Rational Functions – Homework Help

In summary, the rational function y = 1 / x^2-1 has a turning point at (1,0) and is discontinuous at x = -1.
  • #1
zebra1707
107
0
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
 
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  • #2
Which of these are causing problems?
 
  • #3
Hi there

b) to f) is causing the problem

Many thanks P
 
  • #4
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.
 
  • #5
It's time for you to show some work. It is hard to believe that all of these give you trouble.
 
  • #6
Cyosis said:
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.
 
  • #7
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)
 
  • #8
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
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.
 

What is a rational function?

A rational function is a mathematical function that can be written as the ratio of two polynomial functions. It can also be described as a function with a numerator and a denominator, where both are polynomial expressions.

How do I solve rational functions?

To solve a rational function, you need to first simplify it by factoring both the numerator and denominator, and then canceling out any common factors. Next, set the remaining factors equal to zero and solve for the variable. This will give you the x-intercepts of the function.

What is the domain of a rational function?

The domain of a rational function is all real numbers except for the values that make the denominator equal to zero. This is because division by zero is undefined in mathematics.

How do I find the vertical asymptotes of a rational function?

The vertical asymptotes of a rational function are the values that make the denominator equal to zero. To find them, set the denominator equal to zero and solve for the variable. The resulting values will be the equations of the vertical asymptotes.

Can I graph a rational function?

Yes, you can graph a rational function by plotting points, finding the x and y-intercepts, and determining the behavior of the function at the vertical asymptotes. You can also use a graphing calculator to graph the function.

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