Alright, this time:
y = (x^3 - 27)/(x^4-2x^3+9x^2-18x)
How does one find the domain of this?
2006-10-04 12:33:33 · 6 answers · asked by Devin 2 in Science & Mathematics ➔ Mathematics
I know that I have to set the denominator to Zero, but how will I factor that?
2006-10-04 12:41:10 · update #1
To the alphabet guy, I have to derive the domain, y-int, x-int, vertical asymptote, and horizontal asymptote. Then graph.
2006-10-04 12:42:46 · update #2
Thank you everyone.
2006-10-04 12:54:11 · update #3
You need to factor the denominator first. Then you can tell when the denominator = 0. That will be the value(s) that you want your domain to avoid.
x(x^3 - 2x^2 + 9x - 18)
= x[x^2(x - 2) + 9(x - 2)]
=x(x - 2)(x^2 + 9)
Domain: x can be all real numbers except for 0, 2
2006-10-04 12:45:36 · answer #1 · answered by Isaac 2 · 0⤊ 0⤋
Remember that domain is the possible x values. So in this case the restrictions would be that you cannot divide by 0, meaning that whichever values of x make the denominator 0 cannot be in the domain.
The way to find which x values make the denominator 0 is to factor it and set each factor equal to 0. You could factor the denominator to x(x^3 - 2x^2 + 9x - 18). From there you can factor using the Remainder theorum and Rational Root Theorum. For more on these, google them. After loads of work the denominator factors to x(x-2)(x^2 +9). So x cannot equal 0 (because of the x), x cannot equal 2 (because of x-2), and x cannot equal 9i (because of x^2 +9).
There you go. Those values are not in the domain. Now, next time you decide to ask for answers to homework, hopefully you'll reference your math book and not a person on the net who gives you the answer but confuses you to no end.
2006-10-04 19:56:08 · answer #2 · answered by shoutingchimp 3 · 0⤊ 0⤋
Let's tackle the problem of factoring the denominator.
First, remove the x:
x(x^3 - 2x^2 +9x -18).
Now factor the second term by grouping to get
x(x^2(x -2) + 9(x-2) )
Now notice the common factor of x-2 in the 2nd term.
So the denominator factors as
x(x-2)(x^2 + 9).
So the only bad points are x = 0 and 2.
Hope that helps a bit!
2006-10-04 19:51:18 · answer #3 · answered by steiner1745 7 · 0⤊ 0⤋
Because you are dividing the no-no numbers will be those that make the denominator a zero.
Zero will make the denominator zero so it's a no-no for the domain.
Now you can remove the x and you are left with x^3-2x^2+9x-18.
Find the zeros of this equation and you will have more x values that will be no-nos.
and so on...
Ok, you poor thing.
Try x = 2,
and so on again.
2006-10-04 19:38:29 · answer #4 · answered by Dr. J. 6 · 0⤊ 0⤋
The equation
x^3 - 27 = 0 is easy to solve.
Try factoring the denominator.
2006-10-04 19:37:34 · answer #5 · answered by cosmo 7 · 0⤊ 1⤋
usually they give u a domain but if not just make up Ur own( first aks your taecher)
2006-10-04 19:38:47 · answer #6 · answered by Anonymous · 0⤊ 2⤋