The function f(x) = 2x^3 - 15x^2 + 36x + 4 is | vedclass

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(A) Given function is $f(x) = 2x^3 - 15x^2 + 36x + 4$.First,find the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 4) = 6x^2 - 3

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$2$

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(A) Given function is $f(x) = 2x^3 - 15x^2 + 36x + 4$.First,find the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 4) = 6x^2 - 30x + 36$.To find the critical points,set $f'(x) = 0$:$6(x^2 - 5x + 6) = 0$$6(x - 2)(x - 3) = 0$So,the critical points are $x = 2$ and $x = 3$.Now,find the second derivative $f''(x)$ to check for maxima or minima:$f''(x) = \frac{d}{dx}(6x^2 - 30x + 36) = 12x - 30$.Evaluate $f''(x)$ at the critical points:For $x = 2$: $f''(2) = 12(2) - 30 = 24 - 30 = -6$.Since $f''(2) For $x = 3$: $f''(3) = 12(3) - 30 = 36 - 30 = 6$.Since $f''(3) > 0$,the function has a local minimum at $x = 3$.Therefore,the function is maximum at $x = 2$.

The function $f(x) = 2x^3 - 15x^2 + 36x + 4$ is maximum at $x=$ ......

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