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

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(C) To find the local maxima and minima, we first find the derivative of the function $f(x) = 2x^3 - 3x^2 - 12x + 4$.$f'(x) = 6x^2 - 6x - 12$.Setting

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one local maximum and one local minimum

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(C) To find the local maxima and minima, we first find the derivative of the function $f(x) = 2x^3 - 3x^2 - 12x + 4$.$f'(x) = 6x^2 - 6x - 12$.Setting $f'(x) = 0$ for critical points:$6(x^2 - x - 2) = 0 \Rightarrow 6(x - 2)(x + 1) = 0$.Thus, the critical points are $x = 2$ and $x = -1$.Now, we find the second derivative $f''(x) = 12x - 6$.For $x = -1$, $f''(-1) = 12(-1) - 6 = -18 For $x = 2$, $f''(2) = 12(2) - 6 = 18 > 0$, so $x = 2$ is a point of local minima.Therefore, the function has one local maximum and one local minimum.

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

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