设函数f(x)=e^x-e^(-x)①证明：f(x)的导数f'(x)≥2②若对所有x≥0,且a∈(-∞,2]时,证明不等式f(x)≥ax成立怎么做?

Question

设函数f(x)=e^x-e^(-x)
①证明：f(x)的导数f'(x)≥2
②若对所有x≥0,且a∈(-∞,2]时,证明不等式f(x)≥ax成立
怎么做?

Answer 1

1、利用不等式：A＞0时,A＋1/A≥2即可
f'(x)＝e^x＋1/(e^x)≥2
2、当x≥0,a≤2时,令F(x)＝f(x)－ax,F'(x)＝f'(x)－a≥0,所以F(x)在[0,+∞)上单调增加,所以当x≥0时,F(x)＝F(0),而F(0)＝0,所以x≥0时,F(x)≥0,即f(x)≥ax