Stationary points/Maxima and minima problem

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Apricity

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Hi all!
Can someone please help me out here, thanks very much in advance.

Determine the turning points of y=x +1/(x-1)

The 1st and 2nd derivatives are:

y'= 1 - 1/(x-1)^2

y"= 2/(x-1)^3

To find the turning points y'=0 Here is where I'm stuck. I don't know how to proceed from here.

y'=0

1- 1/(x-1)^2 = 0

How do I find the roots?
 
Could someone help me with this one. Thanks.

I need to find the roots of y= [2(x+3)/(x+1)^3] - [2/(x+1)^2] ...What should I do next? Thanks very much!!
 
The general idea on these is to clear the denominators (put the expression over a common denominator), so
y= [2(x+3)/(x+1)^3] - [2/(x+1)^2]=\(\displaystyle \frac{2(x+3)}{(x+1)^3} - \frac{2}{(x+1)^2} \frac{x+1}{x+1}\)
or
y = \(\displaystyle \frac{2(x+3)-2(x+1)}{(x+1)^3}=\frac{4}{(x+1)^3}\)

Now, depending on how you want to do it, you can note that y is zero only at x equal \(\displaystyle \pm\)∞

Or you can do a transformation, a (almost) one to one mapping which maps infinity to zero and -1 to infinity, i.e. let
t = \(\displaystyle \frac{1}{(x+1)}\)
so that
y = 4 t3
and there is a multiple root at t = 0 with multiplicity 3, i.e. x is \(\displaystyle \pm\)∞.
 
Ishuda said:
The general idea on these is to clear the denominators (put the expression over a common denominator), so
y= [2(x+3)/(x+1)^3] - [2/(x+1)^2]=\(\displaystyle \frac{2(x+3)}{(x+1)^3} - \frac{2}{(x+1)^2} \frac{x+1}{x+1}\)
or
y = \(\displaystyle \frac{2(x+3)-2(x+1)}{(x+1)^3}=\frac{4}{(x+1)^3}\)

Yes, I did this too :)

Now, depending on how you want to do it, you can note that y is zero only at x equal \(\displaystyle \pm\)∞

Or you can do a transformation, a (almost) one to one mapping which maps infinity to zero and -1 to infinity, i.e. let
t = \(\displaystyle \frac{1}{(x+1)}\)
so that
y = 4 t3
and there is a multiple root at t = 0 with multiplicity 3, i.e. x is \(\displaystyle \pm\)∞.
Click to expand...

Thanks very much Ishuda. What topic is this ( transformations?) and where can I learn more about it? Thanks again
 
Apricity said:
Thanks very much Ishuda. What topic is this ( transformations?) and where can I learn more about it? Thanks again
Click to expand...

A transformation is just a function, i.e. the function transforms the domain into the range.
 
Apricity said:
I must admit that I don't understand what you mean, but thanks for the help :)
Click to expand...

A transformation or mapping is just another way of saying function. You have the function
y(x) = \(\displaystyle \frac{1}{x - 1}\)

You could also call this a transformation or mapping from the reals minus the value x=1 to the reals minus the value y=0, defined by
y(x) = \(\displaystyle \frac{1}{x - 1}\)
That is, the domain of the function is the reals minus the value x=1 (you can't divide by zero) and the range of the function is the reals minus the value y=0 (there is no real number x for which y is zero).
 
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