# How do you find the local extrema for y=sqrtx/(x-5)?

Jan 27, 2018

There is none.

#### Explanation:

First find the derivative and set the derivative equal to 0:

$y = \frac{\sqrt{x}}{x - 5}$

Using the quotient rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{1}{2 \sqrt{x}} \left(x - 5\right) - \sqrt{x}}{x - 5} ^ 2$

We now need to find the roots of this equation:

$\frac{\frac{1}{2 \sqrt{x}} \left(x - 5\right) - \sqrt{x}}{x - 5} ^ 2 = 0$

$\implies \frac{1}{2 \sqrt{x}} \left(x - 5\right) - \sqrt{x} = 0$

$\implies \frac{1}{2 \sqrt{x}} \left(x - 5\right) = \sqrt{x}$

$\implies x - 5 = 2 x$

$\implies x = - 5$

But there is a problem in finding the $y$ value:

$y = \frac{\sqrt{- 5}}{- 5 - 5}$

Notice that we would have to take the square root of a negative so we do not have a value for $y$. And as such there are no local extrema. Indeed if you plotted the curve for $y$ you would get:

graph{sqrt(x)/(x-5) [-10, 10, -5, 5]}

As a result of the above and a quick look at the graph we can see that there are no turning points and as such there are no local extrema.