已知函数f(x)=ln(x+a)-x2-x在x=0处取得极值.(1)求函数f(x)的单调区间;(2)若关于x

(1)f'(x) = 1/(x + a) - 2x - 1f'(0) = 1/a - 1 = 0,a = 1(2)g(x) = f(x) - (-5x/2 + b) = ln(x + 1) - x² - x + 5x/2 - b = ln(x + 1) - x² + 3x/2 - bg'(x) = 1/(x + 1) - 2x + 3/2 = (-4x² - x + ...