f(x)=f'(1)e^x-1-f(0)x+½x²
因?yàn)閒（x）≥（1/2）x²+ax+b
f'(1)e^x-1-f(0)x+½x²≥（1/2）x²+ax+b
整理得：f'(1)e^x-1-f(0)x≥ax+b
即：ax+b
已知函數(shù)f(x)=f'(1)e^x-1-f(0)x+½x²
已知函數(shù)f(x)=f'(1)e^x-1-f(0)x+½x²
若f（x）≥（1/2）x²+ax+b,求（a+1）b的最大值
若f（x）≥（1/2）x²+ax+b,求（a+1）b的最大值
數(shù)學(xué)人氣：561 ℃時(shí)間：2020-01-28 16:55:18
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