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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: No matter what $c$ is, the function is automatically continuous for every $s\ne 3$.2011-10-08
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    As $s$ approaches $3$ from the left, $f(s)$ approaches $9-c$. Thus we must have $9-c=?$.2011-10-08
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    $\textstyle \lim_{s \to 3}(s^2-c)=3c+8$2011-10-08
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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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