f'(x)=1/2x^(-1/2)
由拉格朗日中值定理得
f(4)-f(1)=(4-1)f'(ξ)
f'(ξ)=(2-1)/3=1/3=1/2ξ^(-1/2)
=> ξ=(2/3)^(-2)=9/4
函数f(x)=√x在区间[1,4]满足拉格朗日中值定理的点ξ
函数f(x)=√x在区间[1,4]满足拉格朗日中值定理的点ξ
数学人气：638 ℃时间：2019-12-18 09:59:36
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