22.
[解](1)h(x)= (-2x+3)(x-2) x∈[1,+∞)
x-2
x∈(-∞,1)
(2) 当x≥1时, h(x)= (-2x+3)(x-2)=-2x2+7x-6=-2(x-
)2+
∴h(x)≤
; 当x<1时, h(x)<-1,
∴当x=
时, h(x)取得最大值是
(3)令 f(x)=sinx+cosx,α=
则g(x)=f(x+α)= sin(x+
)+cos(x+
)=cosx-sinx,
于是h(x)= f(x)·f(x+α)= (sinx+cosx)( cosx-sinx)=cos2x.
另解令f(x)=1+
sinx, α=π,
g(x)=f(x+α)=
1+
sin(x+π)=1-
sinx,
于是h(x)= f(x)·f(x+α)= (1+
sinx)( 1-
sinx)=cos2x.