Hi, I have this function, g(x)= (x^3-x)/(x^3-4x).
Now I need help with the following...
- How to find the zeros for f?
- How to find f(-x)?
- How to find point of discontinuity?

And also, would finding the asymptotes, slant asymptote, and point of discontinuity be easier if I simplify this rational first?:
(2x^3-3x^2-12x+20)/(x^2-5x+6)
I can simplify the denominator as (x+1)(x-6) but that's it...
Is the first step to simplifying the numerator... Factoring out x?

I'd appreciate any help, thanks so much!

2. Originally Posted by bubblegum
Hi, I have this function, g(x)= (x^3-x)/(x^3-4x).
Now I need help with the following...
- How to find the zeros for f?
- How to find f(-x)?
- How to find point of discontinuity?

And also, would finding the asymptotes, slant asymptote, and point of discontinuity be easier if I simplify this rational first?:
(2x^3-3x^2-12x+20)/(x^2-5x+6)
I can simplify the denominator as (x+1)(x-6) but that's it...
Is the first step to simplifying the numerator... Factoring out x?

I'd appreciate any help, thanks so much!
A rational functions zeros are only in the numerator...so you need to solve $\displaystyle x^3-x=0\Rightarrow{x(x^2-1)=0}$..so x=0,x=-1,x=1

since this is odd $\displaystyle f(-x)=-f(x)$
solve the denominator for 0 we get $\displaystyle x(x^2-4)=0$ so x=0,x=2, x=-2 all points of discontinuity

just polynomial divide...DO NOT simplify first to find slant asymptotes