Question

函数ln(x+1)在区间[0,1]上满足拉格朗日中值定理的一批西诺=

Answer 1

令f(x)=ln(x+1)
f'(ε) = 1/(ε+1)
(ln(1+1)-ln(0+1))/(1-0) = f'(ε) = 1/(ε+1)
即 ln2=1/(ε+1)
解得 ε = 1/ln2 -1