Let f(x)=(x-2)^{17}(x+5)^{24}. Then | vedclass

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(C) Given $f(x)=(x-2)^{17}(x+5)^{24}$.To find the critical points, we find the derivative $f'(x)$:$f'(x) = 17(x-2)^{16}(x+5)^{24} + 24(x-2)^{17}(x+5)^

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$f$ has neither a maximum nor a minimum at $x=2$

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(C) Given $f(x)=(x-2)^{17}(x+5)^{24}$.To find the critical points, we find the derivative $f'(x)$:$f'(x) = 17(x-2)^{16}(x+5)^{24} + 24(x-2)^{17}(x+5)^{23}$$f'(x) = (x-2)^{16}(x+5)^{23} [17(x+5) + 24(x-2)]$$f'(x) = (x-2)^{16}(x+5)^{23} [17x + 85 + 24x - 48]$$f'(x) = (x-2)^{16}(x+5)^{23} (41x + 37)$The critical points are $x=2, x=-5, x=-\frac{37}{41}$.Now, examine the sign of $f'(x)$ around $x=2$:Since $(x-2)^{16}$ is always non-negative, the sign of $f'(x)$ depends on $(x+5)^{23}(41x+37)$.For $x$ slightly less than $2$, $(x+5)^{23} > 0$ and $(41x+37) > 0$, so $f'(x) > 0$.For $x$ slightly greater than $2$, $(x+5)^{23} > 0$ and $(41x+37) > 0$, so $f'(x) > 0$.Since the sign of $f'(x)$ does not change as $x$ passes through $2$, $f(x)$ has neither a maximum nor a minimum at $x=2$.

Let $f(x)=(x-2)^{17}(x+5)^{24}$. Then

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