已知函数f(x)=x^2-4x+3.(1)求证:对于任意的t∈[-1,1]都有f(t)≥0恒成立.(2)若t1,t2∈(0,1),且t1^2+t2^2=1满足f(4t1)=f(2t2),求t1.(3)若f(2^x+2^-x+a)<f(1.5)对于任意的x∈[-1,1]恒成立,求a的取值范围.
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![已知函数f(x)=x^2-4x+3.(1)求证:对于任意的t∈[-1,1]都有f(t)≥0恒成立.(2)若t1,t2∈(0,1),且t1^2+t2^2=1满足f(4t1)=f(2t2),求t1.(3)若f(2^x+2^-x+a)<f(1.5)对于任意的x∈[-1,1]恒成立,求a的取值范围.](/uploads/image/z/4862190-30-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dx%5E2-4x%2B3.%281%29%E6%B1%82%E8%AF%81%3A%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8F%E7%9A%84t%E2%88%88%5B-1%2C1%5D%E9%83%BD%E6%9C%89f%28t%29%E2%89%A50%E6%81%92%E6%88%90%E7%AB%8B.%282%29%E8%8B%A5t1%2Ct2%E2%88%88%280%2C1%29%2C%E4%B8%94t1%5E2%2Bt2%5E2%3D1%E6%BB%A1%E8%B6%B3f%284t1%29%3Df%282t2%29%2C%E6%B1%82t1.%283%29%E8%8B%A5f%282%5Ex%2B2%5E-x%2Ba%29%EF%BC%9Cf%281.5%29%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8F%E7%9A%84x%E2%88%88%5B-1%2C1%5D%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.)
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