f(x)=x⁴-x²-2x+6

f’(x)=4x³-2x-2

so here put f’(x)=0

4x³-2x-2=0

2(x-1)(2x²+2x+1)=0

x=1 and other roots are complex.

so we will consider x=1

Hence by second derivative test

f’’(x)>0 so it’s a point of minimum.

f”(x)=12x²-1

f” (1) =12(1)-2

=10>0

At x=1 minimum

So, f (1) =1-1-2(1) +6

=4

Option(B)