\begin{frame}
  \frametitle{Continuity: Intermediate Value Theorem}
  
  \begin{exampleblock}{}
    Show that the following equation
    \begin{talign}
      6 \cdot 3^{-x} = 4 - x
    \end{talign}
    has a solution for $x$ in $[0,1]$.
    \pause\medskip
    
    Define
    \begin{talign}
      6 \cdot 3^{-x} = 4 - x \quad\iff\quad 6 \cdot 3^{-x} + x - 4 = 0
    \end{talign}
    \pause
    The function $f(x) = 6 \cdot 3^{-x} + x - 4$ is a sum and product of continuous functions, 
    and hence continuous.
    \pause\medskip
    
    We have:
    \begin{itemize}
    \pause
      \item $f(0) = \pause 6\cdot 3^0 + 0 - 4 = 2$
    \pause
      \item $f(1) = \pause 6\cdot 3^{-1} + 1 - 4 = -1$
    \end{itemize}   
    \pause
        
    Moreover $-1 < 0 < 2$.
    \pause\medskip

    By the Intermediate Value Theorem
    there exists $x$ in the interval $[0,1]$
    such that $f(x) = 0$.
    This $x$ is a solution of the equation.
  \end{exampleblock}
\end{frame}