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f(x)=f'(1)e^x-1-f(0)x+½;x²; 因為f(x)≥(1/2)x²;+ax+b f'(1)e^x-1-f(0)x+½;x²;≥(1/2)x²;+ax+b 整理得:f'(1)e^x-1-f(0)x≥ax+b 即:ax+b
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