# How do you find the local extremas for f(x)=2x + (5/x) ?

Apr 10, 2018

Local min is at ${p}_{\min} \left(\frac{\sqrt{10}}{2} | 2 \sqrt{10}\right)$ and local max is at ${p}_{\max} \left(- \frac{\sqrt{10}}{2} | - 2 \sqrt{10}\right)$

#### Explanation:

Let $f \left(x\right) = y$
$y = 2 x + \left(\frac{5}{x}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 - \left(\frac{5}{x} ^ 2\right)$

At the local extremas $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$2 - \frac{5}{x} ^ 2 = 0$

Solve for $x$

$2 - \frac{5}{x} ^ 2 = 0 | \cdot {x}^{2}$
2x^2-5=0|+5|:2
${x}^{2} = \frac{5}{2} | \sqrt{}$
$x = \pm \sqrt{\frac{5}{2}} = \pm \frac{\sqrt{10}}{2}$

${x}_{1} = \frac{\sqrt{10}}{2} \mathmr{and} {x}_{2} = - \frac{\sqrt{10}}{2}$

Testing to see which one is minimum and maximum we find the second derivative. If $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} > 0$, it is a local minimum. If $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} < 0$, it is a local maximum.

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{10}{x} ^ 3$
${x}_{2} = - \frac{\sqrt{10}}{2}$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{10}{\frac{\sqrt{10}}{2}} ^ 3 < 0$
Since it is less than $0$ the local maxima is at ${x}_{2} = - \frac{\sqrt{10}}{2}$

When $x = - \frac{\sqrt{10}}{2}$,$y = - 2 \sqrt{10}$.
Therefore local maxima is at ${p}_{\max} \left(- \frac{\sqrt{10}}{2} | - 2 \sqrt{10}\right)$

When $x = \frac{\sqrt{10}}{2}$,$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left(\frac{4}{5}\right) \sqrt{10} > 0$. Hence local minimum is at $x = \frac{\sqrt{10}}{2}$
When $x = \frac{\sqrt{10}}{2}$, $y = 2 \sqrt{10}$
Hence local minimum is at ${p}_{\min} \left(\frac{\sqrt{10}}{2} | 2 \sqrt{10}\right)$