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 $p =1 - \sqrt{5}$.