If f(x) = x^3 + e^{x/2} and g(x) = f^{-1}(x), | vedclass

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(B) Given $f(x) = x^3 + e^{x/2}$.Since $g(x) = f^{-1}(x)$,we have $g(f(x)) = x$.Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot

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$2$

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(B) Given $f(x) = x^3 + e^{x/2}$.Since $g(x) = f^{-1}(x)$,we have $g(f(x)) = x$.Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$.To find $g'(1)$,we set $f(x) = 1$.$x^3 + e^{x/2} = 1$.By inspection,at $x = 0$,$0^3 + e^0 = 0 + 1 = 1$. Thus,$f(0) = 1$.Now,find $f'(x) = 3x^2 + \frac{1}{2}e^{x/2}$.At $x = 0$,$f'(0) = 3(0)^2 + \frac{1}{2}e^0 = 0 + \frac{1}{2} = \frac{1}{2}$.Substituting these values into the derivative formula: $g'(f(0)) \cdot f'(0) = 1 \implies g'(1) \cdot \frac{1}{2} = 1$.Therefore,$g'(1) = 2$.

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

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