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问题: 已知函数f(x)=ax^2+bx+c,当|x|≤1时,|f(x)|≤1,求证:当|x|<=1时,|ax+b|<=2

解答:

把0代入:-1<=c<=1
把1代入:-1<=a+b+c<=1
所以-2<=a+b<=2
把-1代入:-1<=a-b+c<=1
所以-2<=a-b<=2
令g(x)=ax+b,
因为g(x)为单调函数,两端在[-2,2]内
所以-2<=g(x)<=2
所以|ax+b|<=2