设f(x)为连续函数,且满足∫0xf(t)dt=ex+∫0xtf(x-t)dt,则曲线y=f(x)的拐点为_______

admin2020-09-23  26

问题 设f(x)为连续函数,且满足∫0xf(t)dt=ex+∫0xtf(x-t)dt,则曲线y=f(x)的拐点为_______

选项

答案(一3,一2e-3)

解析 令x—t=u,则
   ∫0xtf(x一t)dt=∫x0(x一u)f(u)(一du)=∫0x(x一u)f(u)du,
  则已知等式可化为
    ∫0xf(t)dt=ex+x∫0xf(u)du一∫0xuf(u)du,
  等式两边对x求导,得
   f(x)=ex+∫0xf(u)du,
  取x=0,得,f(0)=1,上式两边对x求导,得
    f’(x)一f(x)=ex
  于是    [f’(x)一f(x)]e-x=1,即[f(x)e-x]’=1,
  等式两边对x积分,得
    f(x)e-x=x+C,  f(x)=ex(x+C),
    由f(0)=1,得C=1,则f(x)=ex(x+1),f’(x)=ex(x+2).f”(x)=ex(x+3),
    由f”(x)=0,得x=-3,且当x<-3时,f"(x)<0,当x>一3时,f"(x)>0,故曲线y=f(x)的拐点为(一3,一2e-3).
转载请注明原文地址:https://kaotiyun.com/show/Lcv4777K
0

随机试题
最新回复(0)