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For what value of the constant $c$ is the function $f$ continuous on $(-\infty,\infty)$ where $$f(s) =\cases{s^2-c &\text{if }s\in(-\infty, 3)\\ cs+8 &\text{if } s \in [3, \infty) }$$

I'm not sure what to do here....can someone tell me how to set this up? It's asking for the value of $c$. All I can think of is to set $s^2-c = cs+8$ but I don't know what $s$!!

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Hint: No matter what $c$ is, the function is automatically continuous for every $s\ne 3$. – Henning Makholm Oct 8 '11 at 3:41
As $s$ approaches $3$ from the left, $f(s)$ approaches $9-c$. Thus we must have $9-c=?$. – André Nicolas Oct 8 '11 at 3:44
$\textstyle \lim_{s \to 3}(s^2-c)=3c+8$ – pedja Oct 8 '11 at 4:08
Hint A function $f$ is continuous at a point $a$ if $\lim_{x\to a}f(x) =f(a)$. So you wish this to be the case for every real $a$. – AD. Oct 8 '11 at 4:33

$f(s)$ is continuous at a point $s=a$, iff $$f(a-)=f(a)=f(a+)$$ where $f(a-)=\lim_{s\to a-} f(s)$ etc. and all the limits exist. In your case, you need to check the continuity at $s=3$ (because the function being polynomial, is continuous elsewhere).

Now, from the definition of $f(s)$, $f(3)=f(3+)=3c+8$ and $$f(3-)=\lim_{s\to3-}f(s)=\lim_{s\to3} (s^2-c)=9-c$$ Therefore, we must have $3c+8=9-c\Rightarrow c=\frac{1}{4}$.

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