MATH 12002 - CALCULUS I 5.1: Calculus of Inverse Functions - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §5.1: Calculus of Inverse Functions

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Inverse Functions

Review algebra of inverse functions in the text, including:

• ne-to-one functions and the horizontal line test;

definition of inverse function, including domain and range; cancellation property (or inverse function property); finding the inverse of a one-to-one function; graph of the inverse of a function.

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Inverse Functions

Definition

A function f with domain D is one-to-one if whenever x1 = x2 in D, we have f (x1) = f (x2).

Definition

If f is a one-to-one function with domain D and range R, then the inverse

• f f is the function f −1 with domain R and range D, defined by

f −1(b) = a ⇐ ⇒ f (a) = b.

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Inverse Functions

Graph of an Inverse

If y = f (x) is a one-to-one function, then the graph of f −1 is the reflection in the line y = x of the graph of f .

Cancellation (Inverse Function Property)

If f is a one-to-one function with domain D and range R, then f −1(f (x)) = x for all x in D and f (f −1(x)) = x for all x in R.

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Continuity and Differentiability

Recall our graphical interpretations of continuity and differentiability: A function f is continuous on an interval if and only if the graph of f has no holes, breaks, or jumps on the interval. A continuous function f is differentiable at x = a if and only if the graph of f

does not have a “sharp corner” at x = a, and does not have a vertical tangent line at x = a.

That is, f is differentiable if its graph is “smooth” and does not have a vertical tangent line. Observe that reflecting a graph across the line y = x cannot introduce a hole, break, jump, or corner in the graph. However, a vertical tangent for the reflected graph will occur at any point where the original graph has a horizontal tangent. Therefore, we have the following result:

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Continuity and Differentiability

Theorem

Let y = f (x) be a one-to-one function. If f is continuous at a and f (a) = b, then f −1 is continuous at b. If f is differentiable at a and f (a) = b, then f −1 is differentiable at b unless f ′(a) = 0. How is the derivative of f −1 related to f and its derivative? Recalling that f (f −1(x)) = x and taking derivatives of both sides, we have f ′(f −1(x)) · (f −1)′(x) = 1. Hence (f −1)′(x) = 1 f ′(f −1(x)) for any x such that f ′(f −1(x)) = 0.

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Examples

1 Find (f −1)′(10) for f (x) = x3 + 2.

By the formula, we have (f −1)′(10) = 1 f ′(f −1(10)). We know that f ′(x) = 3x2. Also, f −1(10) is the number x such that f (x) = 10, that is, x3 + 2 = 10, so x3 = 8 and x =

3

√ 8 = 2. Hence f −1(10) = 2 , and so (f −1)′(10) = 1 f ′(f −1(10)) = 1 f ′(2) = 1 3(22) = 1 12. [Continued →]

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Examples

[Example 1, continued] In this example, we could also find f −1(x) explicitly: y = x3 + 2 x3 = y − 2 x =

3

• y − 2,

and interchanging x and y we have f −1(x) = y =

3

√x − 2. Therefore, (f −1)′(x) = 1 3(x − 2)−2/3 = 1 3( 3 √x − 2)2 , and so (f −1)′(10) = 1 3( 3 √10 − 2)2 = 1 3(

3

√ 8)2 = 1 3(22) = 1 12, confirming our previous result.

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Examples

Recall that we proved the power rule for derivatives,

d dx xn = nxn−1,

• nly for a positive integer n.

2 Prove that if g(x) = x1/n for a positive integer n, then

g′(x) = 1 nx

1 n −1.

The function g(x) = x1/n =

n

√x is the inverse of the function f (x) = xn (with range [0, ∞) if n is even). Hence f (g(x)) = x for x in the domain of g. We have already shown that f ′(x) = nxn−1 since n is a positive integer. Therefore, taking derivatives of both sides of the equation f (g(x)) = x, we obtain f ′(g(x))g′(x) = 1, and so g′(x) = 1 f ′(g(x)) = 1 n(g(x))n−1 = 1 n(x1/n)n−1 = 1 nx1− 1

n

= 1 nx

1 n −1. D.L. White (Kent State University) 9 / 9