In such type of questions

find both the maximum and minimum value to compare the options.

so first,

so,

put f’(x)=0;

⇒ x=±1

Hence by second derivative test

f’’(x)>0 or f”(x)<0

so it’s a point of minimum or maximum respecctively.

f”(-1)=-2<0;

f”(1)=2>0

so x=1 is a point of minimum

and x=-1 is a point of maximum

f(1)=2 is minimum value.

f(-1)=-2 is maximum value.

therefore,

maximum value<minimum value.