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Let f(x)=(1)/(2)x^(3)+3x-4 and let h be the inverse function of f. Notice that f(-2)=-14. h^(')(-14)=

Let f(x)=12x3+3x−4 f(x)=\frac{1}{2} x^{3}+3 x-4 and let h h be the inverse function of f f . Notice that f(−2)=−14 f(-2)=-14 .\newlineh′(−14)= h^{\prime}(-14)=

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Full solution

Q. Let f(x)=12x3+3x−4 f(x)=\frac{1}{2} x^{3}+3 x-4 and let h h be the inverse function of f f . Notice that f(−2)=−14 f(-2)=-14 .\newlineh′(−14)= h^{\prime}(-14)=
  1. Find h′(−14)h'(-14): To find h′(−14)h'(-14), we need to use the formula for the derivative of the inverse function: h′(y)=1f′(x)h'(y) = \frac{1}{f'(x)} where f(x)=yf(x) = y.
  2. Find f′(−2)f'(-2): Since f(−2)=−14f(-2) = -14, we have x=−2x = -2 when y=−14y = -14. Now we need to find f′(−2)f'(-2).
  3. Differentiate f(x)f(x): Differentiate f(x)=12x3+3x−4f(x) = \frac{1}{2}x^3 + 3x - 4 to get f′(x)=32x2+3f'(x) = \frac{3}{2}x^2 + 3.
  4. Plug x=−2x = -2: Now plug x=−2x = -2 into f′(x)f'(x) to find f′(−2)f'(-2): f′(−2)=(32)(−2)2+3=(32)⋅4+3=6+3=9f'(-2) = (\frac{3}{2})(-2)^2 + 3 = (\frac{3}{2})\cdot4 + 3 = 6 + 3 = 9.
  5. Use formula for h′(−14)h'(-14): Use the formula for the derivative of the inverse function: h′(−14)=1f′(−2)=19h'(-14) = \frac{1}{f'(-2)} = \frac{1}{9}.

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