If the function f(x) = x^3 + e^{x/2} and g(x) | vedclass

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(C) Given $f(x) = x^3 + e^{x/2}$.We need to find $g^{\prime}(1)$ where $g = f^{-1}$.First,find $x$ such that $f(x) = 1$.$x^3 + e^{x/2} = 1$.By inspect

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$2$

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(C) Given $f(x) = x^3 + e^{x/2}$.We need to find $g^{\prime}(1)$ where $g = f^{-1}$.First,find $x$ such that $f(x) = 1$.$x^3 + e^{x/2} = 1$.By inspection,if $x = 0$,then $f(0) = 0^3 + e^0 = 0 + 1 = 1$.Thus,$f(0) = 1$,which implies $g(1) = 0$.Using the formula for the derivative of an inverse function: $g^{\prime}(y) = \frac{1}{f^{\prime}(x)}$ where $y = f(x)$.Here $y = 1$,so $x = 0$.$f^{\prime}(x) = \frac{d}{dx}(x^3 + e^{x/2}) = 3x^2 + \frac{1}{2}e^{x/2}$.At $x = 0$,$f^{\prime}(0) = 3(0)^2 + \frac{1}{2}e^0 = 0 + \frac{1}{2} = \frac{1}{2}$.Therefore,$g^{\prime}(1) = \frac{1}{f^{\prime}(0)} = \frac{1}{1/2} = 2$.

If the function $f(x) = x^3 + e^{x/2}$ and $g(x) = f^{-1}(x)$,then the value of $g^{\prime}(1)$ is

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