Let, f(x)=ax3+bx2+cx+d
Diff w. r. t. 'x'
f′(x)=3ax2+2bx+c
Again diff. w. r. t. 'x'
f"(x)=6ax+2b
f"(−1)=0
−6a+2b=0
b=3a
f′(1)=0
3a+6a+c=0
c=−9a
f(1)=−10
−5a+d=−10…(i)
f(−1)=6
11a+d=6…(ii)
From equations (i)−(ii)
We get a=1,d=−5,b=3,c=−9
Then, f(x)=x3+3x2−9x−5
Hence, f(3)=27+27−27−5
f(3)=22