g(x)=x^2-5x+5
=(x-5/2)^2-5/4
所以g(x)>=-5/4
f[(g(x)]
=-[g(x)]^2-[g(x)]
=-[g(x)+1/2]^2+1/4
当g(x)=-1/2时f[g(x)]最大值为1/4
所以f((g(x))的值域是(-∞,1/4]
g(x)=x^2-5x+5
=(x-5/2)^2-5/4
所以g(x)>=-5/4
f[(g(x)]
=-[g(x)]^2-[g(x)]
=-[g(x)+1/2]^2+1/4
当g(x)=-1/2时f[g(x)]最大值为1/4
所以f((g(x))的值域是(-∞,1/4]