let `f(x)=-x^3+(b^3-b^2+b-1)/(b^2+3b+2)` if `x` is `0` to `1` and `f(x)=2x-3` if `x` if `1` to `3`.All possible real values of `b` such that `f (x)` has the smallest value at `x=1` are

Explanation

`Let(b^(3)-b^(2)+b-2)/(b^(2)+3b+2)=a`
If a=0 then `f(x)={-x^(3),0lexle1 lt 1`
`2x-3,1lexle3`

Clearly when a=0, x=1 is point of minima
If `agt0` then graph of `y =-x^(3)`+a shifts upward,then also x=1 remains point of minima.
If `alt0`, then graph of `y=-x^(3)+a` shifts downward .In this case x=1 is not a point of minima.
So for point of minima at x=1 ,`age0`
`(b^(3)-b^(2)+b-1)/(b^(2)+3b+2)ge0`
`(b^(2)+1(b-1))/(b+2(b+1))ge0`
Sing scheme of `(b^(2)+1(b-1))/(b+2(b+1))`
From the sign scheme,b in `(-2,-1)cup[1,00)`