Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.3 Exercises - Page 345: 88

(a) $2$ (b) $\frac{1}{2}$

By Theorem 5.9, we have: $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ Therefore, $(f^{-1})'(-1/2) = \frac{1}{f'(f^{-1}(-1/2))} =\frac{1}{f'(-1)} = \frac{1}{1/2} =2$ And, $(f^{-1})'(1) = \frac{1}{f'(f^{-1}(1))}= \frac{1}{f'(2)} = \frac{1}{2}$