F(x) = x 3-(3a/2) x+b can be known from the derivative forms of f (x) and f (0) = b.

If the derivative of f(x) is equal to 0, x=0 or A can be solved.

By1< a <; It is known that f(x) monotonically increases and decreases on [0, 1].

Therefore, it can be seen that f(0)=b is the maximum, that is, b= 1. So f (x) = x 3-(3a/2) x+1,

F(- 1) or f( 1) is the minimum value. A=4/3 is obtained from f(- 1)=-2.

From f( 1)=-2, a=8/3 (truncation) is obtained.

The last knowledge: f (x) = x 3-2x 2+ 1.

I hope it helps you!