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2.1.3 函数可导与连续的关系
函数y=f(x)在点x0处连续是指
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00074005.jpg?sign=1739605505-oTnQP7dgEPxd1zLAcSTOeJArY82x5ua3-0-f0cf53f81e4fb0c2a6fb5d2aa6b232ce)
函数y=f(x)在点x0处可导是指
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00074006.jpg?sign=1739605505-bCch7tMo4QMnfORvxK3PFu7kOCGUXXGj-0-5fe667c35ad68aa561fc36e0e182c901)
那么,它们之间有什么关系呢?
设函数y=f(x)在点x0处可导,即存在.由具有极限的函数与无穷小的关系知道
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00075001.jpg?sign=1739605505-Vd7cIYnJKun554zmF7dcVOaFR17VWQdJ-0-eaf54e91d2973ecfd22d273f44d6e295)
其中,当Δx→0时,α为无穷小.上式两边同乘以Δx,得
Δy=f′(x0)Δx+αΔx.
由此可见,当Δx→0时,Δy→0.这就是说,函数y=f(x)在点x0处是连续的.
所以,如果函数y=f(x)在点x0处可导,则函数在该点必连续.反之,一个函数在某点连续却不一定在该点可导.
例11 讨论函数
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00075002.jpg?sign=1739605505-p5p8IOlXJAp0ky0RfO6qWiRGqWxSr7rU-0-ee4fbb3fc2fb59c444dcc901673761bf)
在x=1处的可导性与连续性.
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00075003.jpg?sign=1739605505-Torf8Wz8WTdyBf8TdeKIJqhd6u9pftFI-0-7c97ee888e3b38784d67519d67d80e53)
所以f(x)在x=1连续.又因
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00075004.jpg?sign=1739605505-2HkJXgMuWtCvahJPaUH6lpEymF9LToMu-0-22278343e439b4af826e95010635a9bd)
故f(x)在x=1处不可导.
定理 如果函数y=f(x)在点x0处可导,则函数y=f(x)在点x0处连续;反之不真.
例如,函数f(x)=|x|在x=0处连续但不可导.
因此,函数在某点处连续是在该点可导的必要条件,但不是充分条件.
例12 a,b为何值时,函数
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00075005.jpg?sign=1739605505-JaKXFCi2orGv0YoJrxJlm2k5rCgs7psK-0-ec719c76590ef9465b31f8a7c68a104a)
在x=1处可导.
解 ,f(1)=1,由于f(x)在x=1处可导,所以f(x)在x=1处连续,故而a-b=1.
又因为
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00075007.jpg?sign=1739605505-Exj6p7aEOXmc6ySN31mh2nt4OEQ6FC9J-0-0c2d55bd8fa015696a76c5d7dce3d7cb)
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00076001.jpg?sign=1739605505-evwLfrcWCuT6VOwexGB0uOo8cEaZ76ZN-0-45a0994d61813fe730df20919ae6a606)
所以a=-1.
将a=-1代入a-b=1中,解得b=-2.
故当a=-1,b=-2时,函数f(x)在x=1处可导.