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?
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![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?](/uploads/image/z/15117768-0-8.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3De%5Ex-e%5E%28-x%29%E2%91%A0%E8%AF%81%E6%98%8E%EF%BC%9Af%28x%29%E7%9A%84%E5%AF%BC%E6%95%B0f%27%28x%29%E2%89%A52%E2%91%A1%E8%8B%A5%E5%AF%B9%E6%89%80%E6%9C%89x%E2%89%A50%2C%E4%B8%94a%E2%88%88%28-%E2%88%9E%2C2%5D%E6%97%B6%2C%E8%AF%81%E6%98%8E%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28x%29%E2%89%A5ax%E6%88%90%E7%AB%8B%E6%80%8E%E4%B9%88%E5%81%9A%3F)
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?
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?
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?
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
1,f`(x)=e^x+e^(-x) , f``(x)=e^x-e^(-x), 当f``(x)=0,则x=0,f```(x)=e^x+e^(-x), f```(0)=2>0, 所以f`(x)在x=0时取最小值,
f`(x)≥f`(0)=2
2,由题可知:不等式f(x)≥ax => f(x)/x≥a => e^x-e^(-x)/x ≥a
设e^x-e^(-x)/x =...
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1,f`(x)=e^x+e^(-x) , f``(x)=e^x-e^(-x), 当f``(x)=0,则x=0,f```(x)=e^x+e^(-x), f```(0)=2>0, 所以f`(x)在x=0时取最小值,
f`(x)≥f`(0)=2
2,由题可知:不等式f(x)≥ax => f(x)/x≥a => e^x-e^(-x)/x ≥a
设e^x-e^(-x)/x =g(x) ,则要使g(x)≥a在x≥0上衡成立,则g(x)min≥2
求g(x)min:
自己算
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不