The function f(x) = 2x^3 - 3x^2 - 12x + 4 has | vedclass

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(B) Given function: $f(x) = 2x^3 - 3x^2 - 12x + 4$Step $1$: Find the first derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 4) = 6x^2 - 6x

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One maximum and one minimum

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(B) Given function: $f(x) = 2x^3 - 3x^2 - 12x + 4$Step $1$: Find the first derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 4) = 6x^2 - 6x - 12$Step $2$: Find critical points by setting $f'(x) = 0$.$6(x^2 - x - 2) = 0$$6(x - 2)(x + 1) = 0$So,the critical points are $x = 2$ and $x = -1$.Step $3$: Find the second derivative $f''(x)$ to check for maxima/minima.$f''(x) = \frac{d}{dx}(6x^2 - 6x - 12) = 12x - 6$Step $4$: Evaluate $f''(x)$ at critical points.For $x = 2$: $f''(2) = 12(2) - 6 = 24 - 6 = 18 > 0$. Since $f''(2) > 0$,the function has a local minimum at $x = 2$.For $x = -1$: $f''(-1) = 12(-1) - 6 = -12 - 6 = -18 Conclusion: The function has one maximum and one minimum.

The function $f(x) = 2x^3 - 3x^2 - 12x + 4$ has

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