Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 36

Determine the critical numbers of the function $\displaystyle h(p) = \frac{p - 1}{p^2 + 4}$


$
\begin{equation}
\begin{aligned}

h'(p) =& \frac{d}{dp} \left( \frac{p - 1}{p^2 + 4} \right)
\\
\\
h'(p) =& \frac{\displaystyle (p^2 + 4) \frac{d}{dp} (p - 1) - (p-1) \frac{d}{dp} (p^2 + 4)}{(p^2 + 4)^2}
\\
\\
h'(p) =& \frac{(p^2 + 4) (1) - (p -1)(2p) }{(p^2 + 4)^2}
\\
\\
h'(p) =& \frac{p^2 + 4 - (2p^2 - 2p)}{(p^2 + 4)^2}
\\
\\
h'(p) =& \frac{p^2 + 4 - 2p^2 + 2p}{(p^2 + 4)^2}
\\
\\
h'(p) =& \frac{-p^2 + 2p + 4}{(p^2 + 4)^2}

\end{aligned}
\end{equation}
$


Solving for critical numbers


$
\begin{equation}
\begin{aligned}

& h'(p) = 0
\\
\\
& 0 = \frac{-p^2 + 2p + 4}{(p^2 + 4)^2}
\\
\\
& (p^2 + 4)^2 \left[ 0 = \frac{-p^2 + 2p + 4}{\cancel{(p^2 + 4)^2}} \right] \cancel{(p^2 + 4)^2}
\\
\\
& 0 = -p^2 + 2p + 4
\\
\\
& \text{ or }
\\
\\
& p^2 - 2p - 4 = 0

\end{aligned}
\end{equation}
$


Using Quadratic Equation


$
\begin{equation}
\begin{aligned}

p =& \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\\
\\
p =& \frac{-(-2) \pm \sqrt{(-2)^2 - (4)(1)(-4)} }{2(1)}
\\
\\
p =& \frac{2 \pm \sqrt{4 + 16}}{2}
\\
\\
p =& \frac{2 \pm \sqrt{20}}{2}
\\
\\
p =& \frac{2 \pm \sqrt{(4)(5)}}{2}
\\
\\
p =& \frac{2 \pm 2 \sqrt{5}}{2}
\\
\\
p =& \frac{\cancel{2} (1 \pm \sqrt{5})}{\cancel{2}}
\\
\\
p =& 1 \pm \sqrt{5}

\end{aligned}
\end{equation}
$


Therefore, the critical numbers are $p = 1 + \sqrt{5}$ and $p =1 - \sqrt{5}$.