\begin{frame}
  \frametitle{Continuity: Intermediate Value Theorem}
  
  \begin{exampleblock}{}
    Show that there is a root of the equation 
    \begin{talign}
      4x^3 - 6x^2 + 3x - 2 = 0
    \end{talign}
    between $1$ and $2$.
    \pause\medskip
    
    We are looking for number $c$ such that $f(c) = 0$ and $1 < c < 2$.
    \pause\medskip
    
    We have:\\
    \begin{itemize}
    \pause
      \item the function is continuous on the interval since it is a polynomial
    \pause
      \item $f(1) = 4 - 6 + 3 - 2 = -1$
    \pause
      \item $f(2) = 4\cdot 8 - 6\cdot 4 + 3\cdot 2 - 2 = 12$
    \end{itemize}
    \pause
    Moreover $-1 < 0 < 12$. \pause
    Thus we can apply the Intermediate Value Theorem for the interval $[1,2]$
    and $N = 0$.
    \pause\medskip
    
    Hence there exists $c$ in $(1,2)$ such that $f(c) = 0$.
  \end{exampleblock}
\end{frame}