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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=1739380385-axAPp8wkN7hUsqYQSnjwD6yxLJPkJ8gq-0-94ec1e86b2b5236212475546c220c588)
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=1739380385-rxxF4zRJkV9er99nS00YATSvVQboys6k-0-71f338b4c32c9bbad8c197371a2a461a)
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=1739380385-qRH710uWGTotaelQpJL7wGcLAL9BRdiY-0-ac83aeebfda6f51fa801ba4b07b871bc)
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=1739380385-KL0OFZ7eKEjxfyBRd0UrRJ92mUcvFHBI-0-094e23b2081aecee635bab4690ed091a)
在x=1处的可导性与连续性.
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00075003.jpg?sign=1739380385-f3QJB8ZzsTEtkzp7QbEyNpxvTDKuEWeI-0-2870fcbc5b766409890b99c6d619406c)
所以f(x)在x=1连续.又因
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00075004.jpg?sign=1739380385-Xdo4XtSgqSXw3F8PgTdzFkzrVHpHHCg9-0-7004678cbd077b2c807ffcb68bef830f)
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=1739380385-Ze2FAOls2lnGBBoiUMfnwF5x2BvGcCyK-0-3db17eeb6344abb50304428e3ff66798)
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=1739380385-Q26fniGJrToPdjaj1ej8PzrblVT84OtL-0-977f9ecf3f6cddb0a75be7d295f798c4)
![](https://epubservercos.yuewen.com/578B36/14615859905722306/epubprivate/OEBPS/Images/img00076001.jpg?sign=1739380385-ah1qeIFJafBEd1s3mYHtXDq6W1P6Fkhk-0-927d6704f611cda9e6a25167bee39179)
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.