\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}