How do you find the critical numbers for h(p) = (p - 2)/(p^2 + 3) to determine the maximum and minimum?

Dec 4, 2016

The critical points are:

${p}_{1} = 2 - \sqrt{7}$ that is a minimum.
${p}_{2} = 2 + \sqrt{7}$ that is a maximum.

Explanation:

The critical points of a function are the points for which its derivative is null.

$\frac{\mathrm{dh}}{\mathrm{dp}} = \frac{\left({p}^{2} + 3\right) - 2 p \left(p - 2\right)}{{\left({p}^{2} + 3\right)}^{2}} = - \frac{{p}^{2} - 4 p - 3}{{\left({p}^{2} + 3\right)}^{2}}$

The denominator is strictly positive for every $p \in \mathbb{R}$, so we can focus on the numerator:

${p}^{2} - 4 p - 3 = 0$

$p = 2 \pm \sqrt{4 + 3} = 2 \pm \sqrt{7}$

To determine whether this point are maximums or minimums we could calculate the second derivative, but as it is easy to see the sign of $\frac{\mathrm{dh}}{\mathrm{dp}}$ we can proceed this way:

1) Around ${p}_{1} = 2 - \sqrt{7}$ we have:

$\frac{\mathrm{dh}}{\mathrm{dp}} < 0$ for $p < {p}_{1}$

$\frac{\mathrm{dh}}{\mathrm{dp}} > 0$ for $p > {p}_{1}$

so $h \left(p\right)$ decreases until ${p}_{1}$ then increases, so that $2 - \sqrt{7}$ is a minimum.

2) Around ${p}_{2} = 2 + \sqrt{7}$ we have:

$\frac{\mathrm{dh}}{\mathrm{dp}} > 0$ for $p < {p}_{2}$

$\frac{\mathrm{dh}}{\mathrm{dp}} < 0$ for $p > {p}_{2}$

so $h \left(p\right)$ increases until ${p}_{2}$ then decreases, so that $2 + \sqrt{7}$ is a maximum.

graph{(x-2)/(x^2+3) [-10, 10, -1, 1]}