# What are the local extrema of f(x) = 2 x + 3 /x?

Nov 28, 2015

The local extrema are $- 2 \sqrt{6}$ at $x = - \sqrt{\frac{3}{2}}$
and $2 \sqrt{6}$ at $x = \sqrt{\frac{3}{2}}$

#### Explanation:

Local extrema are located at points where the first derivative of a function evaluate to $0$. Thus, to find them, we will first find the derivative $f ' \left(x\right)$ and then solve for $f ' \left(x\right) = 0$.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(2 x + \frac{3}{x}\right) = \left(\frac{d}{\mathrm{dx}} 2 x\right) + \frac{d}{\mathrm{dx}} \left(\frac{3}{x}\right) = 2 - \frac{3}{x} ^ 2$

Next, solving for $f ' \left(x\right) = 0$

$2 - \frac{3}{x} ^ 2 = 0$

$\implies {x}^{2} = \frac{3}{2}$

$\implies x = \pm \sqrt{\frac{3}{2}}$

Thus, evaluating the original function at those points, we get

$- 2 \sqrt{6}$ as a local maximum at $x = - \sqrt{\frac{3}{2}}$
and
$2 \sqrt{6}$ as a local minimum at $x = \sqrt{\frac{3}{2}}$