设f(x)=x2-2ax+2(a∈R),当x∈[-1,+∞)时,f(x)≥a恒成立,求a的取值范围.
设f(x)=x2-2ax+2(a∈R),当x∈[-1,+∞)时,f(x)≥a恒成立,求a的取值范围.
数学人气:158 ℃时间:2019-08-21 10:14:12
优质解答
f(x)=x2-2ax+2=(x-a)2+2-a2f(x)图象的对称轴为x=a为使f(x)≥a在[-1,+∞)上恒成立,只需f(x)在[-1,+∞)上的最小值比a大或等于a即可∴(1)a≤-1时,f(-1)最小,解,解得-3≤a≤-1 (2)a≥-1时...
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