F[g(x)]=g[f(x)] holds for any x constant, then 2(kx+b)+3=k(2x+3)+b means b=3k-3.

g(x)=kx+b=g(x)=kx+3k-3

g( 1)=k+3k-3=4k-3

g(- 1)=-k+3k-3=2k-3

When k>0, g(x) monotonically increases on x ∈ [- 1, 1], the maximum value is g( 1) and the minimum value is g(- 1).

So g( 1)-g(- 1)=2k=2, and k= 1.

When k < 0, g(x) monotonically decreases on x ∈ [- 1, 1], with the maximum value of g(- 1) and the minimum value of g( 1).

So g(- 1)-g( 1)=-2k=2, and k=- 1.

So when k= 1, b=0, and when k=- 1, b=-6.

I did it twice, no problem.