(1)∵f(x+)=x2+=(x+)2-2,∴f(x)=x2-2(x≠0);
(2)设t=x-2,则x=t+2,代入得:f(t)=(t+2)2+3(t+2)+1=t2+7t+11, ∴f(x)=x2+7x+11;
(3)由题意设f(x)=ax+b, ∵3f(x+1)-2f(x-1)=2x+17, ∴3a(x+1)+3b-2a(x-1)-2b=2x+17,即ax+5a+b=2x+17, 则a=2且5a+b=17,解得a=2,b=7; ∴f(x)=2x+7.
(4)∵f(x)=x2-1,g(x)=(x≥-1), ∴f[g(x)]]=x+1-1=x(x≥-1), ∵x2-1≥-1, ∴g[f(x)]==|x|,且定义域是[-1,+∞). |