Question
The function $f(x) = 2x^3 - 3x^2 - 12x + 4,$ has:
✓
Answer
We have, $f(x) = 2x^3 - 3x^2 - 12x + 4$
$\Rightarrow f'(x) = 6x^2 - 6x - 12$
$\Rightarrow f'(x) = 6(x^2 - x - 2)$
$\Rightarrow f'(x) = 6(x + 1)(x - 2)$
Find the critical points by equating $f'(x)$ to $0.$
$\therefore f'(x) = 0$
$\Rightarrow 6(x + 1)(x - 2) = 0$
$\Rightarrow x = -1$ and $x = +2$
From the above number line, we can conclude that, $x = -1$ is point of local maxima and $x = 2$ is point of local minima.
Thus, $f(x)$ has one maxima and one minima.
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