
What are your options for answering this question?
1) Find the inverse, then find its derivative at x=8.
2) Find where f(x) = 8, find the derivative there,
and reciprocate.
3) Use the formula.
Examples

  (8) if
1. Find f
1
f ( x)  x 3
First, let’s try the first option. Find the inverse by switching x and y
and solving for y:
x  y3
x1/3  y
Find the derivative of f -1:

f 
1
1
1 2 /3
 2/3
 x
3x
3
f
1
( x )  x1/3

f 
1
1
1
1


8  
2/3
3  4  12
3  8 
Examples

  (8) if
1. Find f
1
f ( x)  x 3
Now, let’s try the second option. Find where f(x) = 8, find the
derivative there, and reciprocate.
f(x ) h a s th e p o in t (? ,8 )
f ( 2 )  1 2
f  1 (x ) h a s th e p o in t (8 ,?)

f  1  (8) 
1
12
This is usually the easiest method.
Examples

  (8) if
1. Find f
1
f ( x)  x 3
Now, let’s try to find the derivative of the inverse by using the
formula:
f (x)  x
3
1
g ( x) 
, f g ( x)   0
f g ( x) 
1 
2
f
( x) 
f ( x )  3 x
 
1
2
3 g ( x ) 
We still need to find g (x); but since g (x) is the inverse of f (x), we
know that the x and y values switch or (x, g (x)) → (g (x), x)
 ( 8 )  ( 8 )
g (8 )  f

f 
1
1
1/3
2
1
1
1
1
(8 ) 


2
2 
3  g (8 ) 
3  2 
3  4  12
Examples

  (2) if
2. Find f
1
f ( x)   x  2 x  1
3
f 1  x  has the point  2,?   f  x  has the point  ?, 2 
 f  x  has the point 1, 2 
Since f  1 =  5,
f 
1

1
(2)  
5
Examples

 
3. Find f
1
6
(5) if f ( x)  x  , x  0
x
3
f 
1

2
(5) 
27
Examples

  (e) if
4. Find f
1
f ( x)  x ln x, x  0
f 
1

1
(e) 
2
Examples
5. Find (f –1)'(3) if
f 
1

1
(3) 
4
Derivative of an Inverse Function
In Example 5, note that at the point (2, 3) the slope of the
graph of f is 4 and at the point (3, 2) the slope of the graph of
f –1 is
(see Figure 5.17).
HWQ
f  x   2 x  x  1,
3
2
find  f
1
 13
 
1

 f  13  20
1
HWQ
1
1
If h(2)  3, h  2  , h  3  , find  h1   3
4
3

 h   3   4
1
Examples

 
5. Find F
1
x
1
(0) if F ( x)  
dt, x  2
2
1  4t
2

 F  (0)  17
1
Homework

Inverse Functions

Day 1: P. 347: 1, 3, 9, 11, 71-89 odd

Day 2: Derivatives of Inverses W/S
Homework

Derivatives of Inverses W/S
AP FRQ

2007 #3