设函数f(x)={x2x<10x=12−xx>1f(x)=\begin{cases}x^{2}&x<1\\0&x = 1\\2 - x&x>1\end{cases}f(x)=⎩⎨⎧x202−xx<1x=1x>1,则f(x)f(x)f(x)的( )间断点.
(A)无穷
(B)振荡
(C)跳跃
(D)可去
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