Let S=(-1, infty) and f: S rightarrow R be de | vedclass

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(B) By the Fundamental Theorem of Calculus,$f'(x) = (e^x-1)^{11}(2x-1)^5(x-2)^7(x-3)^{12}(2x-10)^{61}$.To find critical points,set $f'(x) = 0$,which g

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$27$

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(B) By the Fundamental Theorem of Calculus,$f'(x) = (e^x-1)^{11}(2x-1)^5(x-2)^7(x-3)^{12}(2x-10)^{61}$.To find critical points,set $f'(x) = 0$,which gives $x = 0, \frac{1}{2}, 2, 3, 5$.We analyze the sign of $f'(x)$ around these points:- At $x=0$: $f'(x)$ changes from $+$ to $-$,so $f(x)$ has a local maximum at $x=0$.- At $x=\frac{1}{2}$: $f'(x)$ changes from $-$ to $+$,so $f(x)$ has a local minimum at $x=\frac{1}{2}$.- At $x=2$: $f'(x)$ changes from $+$ to $-$,so $f(x)$ has a local maximum at $x=2$.- At $x=3$: $f'(x)$ does not change sign (power is $12$),so it is a point of inflection.- At $x=5$: $f'(x)$ changes from $-$ to $+$,so $f(x)$ has a local minimum at $x=5$.Local maxima occur at $x=0$ and $x=2$. Thus,$p = 0^2 + 2^2 = 4$.Local minima occur at $x=\frac{1}{2}$ and $x=5$. Thus,$q = \frac{1}{2} + 5 = \frac{11}{2}$.Finally,$p^2 + 2q = 4^2 + 2(\frac{11}{2}) = 16 + 11 = 27$.

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