\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}