If the function f(x) = x^5 + e^{x/5} and g(x) | vedclass

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(A) Given $f(x) = x^5 + e^{x/5}$ and $g(x) = f^{-1}(x)$.We need to find the value of $\frac{1}{g'(1 + e^{1/5})}$.By the property of inverse functions,

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$5 + \frac{e^{1/5}}{5}$

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(A) Given $f(x) = x^5 + e^{x/5}$ and $g(x) = f^{-1}(x)$.We need to find the value of $\frac{1}{g'(1 + e^{1/5})}$.By the property of inverse functions,$g'(y) = \frac{1}{f'(x)}$ where $y = f(x)$.Here,$y = 1 + e^{1/5}$.Setting $f(x) = x^5 + e^{x/5} = 1 + e^{1/5}$,we observe that $x = 1$ satisfies the equation because $1^5 + e^{1/5} = 1 + e^{1/5}$.Now,find the derivative $f'(x) = \frac{d}{dx}(x^5 + e^{x/5}) = 5x^4 + \frac{1}{5}e^{x/5}$.At $x = 1$,$f'(1) = 5(1)^4 + \frac{1}{5}e^{1/5} = 5 + \frac{e^{1/5}}{5}$.Since $g'(y) = \frac{1}{f'(x)}$,we have $g'(1 + e^{1/5}) = \frac{1}{f'(1)} = \frac{1}{5 + \frac{e^{1/5}}{5}}$.Therefore,$\frac{1}{g'(1 + e^{1/5})} = f'(1) = 5 + \frac{e^{1/5}}{5}$.

If the function $f(x) = x^5 + e^{x/5}$ and $g(x) = f^{-1}(x)$,then the value of $\frac{1}{g'(1 + e^{1/5})}$ is

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