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

The correct answer is Option (3) → one maxima and one minima ##

We have $f(x) = 2x^3 - 3x^2 - 12x + 4$

$∴f'(x) = 6x^2 - 6x - 12$

Now, $f'(x) = 0 \Rightarrow 6(x^2 - x - 2) = 0$

$\Rightarrow 6(x + 1)(x - 2) = 0$

$x = -1 \text{ and } x = +2$

On number line for $f'(x)$, we get:

Hence $x = -1$ is point of local maxima and $x = 2$ is point of local minima.

So, $f(x)$ has one maxima and one minima.