Rational Functions – Homework Help

[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.

Homework Statement

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)

Homework Equations

The Attempt at a Solution

Needing some guidence about where to start and setting out.

Cheers P

 

[Cyosis]

Which of these are causing problems?

 

[zebra1707]

Hi there

b) to f) is causing the problem

Many thanks P

 

[The Chaz]

b) Set the derivative = 0
c) undefined => discontinuous
d) google end behavior/horizontal asymptotes
e) i.e. "show that this is an even function" (google "even function")
f) google evaluating functions.

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.

 

[Cyosis]

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

 

[zebra1707]

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

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.

 

[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... ?

What about
c)
d)
e)
f)

 

[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

 

[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.