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\begin{frame}
\frametitle{1st Midterm Exam - Review}

\begin{exampleblock}{}
For what value of $k$ is the following function continuous?
\begin{talign}
f(x) =
\begin{cases}
x^2 + 2k &\text{for $x < 2$}\\
3^x - k &\text{for $x \ge 2$}
\end{cases}
\end{talign}
\pause
For any $k$, the function is continuous at all $x \ne 2$ since
\begin{itemize}
\item $x^2 + 2k$ is continuous, and
\item $3^x - k$ is continuous.
\end{itemize}
(Both are compositions of continuous functions)
\pause\medskip

At point $x = 2$ we have:
\begin{talign}
\lim_{x\to2^-} f(x) &=  \mpause[1]{\lim_{x\to2^-} x^2 + 2k =}\mpause[2]{ 4 + 2k}\\
\lim_{x\to2^+} f(x) &=  \mpause[3]{\lim_{x\to2^+} 3^x - k =}\mpause[4]{ 9 - k}\\
f(2) &= \mpause[5]{3^2 - k =}\mpause[6]{ 9 - k}
\end{talign}
\pause[10]
We have continuity at $2$ if $4 + 2k = 9 - k$. \pause Thus $k = \frac{5}{3}$.
\end{exampleblock}
\end{frame}