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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 $s$ is...help!!

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    **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 $a$.2011-10-08

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$f(s)$ 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 $3c+8=9-c\Rightarrow c=\frac{1}{4}$.