# How do you find the local extrema for f(x) = x sqrt( x - 3 )?

Oct 21, 2016

There is only a minimum at $\left(3 , 0\right)$

#### Explanation:

To find the domain of the function, $\sqrt{x - 3}$
$x - 3 \ge 0$ $\implies$ $x \ge 3$
If $f ' \left(x\right) = 1 \cdot \sqrt{x - 3} + \frac{x}{2 \sqrt{x - 3}} = \frac{2 \left(x - 3\right) + x}{2 \sqrt{x - 3}}$
$= \frac{3 x - 6}{2 \sqrt{x - 3}}$
$f ' \left(x\right) = 0$ for $x = 2$

This is incomposible since x>=3 This is easy to understand from the graph graph{x*sqrt(x-3) [-20, 20, -10, 10]}