已知函数f(x)满足满足f(x)=f'(1)e^(x-1)-f(0)x+(1/2)x²1.求f(x)的解析式及单调区间.解析式为f(x)=e^x-x+1/2x²单调递增区间为(0,+∞),单调递减区间为(﹣∞,0)2.若f(x)≥1/2x²+ax+b,求(a+1)b的最
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已知函数f(x)满足满足f(x)=f'(1)e^(x-1)-f(0)x+(1/2)x²1.求f(x)的解析式及单调区间.解析式为f(x)=e^x-x+1/2x²单调递增区间为(0,+∞),单调递减区间为(﹣∞,0)2.若f(x)≥1/2x²+ax+b,求(a+1)b的最
已知函数f(x)满足满足f(x)=f'(1)e^(x-1)-f(0)x+(1/2)x²
1.求f(x)的解析式及单调区间.
解析式为f(x)=e^x-x+1/2x²
单调递增区间为(0,+∞),单调递减区间为(﹣∞,0)
2.若f(x)≥1/2x²+ax+b,求(a+1)b的最大值.
关键是第二问有没有简单的方法?我看了标答太麻烦了,而且那样讨论容易出错
已知函数f(x)满足满足f(x)=f'(1)e^(x-1)-f(0)x+(1/2)x²1.求f(x)的解析式及单调区间.解析式为f(x)=e^x-x+1/2x²单调递增区间为(0,+∞),单调递减区间为(﹣∞,0)2.若f(x)≥1/2x²+ax+b,求(a+1)b的最
晕,你都有标答了.还问,这种题目应该没有太简单的方法
即g(x)=e^x -(a+1)x-b≥0恒成立
然后就是求g(x)的最小值≥0即可
当然需要分类讨论,我的能力,无法分离出常数.
但是a+1≤0时 ,显然不满足,
只需要讨论a+1>0的时候.
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