Answer :
The critical points of the given function are p = 2 ± √13.
What are functions?
A relation between a collection of inputs and outputs is known as a function.
A function is, to put it simply, a relationship between inputs in which each input is connected to precisely one output.
Each function has a range, codomain, and domain.
Critical function:
The key points are locations inside the function's domain where the function's nature changes.
The function stops increasing or decreasing at these places.
The function is: [tex]h(p)=\frac{p-2}{p^2+9}[/tex]
The derivative test will be applied here:
[tex]\begin{aligned}& \therefore h^{\prime}(p)=0 \\& \Rightarrow \frac{d}{d p}\left(\frac{p-2}{p^2+9}\right)=0 \\& \Rightarrow \frac{\frac{d}{d p}(p-2)\left(p^2+9\right)-\frac{d}{d p}\left(p^2+9\right)(p-2)}{\left(p^2+9\right)^2}=0 \\& \Rightarrow \frac{1 \cdot\left(p^2+9\right)-2 p(p-2)}{\left(p^2+9\right)^2}=0 \\& \Rightarrow \frac{-p^2+4 p+9}{\left(p^2+9\right)^2}=0 \\& \Rightarrow-p^2+4 p+9=0\end{aligned}[/tex]
Then,
[tex]\begin{aligned}& \Rightarrow p_{1,2}=\frac{-4 \pm \sqrt{4^2-4(-1) 9}}{2(-1)} \\& \Rightarrow p_{1,2}=\frac{-4 \pm \sqrt{52}}{-2} \\& \Rightarrow p_{1,2}=-\frac{-4 \pm \sqrt{52}}{2} \\& \Rightarrow p_{1,2}=-\frac{-4 \pm 2 \sqrt{13}}{2} \\& \Rightarrow p_{1,2}=2 \pm \sqrt{13}\end{aligned}[/tex]
Therefore, the critical points of the given function are p = 2 ± √13.
Know more about functions here:
https://brainly.com/question/29523308
#SPJ4