Question

Give the following for the function : f(x) = (x^4+3x^2) / (x^2-1)
Domain:
Range:
x intercept:
y intercept:

Answer 1

Factor Everything. You need to show your work. I can do this all day, but you won't learn anything.

Answer 2

What do you do after you factor everything?
(x^2(x^2+3)) / (x+1)(x-1)

Answer 3

You start answering the questions.

Domain: With no common factor in the numerator, you can just look at the denominator for exclusions.

Range: This can be a little tricky. There may be odd gaps. How do you propose we approach this?

x intercept: Substitute y = 0 and solve for x. You should get x = 0 ONLY on this one.

y intercept: Substitute x = 0 if that is in the Domain.

Answer 4

I still do not understand the domain and range part. I do understand that y intercept is (0,0). I don't understand how x intercept is x=0.
x^2(x^2+3) = 0

x^2 = 0
x=0

x^2+3 = 0
x^2=-3
x= \[\sqrt{-3}\]

Answer 5

Domain: What do you see in the denominator? Anything dangerous down there? Is x = 2 okay? How about x = 1?

Answer 6

x=2 would be okay. x=1 would be not okay since it'll be undefined.

Answer 7

and x=-1

Answer 8

Ah.. so the domain would be (-∞,-1)U(-1,1)U(1,∞).
How would I find the range ?

Answer 9

Very good. Let's do the Range last. For y = 0, the x-intercepts, we have your previous work. only x = 0 makes this thing zero (0). (0,0) is the x-intercept.

Answer 10

I see. I understand.
Let us move onto the range.

Answer 11

y-intercept x = 0 produces y = 0. Thus (0,0) is also the y-intercept.

Answer 12

Right. but how does that explain the range ?

Answer 13

It doesn't. I was just getting that loose end out of the way.

For the Range, we must understand the non-vertical asymptotes. Do you know what sort of non-vertical asymptote this one has?

Answer 14

What do you mean by non-vertical asymptote ?
Do you mean slant asymptote ?

Answer 15

If the degree of the numerator were ONE (1) greater then the degree of the denominator, yes, "slant" or "oblique" asymptote. In this case, the difference is two (2), so we'll get a quadratic, parabolic, or 2nd degree asymptote.

Long division suggests \(f(x) = x^{2} + 4 + \dfrac{4}{x^{2}-1}\), and this exposes the exact nature of the asymptotic behavior. f(x) approaches \(g(x) = x^{2} + 4\) as x increases without bound in either the negative or positive direction.

A quick look at g(x) shows it's minimum value is 4. Now we have a hint at the Range. It is something in the neighborhood of \((-\infty,0]\cup [4,\infty)\), but that's still too much. f(x) isn't that close to g(x) for small values of x.

Oddly, there may be a hint in the numerator. Recall the effort to find an x-intercept? \(x^{2} + 3 = 0\). This suggested \(x = \sqrt{3}i\;or\;x=-\sqrt{3}i\) as zeros. This may be a clue where to look for the minimum values of f(x) - not at \(x = \sqrt{3}i\;or\;x=-\sqrt{3}i\), but \(x = \sqrt{3}\;or\;x=-\sqrt{3}\). See if you can get f(x) to go any lower around \(\sqrt{3}\). If not, we have it.

Range: \((-\infty,0]\cup [9,\infty)\)