Question

这个函数怎么求导？求详细

Answer 1

1. f'(x)=e^(x-1)-2ax+2a-1

   a=0→f'(x)=e^(x-1)-1

   f(2)=e-2 f'(2)=e-1

   ∴切线：y-f(2)=f'(2)(x-2)→y=(e-1)x-e

2. f(x)=e^(x-1)-ax²+(2a-1)x-a

   f(1)=1-a+2a-1-a=0

   f'(x)=e^(x-1)-2ax+2a-1

   f'(1)=1-2a+2a-1=0

   f''(x)=e^(x-1)-2a

   当a≤0时，f''(x)≥0→f'(x)单调递增→f'(x)≥f'(1)=0→f(x)单调递增→f(x)≥f(1)=0

   a∈(-∞,0]