因?yàn)椋?^(x^2)<=2^x
所以x^2<=x
即0<=x<=1.
f(x)=-(2^x)^2+(1/2)x^2-m
=-(2^x-1/4)^2+(1/16)-m
設(shè)2^x=t,則有1<=t<=2.
f(x)=-(t-1/4)^2+(1/16)-m
開(kāi)口向下,對(duì)稱(chēng)軸t=1/4,區(qū)間[1,2]在對(duì)稱(chēng)軸的右方,要使f(x)>=0,則有：
f(2)>=0
代入即得到：
m<=-3.