Let f(x) = 3x^3 + 4e^x and g(x) = f^{-1}(x). | vedclass

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(D) Given $f(x) = 3x^3 + 4e^x$ and $g(x) = f^{-1}(x)$.We know that $g'(y) = \frac{1}{f'(x)}$ where $y = f(x)$.First, find $x$ such that $f(x) = 4$: $3

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$\frac{1}{4}$

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(D) Given $f(x) = 3x^3 + 4e^x$ and $g(x) = f^{-1}(x)$.We know that $g'(y) = \frac{1}{f'(x)}$ where $y = f(x)$.First, find $x$ such that $f(x) = 4$: $3x^3 + 4e^x = 4$. By inspection, $x = 0$ satisfies this equation since $3(0)^3 + 4e^0 = 0 + 4(1) = 4$.Now, find $f'(x) = \frac{d}{dx}(3x^3 + 4e^x) = 9x^2 + 4e^x$.At $x = 0$, $f'(0) = 9(0)^2 + 4e^0 = 0 + 4 = 4$.Therefore, $g'(4) = \frac{1}{f'(0)} = \frac{1}{4}$.

Let $f(x) = 3x^3 + 4e^x$ and $g(x) = f^{-1}(x)$. The value of $g'(4)$ is equal to...

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