题目内容
x≠0,求证ex＞1+x
证明：令f(x)=ex-1-x,
f(0)=e0-1-0=0,f′(x)=ex-1.
①当x＞0时，f′(x)=ex-1＞0，即f(x)在(0，+∞)上为增函数，∴f(x)＞f(0),即ex-1-x＞0,即ex＞1+x.
②当x＜0时，f′(x)=ex-1＜0即f(x)在(-∞,0)上为减函数，∴f(x)＞f(0).
即ex-1-x＞0，即ex＞1+x.综上可知：x≠0时，ex＞1+x.
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题目内容
x≠0,求证ex＞1+x
证明：令f(x)=ex-1-x,
f(0)=e0-1-0=0,f′(x)=ex-1.
①当x＞0时，f′(x)=ex-1＞0，即f(x)在(0，+∞)上为增函数，∴f(x)＞f(0),即ex-1-x＞0,即ex＞1+x.
②当x＜0时，f′(x)=ex-1＜0即f(x)在(-∞,0)上为减函数，∴f(x)＞f(0).
即ex-1-x＞0，即ex＞1+x.综上可知：x≠0时，ex＞1+x.