f（x）=2^x+1/2^|x|
x≥0時(shí),
f(x)=2^x+1/2^x
設(shè)t=2^x≥1,
f(x)=t+1/t≥2 【均值定理】
當(dāng)且僅當(dāng)t=1時(shí)取等號(hào)
且f(x)為增函數(shù)
√2/2≤f(x）恒成立
由f(x)=2^x+1/2^x=17/4
==>t^2-17/4t+1=0
==>t=1/4(舍去）,或t=4,
==>x=2
∴2^x+1/2^x≤17/4
==>0≤x≤2
即√2/2≤f(x)≤17/4解得0≤x≤2
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