h(x) = \begin{cases} x^{2} & \text{if $x\le5$} \\ x+k & \text{if $x>\ 5$} \\ \end{cases}
Answer choices are A. k=20 B.k=-5 C. k=5 D. k=30
$\endgroup$ 22 Answers
$\begingroup$$$\lim_{x\to 5-}x^2=\lim_{x\to 5+}x+k$$$$5^2=5+k$$$$k=20$$
$\endgroup$ 0 $\begingroup$Note that a function $f(x)$ is continuous if $\lim_{x\to a}f(x)=f(a)$, means that the global limit as $x$ approches $a$ exists and it's exactly equal to $f(a)$.
The two one-sided limits exist and are equal if the limit of $f(x)$ as $x$ approaches $a$ exists.
So our condition is: $$\lim_{x\to 5+}h(x)=\lim_{x\to 5-}h(x)=h(5)$$
then it's easy to get the solution as: $$\lim_{x\to 5+}x+k=\lim_{x\to 5-}x^2=5^2$$
so $$k=20$$
and you can redefine your continous function in $R$:
$$h(x) = \begin{cases} x^{2} & \text{if $x\le5$} \\ x+20 & \text{if $x>\ 5$} \\ \end{cases}$$
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