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

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(C) Given $f(x) = x^3 + e^{\frac{x}{2}}$.First,find the value of $x$ such that $f(x) = 1$.$x^3 + e^{\frac{x}{2}} = 1$.By inspection,if $x = 0$,then $f

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$2$

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(C) Given $f(x) = x^3 + e^{\frac{x}{2}}$.First,find the value of $x$ such that $f(x) = 1$.$x^3 + e^{\frac{x}{2}} = 1$.By inspection,if $x = 0$,then $f(0) = 0^3 + e^0 = 0 + 1 = 1$.Thus,$f(0) = 1$,which implies $g(1) = 0$.We know that $g(f(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule:$g^{\prime}(f(x)) \cdot f^{\prime}(x) = 1$.To find $g^{\prime}(1)$,we set $x = 0$:$g^{\prime}(f(0)) \cdot f^{\prime}(0) = 1 \Rightarrow g^{\prime}(1) \cdot f^{\prime}(0) = 1$.Now,calculate $f^{\prime}(x)$:$f^{\prime}(x) = 3x^2 + \frac{1}{2} e^{\frac{x}{2}}$.$f^{\prime}(0) = 3(0)^2 + \frac{1}{2} e^0 = 0 + \frac{1}{2} = \frac{1}{2}$.Therefore,$g^{\prime}(1) = \frac{1}{f^{\prime}(0)} = \frac{1}{1/2} = 2$.

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

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