Question

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

Answer 1

There is none.

Explanation:

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

`y=sqrt[x]/(x-5)`

Using the quotient rule.

`dy/dx=(1/(2sqrt(x))(x-5)-sqrt(x))/(x-5)^2`

We now need to find the roots of this equation:

`(1/(2sqrt(x))(x-5)-sqrt(x))/(x-5)^2=0`

`implies 1/(2sqrt(x))(x-5)-sqrt(x)=0`

`implies 1/(2sqrt(x))(x-5)=sqrt(x)`

`implies x-5=2x`

`implies x = -5`

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

`y=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.