f′′(x)=6(x−1). Integrating we get f′(x)=3x2−6x+c Slope at (2,1)=f′(2)=c=3 [ As slope of tangent at (2,1) is 3 ] ∴f′(x)=3x2−6x+3=3(x−1)2 Integrating again, we get f(x)=(x−1)3+D The curve passes through (2,1) ⇒1=(2−1)3+D⇒D=0 ∴f(x)=(x−1)3