设函数f(x)={ex−1x,x<0k,x=05x+1,x>0f(x)=\begin{cases} \dfrac{e^{x}-1}{x}, & x < 0\\ k, & x = 0\\ 5x + 1, & x > 0 \end{cases}f(x)=⎩⎨⎧xex−1,k,5x+1,x<0x=0x>0,若函数f(x)f(x)f(x)在其定义域内连续,求kkk的值.
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