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

  \begin{exampleblock}{}
    Prove that the equation 
    \begin{talign}
      2\cos(\frac{x\pi}{2}) = 2^x 
    \end{talign}
    has a solution for $x$ on the interval $[0,1]$.
    \pause\medskip
    
    Define $f(x) = 2\cos(\frac{x\pi}{2}) - 2^x$:
    \begin{talign}
      2\cos(\frac{x\pi}{2}) = 2^x &&\iff&&
      f(x) = 0
    \end{talign}
    \pause
    We have:
    \begin{itemize}
    \pause
      \item $f(x)$ is defined on $[0,1]$
    \pause
      \item $f(x)$ is continuous on its domain since it is a composition,
        product, division and multiplication of continuous functions
    \pause
      \item $f(0) = \pause 1$ \pause and $f(1) = \pause -2$
    \end{itemize}
    \pause
    Since $0$ is between $-2$ and $1$, by the Intermediate Value Theorem, \pause
    there exists $c$ in $(0,1)$ such that $f(c) = 0$. \pause
    This $c$ is a solution of the equation.
  \end{exampleblock}
\end{frame}