The function

Solution:

The given function is $f(x)=2 x^{3}-3 x^{2}-12 x+4$

$f(x)=2 x^{3}-3 x^{2}-12 x+4$

Differentiating both sides with respect to x, we get

$f^{\prime}(x)=6 x^{2}-6 x-12$

$\Rightarrow f^{\prime}(x)=6\left(x^{2}-x-2\right)$

 

$\Rightarrow f^{\prime}(x)=6(x+1)(x-2)$

For maxima or minima,

$f^{\prime}(x)=0$

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

$\Rightarrow x+1=0$ or $x-2=0$

 

$\Rightarrow x=-1$ or $x=2$

Now,

$f^{\prime \prime}(x)=12 x-6$

At $x=-1$, we have

$f^{\prime \prime}(-1)=12 \times(-1)-6=-12-6=-18<0$

So, $x=-1$ is the point of local maximum.

At x = 2, we have

$f^{\prime \prime}(2)=12 \times 2-6=24-6=18>0$

So, $x=2$ is the point of local minimum.

Thus, the given function $f(x)=2 x^{3}-3 x^{2}-12 x+4$ has one point of local maximum and one point of local minimum.

Hence, the correct answer is option (c).