If f(x) = exp(2x^3 + 3x^2 + 6x) and g(x) is t | vedclass

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(D) Given $f(x) = e^{2x^3 + 3x^2 + 6x}$.Since $g(x)$ is the inverse of $f(x)$,we have $g(f(x)) = x$.Differentiating both sides with respect to $x$,we

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$\frac{1}{18e^{11}}$

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(D) Given $f(x) = e^{2x^3 + 3x^2 + 6x}$.Since $g(x)$ is the inverse of $f(x)$,we have $g(f(x)) = x$.Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$,which implies $g'(f(x)) = \frac{1}{f'(x)}$.First,find $f'(x)$:$f'(x) = e^{2x^3 + 3x^2 + 6x} \cdot \frac{d}{dx}(2x^3 + 3x^2 + 6x) = e^{2x^3 + 3x^2 + 6x} \cdot (6x^2 + 6x + 6)$.We need to find $g'(e^{11})$. Set $f(x) = e^{11}$,so $e^{2x^3 + 3x^2 + 6x} = e^{11}$.This implies $2x^3 + 3x^2 + 6x = 11$,or $2x^3 + 3x^2 + 6x - 11 = 0$.By inspection,$x = 1$ is a root because $2(1)^3 + 3(1)^2 + 6(1) - 11 = 2 + 3 + 6 - 11 = 0$.Thus,$f(1) = e^{11}$.Using the formula $g'(f(1)) = \frac{1}{f'(1)}$:$f'(1) = e^{11} \cdot (6(1)^2 + 6(1) + 6) = e^{11} \cdot (6 + 6 + 6) = 18e^{11}$.Therefore,$g'(e^{11}) = \frac{1}{18e^{11}}$.

If $f(x) = \exp(2x^3 + 3x^2 + 6x)$ and $g(x)$ is the inverse function of $f(x)$,then the value of $g'(e^{11})$ is -

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