Question

Suppose f is a one-to-one, differentiable function and its inverse function f^-1 is also differentiable. One can show, using implicit differentiation (do it!), that (f^-1)'(x) = 1 / f'(f^-1(x)) Find (f^-1)'(-3) if f(1) = -3 and f'(1) = 3/7. (f^-1)'(-3) =

          Suppose f is a one-to-one, differentiable function and its inverse function f^-1 is also differentiable. One can show, using implicit differentiation (do it!), that
(f^-1)'(x) = 1 / f'(f^-1(x))
Find (f^-1)'(-3) if f(1) = -3 and f'(1) = 3/7.
(f^-1)'(-3) =

Suppose f is a one-to-one, differentiable function and its inverse function f^-1 is also differentiable. One can show, using implicit differentiation (do it!), that
(f^-1)'(x) = 1 / f'(f^-1(x))
Find (f^-1)'(-3) if f(1) = -3 and f'(1) = 3/7.
(f^-1)'(-3) =

Added by Michelle A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Vincenzo Zaccaro

02/09/2022

Step 1

Step 1: Given that \(f(1) = -3\) and \(f'(1) = \frac{3}{7}\), we know that \(f^{-1}(-3) = 1\). Show more…

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Suppose f is a one-to-one, differentiable function and its inverse function f^-1 is also differentiable. One can show, using implicit differentiation (do it!), that (f^-1)'(x) = 1 / f'(f^-1(x)) Find (f^-1)'(-3) if f(1) = -3 and f'(1) = 3/7. (f^-1)'(-3) =