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Let f(x)=x^(3)+2x-1 and let g be the inverse function of f. Notice that f(2)=11. g^(')(11)=

Let f(x)=x3+2x−1 f(x)=x^{3}+2 x-1 and let g g be the inverse function of f f . Notice that f(2)=11 f(2)=11 .\newlineg′(11)= g^{\prime}(11)=

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Q. Let f(x)=x3+2x−1 f(x)=x^{3}+2 x-1 and let g g be the inverse function of f f . Notice that f(2)=11 f(2)=11 .\newlineg′(11)= g^{\prime}(11)=
  1. Identify Inverse Function: Since f(2)=11f(2)=11, we know that g(11)=2g(11)=2 because gg is the inverse of ff.
  2. Calculate Derivative of Inverse: To find g′(11)g'(11), we use the formula for the derivative of the inverse function: g′(f(x))=1f′(g(x)).g'(f(x)) = \frac{1}{f'(g(x))}.
  3. Find Derivative of Original Function: First, we need to find f′(x)f'(x), which is the derivative of f(x)=x3+2x−1f(x)=x^3+2x-1. f′(x)=3x2+2f'(x) = 3x^2 + 2.
  4. Substitute into Derivative: Now we substitute xx with g(11)g(11), which is 22, into f′(x)f'(x) to get f′(g(11))f'(g(11)).\newlinef′(g(11))=f′(2)=3(2)2+2=12+2=14f'(g(11)) = f'(2) = 3(2)^2 + 2 = 12 + 2 = 14.
  5. Calculate Final Derivative: Finally, we calculate g′(11)g'(11) using the formula g′(f(x))=1f′(g(x))g'(f(x)) = \frac{1}{f'(g(x))}.\newlineg′(11)=1f′(2)=114.g'(11) = \frac{1}{f'(2)} = \frac{1}{14}.

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