The function f(x) = 2x^3 - 21x^2 + 36x + 7 ha | vedclass

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(B) Given function is $f(x) = 2x^3 - 21x^2 + 36x + 7$.First,find the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(2x^3 - 21x^2 + 36x + 7) = 6x^2 - 4

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$1$

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(B) Given function is $f(x) = 2x^3 - 21x^2 + 36x + 7$.First,find the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(2x^3 - 21x^2 + 36x + 7) = 6x^2 - 42x + 36$.Set $f'(x) = 0$ to find critical points:$6(x^2 - 7x + 6) = 0$$6(x - 1)(x - 6) = 0$So,$x = 1$ and $x = 6$.Now,find the second derivative $f''(x)$:$f''(x) = \frac{d}{dx}(6x^2 - 42x + 36) = 12x - 42$.Check the sign of $f''(x)$ at critical points:For $x = 1$: $f''(1) = 12(1) - 42 = -30$. Since $f''(1) For $x = 6$: $f''(6) = 12(6) - 42 = 72 - 42 = 30$. Since $f''(6) > 0$,the function has a local minimum at $x = 6$.Therefore,the function has a local maximum at $x = 1$.

The function $f(x) = 2x^3 - 21x^2 + 36x + 7$ has a local maximum at $x = $ ........

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