# Rational Functions

1. May 17, 2010

### zebra1707

Hi there

I have a Rational function y = 1 / x^2-1 . I have a good idea what the graph looks it, it will have vertical asymptotes at -1 and 1 and I can work out the y intercept (-1 concave down). However I'm not sure about the other parts to the question.

1. The problem statement, all variables and given/known data

dy/dx = - 2x/(x-1)^2(x+1)^2

The other parts of the question are as follows:

b) Find any turning points and determine their nature
c) For what values of x is the curve discontinous
d) How does the curve behave when x(arrow)(infinity) and when x(arrow)(-infinity)
e) Show f(x) = f(-x)
f) Find f(2) and f(-2)

2. Relevant equations

3. The attempt at a solution

Needing some guidence about where to start and setting out.

Cheers P

2. May 17, 2010

### Cyosis

Which of these are causing problems?

3. May 17, 2010

### zebra1707

Hi there

b) to f) is causing the problem

Many thanks P

4. May 17, 2010

### The Chaz

b) Set the derivative = 0
c) undefined => discontinuous
e) i.e. "show that this is an even function" (google "even function")

Have you tried part f)????? It seems strange that you could take the derivative of a function (using chain/quotient rule) and NOT know how to plug in 2.

5. May 18, 2010

### Cyosis

It's time for you to show some work. It is hard to believe that all of these give you trouble.

6. May 18, 2010

### zebra1707

Part b)

I use the f '(x) to find the turning point and determine their nature.

I think that the second derivative is - (2x/(x^2-1) or - (2x/(x-1)^2(x+1)2

Therefore the stationary points where dy/dx = 0 ie x = +1 and -1

Therefore when x = +1 y + 0 and when x = -1 y = 0

Therefore (1,0) and (0,-1) are the stationary points ( this does not look right)

To determine the Nature of these stationary points we examine the sign of the gradient function which is - (2x/(x^2-1)

which we have seen has a zero value when x = -1 or x = 1

Or should I use the second derivative??

Im getting bogged down and confused.

7. May 18, 2010

### The Chaz

I have never used the exact phrase "turning points" in this context, but it probably means "relative extrema" (i.e. relative max/min).
These can occur only where the derivative ("gradient") is zero. You have found the derivative in your original post. It is a fraction. A fraction is zero when its NUMERATOR is zero. Should be pretty straightforward.

Yes, there is a second derivative test, but don't do that. The critical x-value will divide the x-axis into two parts. Test points on both intervals to see if the derivative is postive or negative there.

For example: g(x) = x^2 -2x + 348923487345
g(x) = 2x - 2.
Equate to zero...
0 = 2x -2
x = 1 is the critical point.
Test a number greater than 1, say 2. g(2) = 2 > 0 so g increases on (1, inf).
Test a number less than 1, say 0. g`(0) = -2 <0 so g decreases on (-inf, 1).
At x = 1, g goes from decreasing to increasing, so this is a relative MINIMUM.
Still surprised that you wouldn't know this... ???

c)
d)
e)
f)

8. May 18, 2010

### zebra1707

Im finding the rational functions diffcult for some reason, sorry just not getting it at the moment.

Many thanks, I will try and nut it myself. Cheers P

9. May 18, 2010

### The Chaz

c) (dis)continuity is intimately linked with the DOMAIN of the function. And there are relatively few restrictions on the domain of real-valued functions (can't divide by zero, can't take an even root of a negative number, can't take a log of a non-positive argument, etc)

d) - f) I just did a google search for these terms, and reiterate my suggestion that you do the same.