Take the extreme value at X= 1 and get: f (1) =1+a+b+a ∧ 2 =10①.

f′( 1)= 3+2a+b = 0②

The solutions are A 1 = 4, b1=-1,A2 =-3, B2 = 3.

∫δ> 0 in ② means δ = 4a∧ 2-12b: 0.

∴a2=-3,b2=3.

∴f(x)=x∧3+4x∧2- 1 1x+ 16

∴f(2)=8+ 16-22+ 16= 18

PS: You may have solved the equation wrong.

By the way, explain why it is not δ≥ 0, because if δ = 0, when A2 =-3 and B2 = 3, the minimum value of the derivative is 0, the lowest point of the derivative image is on the X axis, and the image is above the X axis. The whole function is monotonically increasing, which does not conform to the cubic function image, so δ ≠ 0.

I hope to adopt. I am a sophomore in science, and I learned it a few months ago.