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数学
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设f(x)=x+1,x0,试判断f(x)在x=0处的连续性和可导性
人气:268 ℃ 时间:2020-07-05 14:32:39
解答
当x0时,f+(0)=1,f′(x)=-1
应为f-(0)=f+(0)=f(0)
∴函数连续,函数不可导如果y=f(x)在(a,b)内可导并且在A+和B-处的导数都存在,则称y=f(x)在闭区间[a,b]上可导。 如果函数y=f(x)在点x处可导,则函数y=f(x)在点X处连续,反之,函数y=f(x)在点x处连续,但函数y=f(x)处不一定可导! 函数在点X处可导的充要条件是函数在点X处的左导数和右导数都存在并且相等。
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