Question

已知f(x)=1+log₂x(1≤x≤4),求函數(shù)g(x)=［f(x)］²+f(x²)的最大值和最小值.

Answer 1

當(dāng)x=1時(shí)，g(x)_min=2,當(dāng)x=2時(shí)，g(x)_max=7.

g(x)=［f(x)］²+f(x²)=(1+log₂x)²+1+log₂x²=1+2log₂x+log₂²x+1+2log₂x=log₂²x+4log₂x+2
=(log₂x+2)²+2-4=(log₂x+2)²-2,由于f(x)的定義域?yàn)椋?，4］,則g(x)的定義域?yàn)椋?，2］,于是當(dāng)x=1時(shí)，g(x)_min=2,當(dāng)x=2時(shí)，g(x)_max=7.