Question

對(duì)定義域分別是D_f、D_g的函數(shù)y=f(x) 、y=g(x),

                   f(x)?g(x)   當(dāng)x∈D_f且x∈D_g

 規(guī)定: 函數(shù)h(x)=   f(x)        當(dāng)x∈D_f且xD_g

                   g(x)       當(dāng)xD_f且x∈D_g

（1）若函數(shù)f(x)=-2x+3,x≥1; g(x)=x-2,x∈R,寫出函數(shù)h(x)的解析式;

（2）求問題(1)中函數(shù)h(x)的最大值;

若g(x)=f(x+α), 其中α是常數(shù),且α∈[0,π],請(qǐng)?jiān)O(shè)計(jì)一個(gè)定義域?yàn)镽的函數(shù)y=f(x),及一個(gè)α的值,使得h(x)=cos2x,并予以證明.

Answer 1

解： (1)h(x)=  (-2x+3)(x-2)   x∈[1,+∞)

                   x-2            x∈(-∞,1)

(2) 當(dāng)x≥1時(shí), h(x)= (-2x+3)(x-2)

=-2x²+7x-6

=-2(x-)²+        ∴h(x)≤; 

當(dāng)x<1時(shí), h(x)<-1, ∴當(dāng)x=時(shí), h(x)取得最大值是

(3)[解法一]令 f(x)=sinx+cosx,α=

則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.

[解法二]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.