SOLUTION: Good day, I would like help with finding the intervals of increase and decrease of this function and its local minimum and maximum if it has them. Since it is a rational function I

Algebra ->  Rational-functions -> SOLUTION: Good day, I would like help with finding the intervals of increase and decrease of this function and its local minimum and maximum if it has them. Since it is a rational function I      Log On


   

Question 810983: Good day, I would like help with finding the intervals of increase and decrease of this function and its local minimum and maximum if it has them. Since it is a rational function I used the quotient rule to derive it but I don't think I'm getting it right. It is y = x/(x-1)^2.
Thank you.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I see two ways to calculate the derivative, and I like that,
because I know I make mistakes, so if both answers agree I'm reassured,
and if they don not agree, I try to find and fix errors.
(I will not say how it went this time).
This is what I ended up with:

Using the quotient rule


The other way:
.
So y=%28x-1%29%5E%28-1%29%2B%28x-1%29%5E%28-2%29 is another expression for the function,
and it could be useful for derivative calculation and more.
Now I will calculate the derivative of y=%28x-1%29%5E%28-1%29%2B%28x-1%29%5E%28-2%29


With both answers in agreement, now I am pretty sure that
dy%2Fdx=-%28x%2B1%29%2F%28x-1%29%5E3 is the derivative.
dy%2Fdx=-%28x%2B1%29%2F%28x-1%29%5E3 changes sign at x=1 and at x=-1 .
For x=0, where y=0 ,
dy%2Fdx=-1%2F-1=1 so dy%2Fdx%3E0 and the function is increasing at that point,
and at all points with -1%3Cx%3C1 .
The denominator of dy%2Fdx=-%28x%2B1%29%2F%28x-1%29%5E3 changes sign at x=1 ,
which we know is a vertical asymptote,
so dy%2Fdx changes sign at x=1 .
For x%3E1 dy%2Fdx%3C0 and the function starts decreasing after x=1 , (but the function does not change sign there).
The numerator of dy%2Fdx=-%28x%2B1%29%2F%28x-1%29%5E3 changes sign at x=-1 ,
so for x%3C-1 dy%2Fdx%3C0 and the function is decreasing.
Since the function decreases for x%3C-1 ,
increases for -1%3Cx%3C1 ,
and for x=-1 the function exists, with y=-1%2F4 ,
the function has a local minimum at x=-1 .
graph%28300%2C300%2C-10%2C10%2C-1%2C5%2Cx%2F%28x-1%29%5E2%29 , and zooming in graph%28300%2C300%2C-5%2C5%2C-0.3%2C0.7%2Cx%2F%28x-1%29%5E2%29