Question

Exercise 5.2.12 (Inverse functions). If f : [a, b] ? R is one-to-one, then there exists an inverse function f^{-1} defined on the range of f given by f^{-1}(y) = x where y = f(x). In Exercise 4.5.8 we saw that if f is continuous on [a, b], then f^{-1} is continuous on its domain. Let's add the assumption that f is differentiable on [a, b] with f'(x) ? 0 for all x ? [a, b]. Show f^{-1} is differentiable with (f^{-1})'(y) = 1/f'(x) where y = f(x).

          Exercise 5.2.12 (Inverse functions). If f : [a, b] ? R is one-to-one, then there exists an inverse function f^{-1} defined on the range of f given by f^{-1}(y) = x where y = f(x). In Exercise 4.5.8 we saw that if f is continuous on [a, b], then f^{-1} is continuous on its domain. Let's add the assumption that f is differentiable on [a, b] with f'(x) ? 0 for all x ? [a, b]. Show f^{-1} is differentiable with
(f^{-1})'(y) = 1/f'(x) where y = f(x).

Exercise 5.2.12 (Inverse functions). If f : [a, b] ? R is one-to-one, then there exists an inverse function f^-1 defined on the range of f given by f^-1(y) = x where y = f(x). In Exercise 4.5.8 we saw that if f is continuous on [a, b], then f^-1 is continuous on its domain. Let's add the assumption that f is differentiable on [a, b] with f'(x) ? 0 for all x ? [a, b]. Show f^-1 is differentiable with
(f^-1)'(y) = 1/f'(x) where y = f(x).

Added by Samantha H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ben Blakesley

03/08/2022

Step 1

** Show more…

Show all steps

Thanks for your feedback!

If f : [a, b] → R is one-to-one, then there exists an inverse function f^{-1} defined on the range of f given by f^{-1}(y) = x where y = f(x). In Exercise 4.5.8 we saw that if f is continuous on [a, b], then f^{-1} is continuous on its domain. Let's add the assumption that f is differentiable on [a, b] with f'(x) ≠ 0 for all x ∈ [a, b]. Show f^{-1} is differentiable with (f^{-1})'(y) = 1/f'(x) where y = f(x).