Let f(x)= 2x³ – 3x² – 12x + 4

Applying the first derivative we get

Applying the sum rule of differentiation, so we get

Applying the derivative,

⇒ f' (x)=6x²-6x-12

Putting f’(x)=0,we get critical points as

6x²-6x-12=0

⇒ 6(x²-x-2)=80

⇒ x²-x-2=0

Splitting the middle term we get

⇒ x²-2x+x-2=0

⇒ x(x-2)+1(x-2)=0

⇒ (x-2)(x+1)=0

⇒ x-2=0 or x+1=0

⇒ x=2 or x=-1

Now we will find the value of f(x) at x=-1, 2

f(x)= 2x³ – 3x² – 12x + 4

f(-1)= 2(-1)³ – 3(-1)² – 12(-1) + 4=-2-3+12+4=11

f(2)= 2(2)³ – 3(2)² – 12(2) + 4 =16-12-24+4=-16

Hence from above we find that x=-1 is point of local maxima and the maximum value of f(x) is 11.

Whereas x=2 is point of local minima and the minimum value of f(x) is -16.

So the correct option is option C.

Hence the given function 2x³ – 3x² – 12x + 4has one maxima and one minima