Practicing Success
Practicing Success
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If the function $f(x)=\left\{\begin{matrix}\frac{x^3-(k+4)x+2k}{x-3},&x≠3\\8,&x=3\end{matrix}\right.$ is continuous at x = 3, then k is: |
3 -3 -15 15 |
15 |
$\underset{x→3}{\lim}\frac{x^3-(k+4)x+2k}{x-3}=\underset{x→3}{\lim}\frac{3x^2-(k+4)}{1}=23-k$ ⇒ 23 - k = 8 ⇒ k = 15 |
