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Exercise 10

Avg difficulty 2.5

Let h(x)=x+sin⁡(x)h(x) = x + \sin(x).

  1. Sketch a graph of y=h(x)y = h(x) and explain why hh must be invertible.

  2. Explain why it does not appear to be algebraically possible to determine a formula for h−1h^{-1}.

  3. Observe that the point (π2,π2+1)(\frac{\pi}{2}, \frac{\pi}{2} + 1) lies on the graph of y=h(x)y = h(x). Determine the value of (h−1)′(π2+1)(h^{-1})'(\frac{\pi}{2} + 1).

Boelkins, Active Calculus, §2.6 Derivatives of inverse functions, Exercise 10

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.