f(x)=e^x*(ax+b)-x^2-4x
則,f'(x)=e^x*(ax+b)+e^x*a-2x-4
所以,f'(0)=b+a-4
已知在(0,f(0))處的切線為y=4x+4
所以,f'(0)=4
===> a+b=8
又點(diǎn)(0,f(0))在切線上,所以：f(0)=4
而,f(0)=b
所以,a=b=4
那么,f'(x)=4(x+2)e^x-2(x+2)=2(x+2)*(2e^x-1)
當(dāng)f'(x)=0時(shí)有：x=-2,或者x=-ln2
當(dāng)x＞-ln2時(shí),f'(x)＞0,f(x)遞增
當(dāng)-2＜x＜-ln2時(shí),f'(x)＜0,f(x)遞減
當(dāng)x＜-2時(shí),f'(x)＞0,f(x)遞增
所以,f(x)有極大值f(-2)=0函數(shù)求導(dǎo)啊，這個(gè)不會(huì)么？！