# Find two parameters so that a function is continuous at a given point

Let

$$f(x) = \begin{cases} 3x & x < 3 \\ a & x = 3 \\ x^2 + b & x > 3 \end{cases}$$

Find $a, b$ so that $f(x)$ is continuous at $x=3$, then prove that $f(x)$ is continuous at $x=3$.

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What have you tried? –  Chris Eagle Oct 3 '12 at 17:19

$f(x)$ is continuous at $x=3$ if and only if $\lim_{x\rightarrow3^{-}}f(x)=\lim_{x\rightarrow3^{+}}f(x)=f(3)$, i.e. $\lim_{x\rightarrow3^{-}}3x=\lim_{x\rightarrow3^{+}}x^{2}+b=a$, so $9=9+b=a$, it means that $a=9$ and $b=0$.

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how did you get to define a=9 –  carolina nunez Oct 3 '12 at 17:56
@carolina nunez: I think it's obvious. Just observed and be more careful. –  Alfred Chern Oct 3 '12 at 18:08

The critical point is $x=3$. At this point you want $a = 3x = x^2 + b$ for $f$ to be continuous.

If you draw it, say for $x \in [0,5]$, you will immediately see which value $a = f(3)$ has to be in order for $f$ to be continuous. You really want to draw a picture.

Hope this helps.

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why is a= 3x if it says the function is a if x=3 –  carolina nunez Oct 3 '12 at 17:48
@carolinanunez First your function is $3x$ then it is $a$ and then it is $x^2 + b$. It is $a$ where $x$ is $3$. For it to be continuous you want $3x = a = x^2 + b$ (try to draw the function for $x \in [0,5]$) –  Rudy the Reindeer Oct 3 '12 at 17:52
@carolinanunez When you draw it you will immediately see which value you want $a = f(3)$ to be so that $f$ is continuous. –  Rudy the Reindeer Oct 3 '12 at 17:53
ugh, i feel like my prof is doing a poor job. I dont even know how to graph it. –  carolina nunez Oct 4 '12 at 0:59