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求f(x)=-x^2+2ax在0
人气:199 ℃ 时间:2020-05-17 02:29:36
解答
f(x)=-x^2+2ax
0<=x<=1
当0<=x<=1<=a,
f(x)min=0,
f(x)max=2a-1
当a<=0<=x<=1
f(x)min=2a-1,
f(x)max=0
当0<=a<=1/2
f(x)min=2a-1,
f(x)max=-a^2+2a^2=a^2
当1/2<=a<=1
f(x)min=0,
f(x)max=-a^2+2a^2=a^2
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