If f(x) and g(x) are functions satisfying f(g | vedclass

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(C) Given $f(g(x)) = x^3 + 3x^2 + 3x + 4$. We can rewrite this as $f(g(x)) = (x+1)^3 + 3$. Given $f(x) = (\ln x)^3 + 3$. Comparing $f(g(x)) = (\ln g(x

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$1$

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(C) Given $f(g(x)) = x^3 + 3x^2 + 3x + 4$. We can rewrite this as $f(g(x)) = (x+1)^3 + 3$. Given $f(x) = (\ln x)^3 + 3$. Comparing $f(g(x)) = (\ln g(x))^3 + 3$ with $f(g(x)) = (x+1)^3 + 3$,we get $\ln g(x) = x+1$. Therefore,$g(x) = e^{x+1}$. To find the slope of the tangent at $x = -1$,we find the derivative $g'(x)$. $g'(x) = \frac{d}{dx}(e^{x+1}) = e^{x+1}$. At $x = -1$,$g'(-1) = e^{-1+1} = e^0 = 1$. Thus,the slope of the tangent is $1$.

If $f(x)$ and $g(x)$ are functions satisfying $f(g(x)) = x^3 + 3x^2 + 3x + 4$ and $f(x) = (\ln x)^3 + 3$,then the slope of the tangent to the curve $y = g(x)$ at $x = -1$ is:

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