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

1 Answer
Jan 27, 2018

Answer:

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.