\begin{frame}
  \frametitle{2nd Midterm Exam - Review}

  \begin{exampleblock}{}
    Is there a line that is tangent to both curves $f$ and $g$?\vspace{-1.2ex}
    \begin{align*}
      f(x) = x^2 && g(x) = x^2 - 2x + 2
    \end{align*}\vspace{-3.5ex}
    \pause
    
    We compute the tangent to $f$ at $(a,f(a))$:\vspace{-1.2ex}
    \begin{align*}
      f'(a) = 2a &\mpause[1]{\quad\implies\quad y-a^2 = 2a(x-a)}\\
      &\mpause[2]{\quad\implies\quad y = (2a)x + (- a^2)}
    \end{align*}\vspace{-3.5ex}
    \pause\pause\pause
    
    We compute the tangent to $g$ at $(b,g(b))$:\vspace{-1.2ex}
    \begin{align*}
      g'(b) = 2b-2 &\mpause[1]{\quad\implies\quad  y-(b^2-2b+2) = (2b-2)(x-b)} \\
      &\mpause[2]{\quad\implies\quad  y = (2b - 2)x + (- b^2 + 2)}
    \end{align*}\vspace{-3.5ex}
    \pause\pause\pause
    
    The tangents are equal if:\vspace{-1.2ex}
    \begin{align*}
      2a &= 2b - 2 \mpause[1]{\quad\implies\quad a = b-1 \mpause[5]{\;\implies\; \alert{a = 1/2}}}\\
      -a^2 &= -b^2 + 2 \mpause[2]{\quad\implies\quad -(b-1)^2 = -b^2 + 2}\\
      &\mpause[3]{\implies\; -(b^2-2b+1) = -b^2 + 2}
      \mpause[4]{\;\implies\; \alert{b = 3/2}}
    \end{align*}\vspace{-3.5ex}
    \pause\pause\pause\pause\pause\pause
    
    Thus the line $y = x - 1/4$ is tangent to both curves.
  \end{exampleblock}
\end{frame}