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使用零點存在定理啊!構造F(x)=e^x-x-2,第一步求導,證明F(x)的導數F'(x)=e^x-1在(0,2)上恒大於零,即F(x)在(0,2)上單調遞增;第二步可求得F(0).F(2)< 0,於是有零點定理的必至少有一根.
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