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Calculus 1  Problems & Solutions  –  Chapter 2  –  Section 2.3.4

 

2.3.4
Differentiation Of Inverse Functions

 

 

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Review

 

1. One-To-One Functions

 

 

Fig. 1.1

 

f is one-to-one.

 

Fig. 1.2

 

g isn't one-to-one.

 

function, one x corresponds to exactly one y. Now we also have that one y corresponds to exactly one x. Thus, f is said
to be one-to-one. A horizontal line cuts the graph of f at at most one point.

 

Note that the implication:

 

 

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2. Inverse Functions

 

Let y = f(x) be a function. We see that f maps x to y. For each y in range ( f ) there correspond one or more x in
dom( f ) where y = f(x). This correspondence maps y back to x. So its action is the reverse of that of f. Thus it's called
the inverse relation of f.

 

 

ie, x is the image of y by f–1 iff y is the image of x by f, or f–1 maps y to x iff f maps x to y.

 

We get a new function, f –1. Usually we treat functions as mapping x to y. Now, f –1 is a function in its own right. Hence
let's take it out of the original situation where it's treated as mapping y to x and put it with all other functions and treat it
as mapping x to y. It follows that we usually write y = f–1(x), rather than x = f –1( y), to show the mapping action of f –1.
Therefore, given that f is a one-to-one function, the definition of the inverse function f–1 of f is usually presented as:

 

 

 

 

ie, y is the image of x by f–1 iff x is the image of y by f, or f–1 maps x to y iff f maps y to x.

 

We say that a function is invertible if it has its inverse function, ie, if its inverse relation is also a function. We've seen
that if a function is one-to-one, then it's invertible. If a function y = f(x) isn't one-to-one, then there exist two different x1
and x2 such that f(x1) = f(x2) = y1. Clearly its inverse relation isn't a function, because y1 would have two different
images: x1 and x2. So if a function isn't one-to-one then it isn't invertible. Thus a function is invertible iff it's one-to-one.

 

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3. Finding Inverse Functions

 

Example 3.1

 

Let f(x) = 2x + 1.
a.  Show that f is invertible.
b.  Find its inverse, f –1.

 

Solution
a.  If f(x1) = f(x2), then 2x1 + 1 = 2x2 + 1, yielding x1 = x2. So f is one-to-one, thus invertible.

 

b.  Let y = f –1(x). Then x = f( y) = 2y + 1, yielding y = (x – 1)/2. As a consequence:

 

   

EOS

 

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4. Graphs Of Inverse Functions

 

As shown in Example 3.1, the inverse of f(x) = 2x + 1 is f –1(x) = (x – 1)/2. The graphs of f and f –1 are sketched in Fig.
4.1. They're mirror images of each other in the line y = x.

 

We now show that the graph of the inverse f –1 of any invertible function f is the mirror image of that of f in the line y =
x. See Fig. 4.2. The graph of y = f(x) is the same as that of x = f –1( y) . Since the function x = f –1( y) maps y to x, its

 

Fig. 4.1

 

Graphs of y = 2x + 1 and y = (x – 1)/2 are mirror images of
each other in the line y = x.

 

Fig. 4.2

 

Graphs of y = f(x) and y = f–1(x) are mirror images of each other
in the line y = x.

 

domain is on the y-axis and its range on the x-axis. Move its domain to the x-axis and its range to the y-axis, and we get
the domain on the x-axis and the range on the y-axis of the function y = f –1(x). This movement is done by flipping the
plane of the axes over about the line y = x, taking the domain, range, and graph of x = f –1( y) along but leaving the
axes intact. The graph of x = f –1( y) then becomes that of y = f –1(x). Thus, the graph of y = f –1(x) is the mirror image
of that of x = f –1( y), hence of that of y = f(x), in the line y = x.

 

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5. Differentiation Of Inverse Functions

 

Consider the inverse f –1 of an invertible function f. The graph of y = f –1(x) is sketched in Fig. 5.1. The graph of x = f( y)
is the same as that of y = f –1(x). If y changes twice as fast as x does, then x changes half as fast as y does. The rate of

 

Fig. 5.1

 

Rate of change of y with respect to x is reciprocal of that of
x with respect to y.

 

change of y with respect to x is the reciprocal of the rate of change of x with respect to y. Now, y is the function of x by
f –1, and x is the function of y by f. So the rate of change of y with respect to x is ( f –1)'(x) and the rate of change of x
with respect to y is f '( y). Thus, ( f –1)'(x) = 1/f '( y) = 1/f '( f –1(x)).

 

 

 

 

 

Using The Leibniz Notation

 

Using the Leibniz notation for ( f –1)'(x) = 1/f '( f –1(x)) we obtain dy/dx = 1/f '( y) =1/(dx/dy):

 

 

                                                                                                                                                

 

 

The notation dy/dx, again, appears to be a normal fraction. This formula states that the rate of change of y with respect
to x is the reciprocal of the rate of change of x with respect to y.

 

Example 5.1

 

 

Solution

EOS

 

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Problems & Solutions

 

1.  Suppose f is an invertible function with f(1) = – 3 and f(5) = 6. Find:
     a.  f –1(–3).
     b.  f( f –1(6)).
     c.  f –1( f(1)).

 

Solution

 

a.  Since f(1) = – 3 we have f–1(– 3) = 1.

 

b.  Since f(5) = 6 we have f–1(6) = 5. So f( f–1(6)) = f(5) = 6.

 

c.  f–1( f(1)) = f–1(– 3) = 1.

 

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2.  Let f(x) = x3 + 1. Find the derivative of the inverse of f at x = 9 in two ways:
     a.  By determining the inverse function and differentiating it.
     b.  By using the formula for the derivative of an inverse function.

 

Solution

 

a.  Let y = f –1(x). Then x = f( y) = y3 + 1. So f –1(x) = y = (x – 1)1/3. Thus:

 

 

Remark

 

 

However, this alternative may be confusing, because the question uses x, not y, as the independent variable for f –1 (it
says “at x = 9”, not “at y = 9”).

 

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3.  Let f(x) = xn where x > 0 is a positive real number and n > 0 is a positive integer.
     a.  Show that f –1(x) = x1/n.
     b.  Differentiate f –1 by using the power rule and using the Leibniz notation.
     c.  Differentiate f –1 by using the formula for the derivative of an inverse function and using the Leibniz notation.

 

Solution

 

a.  Let y = f –1(x). Then x = f( y) = yn, so that f –1(x) = y = x1/n.

 

 

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4.  Let g(x) = (x – 1)/(x + 2). Calculate the derivative of the inverse of g at x = 0 in two ways:
     a.  By determining the inverse function and differentiating it.
     b.  By using the formula for the derivative of an inverse function.

 

Solution

 

 

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5.  Suppose f is a one-to-one differentiable function satisfying f '(x) = 1/x. Let y = f –1(x). Prove that dy/dx = y.

 

Solution

 

 

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