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Let h(x)=5-x-x^(3) and let f be the inverse function of h. Notice that h(-1)=7. f^(')(7)=

Let h(x)=5−x−x3 h(x)=5-x-x^{3} and let f f be the inverse function of h h . Notice that h(−1)=7 h(-1)=7 .\newlinef′(7)= f^{\prime}(7)=

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Q. Let h(x)=5−x−x3 h(x)=5-x-x^{3} and let f f be the inverse function of h h . Notice that h(−1)=7 h(-1)=7 .\newlinef′(7)= f^{\prime}(7)=
  1. Find h′(x)h'(x): Find h′(x)h'(x), the derivative of h(x)h(x).\newlineh′(x)=−1−3x2h'(x) = -1 - 3x^2
  2. Evaluate h′(−1)h'(-1): Evaluate h′(−1)h'(-1) since h(−1)=7h(-1)=7 and we need the derivative of the inverse function at h(−1)h(-1).h′(−1)=−1−3(−1)2=−1−3(1)=−1−3=−4h'(-1) = -1 - 3(-1)^2 = -1 - 3(1) = -1 - 3 = -4
  3. Use inverse function property: Use the property of inverse functions that (f−1)′(b)=1f′(a)(f^{-1})'(b) = \frac{1}{f'(a)} where f(a)=bf(a) = b. Here, f−1f^{-1} corresponds to h−1h^{-1} and b=7b=7, so we need to find 1h′(−1)\frac{1}{h'(-1)}. (f−1)′(7)=1h′(−1)=1−4=−14(f^{-1})'(7) = \frac{1}{h'(-1)} = \frac{1}{-4} = -\frac{1}{4}

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