How do you find the critical points for  x+2x^-1 and the local max and min?

Apr 7, 2018

Critical pointss are at $\left(- \sqrt{2} , - 2 \sqrt{2}\right)$ and $\left(\sqrt{2} , 2 \sqrt{2}\right)$. At $\left(- \sqrt{2} , - 2 \sqrt{2}\right)$ we have a local maxima and at $\left(\sqrt{2} , 2 \sqrt{2}\right)$ we have a local minima.

Explanation:

Critical points for a function $y = f \left(x\right)$ appear where $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$. At these points we have local maxima and minima too. The maxima is when $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} < 0$ and minima is when $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} > 0$.

As $y = x + 2 {x}^{- 1}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{2}{x} ^ 2$ and $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{6}{x} ^ 3$

$\frac{\mathrm{dy}}{\mathrm{dx}}$ is zero when $1 - \frac{2}{x} ^ 2 = 0$

or ${x}^{2} - 2 = 0$ i.e. $x = \pm \sqrt{2}$, hence we have critical points at $x = - \sqrt{2}$ (where $f \left(x\right) = - \sqrt{2} - \frac{2}{\sqrt{2}} = - 2 \sqrt{2}$) and at $x = \sqrt{2}$ (where $f \left(x\right) = \sqrt{2} + \frac{2}{\sqrt{2}} = 2 \sqrt{2}$) and hence coordinates are $\left(- \sqrt{2} , - 2 \sqrt{2}\right)$ and $\left(\sqrt{2} , 2 \sqrt{2}\right)$.

At $x = - \sqrt{2}$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{6}{2 \sqrt{2}} < 0$ hence we have a local maxima

and at $x = \sqrt{2}$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{6}{2 \sqrt{2}} > 0$ hence we have a local minima.

graph{x+2x^(-1) [-10, 10, -5, 5]}