求讲解

首先令x=0,可得f''(0)=0
f''(x)+p(x)f'(x)+xf(x)=x+x²,两边对x求导得
f'''(x)+p'(x)f'(x)+p(x)f''(x)+f(x)+xf'(x)=1+2x,再令x=0,得
f'''(0)=0,上式两边再对x求导得
f''''(x)+p''(x)f'(x)+p'(x)f''(x)+p'(x)f''(x)+p(x)f'''(x)+f'(x)+f'(x)+xf''(x)=2
再令x=0,得f''''(0)=2>0,∴x=0是f(x)的极小值点
这是∵由f'''(0)=0,f''''(0)>0可知,x=0是f''(x)的极小值点
∴x>0或xf''(0)=0,∴f'(x)在x>0和x