Leny06
幼苗
共回答了15个问题采纳率：86.7% 举报
f(x) = (x^2-x-1/a) e^(ax)
f '(x) = ( 2x-1) e^(ax) + (x^2-x-1/a) e^(ax) * a = e^(ax) [ a x^2 + (2-a) x - 2 ]
设 g(x) = a x^2 + (2-a) x - 2 = a (x + 2/a) ( x - 1)
g(x) = 0 => 驻点x1 = 1, x2 = -2/a (a 0, f(x)单调增加；
x 1, g '(x)
1年前
2
