分析:f(x)=
,當(dāng)x>1時(shí),x-1>0,g(x)=x
2f(x-1)=x
2>1;當(dāng)x=1時(shí),g(x)=x
2f(x-1)=0;當(dāng)0<x<1時(shí),g(x)=x
2f(x-1)=-x
2∈(-1,0);當(dāng)x<0時(shí),g(x)=x
2f(x-1)=-x
2<0.由此能求出函數(shù)g(x)=x
2f(x-1)的值域.
解答:解:∵f(x)=
,
∴當(dāng)x>1時(shí),x-1>0,f(x-1)=1,
g(x)=x
2f(x-1)=x
2>1;
當(dāng)x=1時(shí),x-1=0,f(x-1)=0,
g(x)=x
2f(x-1)=0;
當(dāng)0<x<1時(shí),x-1<0,f(x-1)=-1,
g(x)=x
2f(x-1)=-x
2∈(-1,0);
當(dāng)x<0時(shí),x-1<0,f(x-1)=-1,
g(x)=x
2f(x-1)=-x
2<0.
綜上所述,函數(shù)g(x)=x
2f(x-1)的值域(-∞,0]∪(1,+∞).
故選C.
點(diǎn)評(píng):本題考查函數(shù)的性質(zhì)和應(yīng)用,是基礎(chǔ)題.解題時(shí)要認(rèn)真審題,仔細(xì)解答,注意分類討論思想的合理運(yùn)用.