Let f(x) = x^5 + 2e^{x/4} for all x in mathbb | vedclass

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(A) Given $(g \circ f)(x) = x$. By the chain rule,$g'(f(x)) \cdot f'(x) = 1$.We need to find $g'(2)$. Let $f(x) = 2$.$x^5 + 2e^{x/4} = 2$. By inspecti

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$16$

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(A) Given $(g \circ f)(x) = x$. By the chain rule,$g'(f(x)) \cdot f'(x) = 1$.We need to find $g'(2)$. Let $f(x) = 2$.$x^5 + 2e^{x/4} = 2$. By inspection,$x = 0$ satisfies this equation $(0^5 + 2e^0 = 2)$.Thus,$g'(f(0)) \cdot f'(0) = 1$,which implies $g'(2) = \frac{1}{f'(0)}$.Now,$f'(x) = 5x^4 + 2 \cdot \frac{1}{4} e^{x/4} = 5x^4 + \frac{1}{2} e^{x/4}$.At $x = 0$,$f'(0) = 5(0)^4 + \frac{1}{2} e^0 = 0 + \frac{1}{2} = \frac{1}{2}$.Therefore,$g'(2) = \frac{1}{1/2} = 2$.The value of $8g'(2) = 8 	imes 2 = 16$.

Let $f(x) = x^5 + 2e^{x/4}$ for all $x \in \mathbb{R}$. Consider a function $g(x)$ such that $(g \circ f)(x) = x$ for all $x \in \mathbb{R}$. Then the value of $8g'(2)$ is:

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