Given: f(x+1)=x3+3x2−4x+2f(x + 1) = x^{3} + 3x^{2} - 4x + 2f(x+1)=x3+3x2−4x+2.
f(2)=f(x+1) ⟹ x+1=2;x=1f(2) = f(x + 1) \implies x + 1 = 2; x = 1f(2)=f(x+1)⟹x+1=2;x=1
f(2)=13+3(1)2−4(1)+2=1+3−4+2=2f(2) = 1^{3} + 3(1)^{2} - 4(1) + 2 = 1 + 3 - 4 + 2 = 2f(2)=13+3(1)2−4(1)+2=1+3−4+2=2
