Question

Determine the values of C that makes the following function continuous

F(x)=3x+c, when x is less than or equal to 2
F(x)=C^2-2x+4, when x is greater than 2
..I got it down to 6-c=c^2 but I don't know how to simplify that.

Answer 1

Skip the first part (until you see red) if you don't want to read it; I'm just recreating what your work probably looks like.
\[f(x)=\begin{cases}3x+c&\text{if $x\le2$}\\
c^2-2x+4&\text{if $x>2$}\end{cases}\]
For f to be continuous at x = 2, the following must be satisfied:
\[\lim_{x\to2^+}f(x)=\lim_{x\to2^-}f(x)\]
The one-sided limits are as follows:
\[\begin{align*}\lim_{x\to2^+}f(x)&=\lim_{x\to2^+}\left(c^2-2x+4\right)\\
&=c^2-4+4\\
&=c^2\end{align*}\]
and
\[\begin{align*}\lim_{x\to2^-}f(x)&=\lim_{x\to2^-}\left(3x+c\right)\\
&=\color{red}{6+c}\\\end{align*}\]
So you made a small mistake when you evaluated the second limit. Now, setting the limits equal to each other,
\[c^2=6+c\\
c^2-c-6=0\]
Have you tried the quadratic formula?