Answer
$c=1-\sqrt5$ and $c=\sqrt5+1$ are critical numbers of $h$.
Work Step by Step
How to find the critical numbers of a function $f$ according to definition
1) Find all numbers of $c$ satisfying either $f'(c)=0$ or $f'(c)$ does not exist.
2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not.
- If $c_0$ lies in $D_f$, $c_0$ is a critical number of $f$.
- If $c_0$ does not lie in $D_f$, $c_0$ is not a critical number of $f$.
$$h(p)=\frac{p-1}{p^2+4}$$
We see that equation $p^2+4\neq0$ for all values of $p$, for $p^2+4\gt0$ for all $p$, so the denominator here cannot be $0$. Therefore, $$D_h=R.$$
1) Now, we find $h'(p)$
$$h'(p)=\frac{1(p^2+4)-(p-1)2p}{(p^2+4)^2}$$
$$h'(p)=\frac{p^2+4-2p^2+2p}{(p^2+4)^2}$$
$$h'(p)=\frac{-p^2+2p+4}{(p^2+4)^2}$$
We find $h'(c)=0$, $$h'(c)=0$$
$$-c^2+2c+4=0$$
$$c=1-\sqrt5\hspace{0.5cm}or\hspace{0.5cm}c=\sqrt5+1$$
There are no values of $c$ that makes $h'(c)$ not exist, for $(c^2+4)^2$ cannot equal $0$.
2) Examine whether $c$ lies in $D_h$ or not.
We see that both $c=1-\sqrt5$ and $c=\sqrt5+1\in R$, so both lie in $D_h$.
We conclude that $c=1-\sqrt5$ and $c=\sqrt5+1$ are critical numbers of $h$.
