Question

How do I find the range of a rational function?

f(x)=(2)/(x^2-2x-3)

PLEASE DON'T ANSWER, JUST AN EXPLANATION OF HOW I WOULD DO IT

Answer 1

Range is your set of y-values (f(x)) that you get when you plug in x values within the domain.

Answer 2

\[f(x)=\frac{ 2 }{ x^2-2x-3 }\]

Answer 3

Yes, I know. So how would I find that in this situation? I watched a couple of videos, and they said to find the domain of the inverse of the function, but the way they found the inverse does not make sense to me :/

Answer 4

Well how would you find the domain of this?

Answer 5

f(x)=2/x^2-2x3
f(x)=2/(x-3)(x+1)

So the domain is (-∞, 3) U (3, ∞), (-∞, -1) U (-1, ∞) right?

Answer 6

I struggled with this area as well so I may have just completely missed the mark here. :/

Answer 7

You'll likely need to use calculus (derivatives) to find the domain of a non simple rational function.

Answer 8

How would I do that? Sorry for the silly question

Answer 9

https://www.wolframalpha.com/input/?i=plot+2%2F(x%5E2+-+2+x+-+3)+for+x%3D-10+to+x%3D10

Answer 10

It's not easy to find the domain of the inverse of a rational function like this one...

You didn't even answer if you knew calculus...?

Answer 11

x=3 is part of the range

Answer 12

I am in pre calculus

Answer 13

https://www.wolframalpha.com/input/?i=solve+2%2F(x%5E2+-+2+x+-+3)%3D3

Answer 14

^it's not @eliesaab

Well then idk what methods you've learned. You could find the inverse but it requires completing the square or quadratic formula.

Answer 15

There are two values for x so that f(x)=3

Answer 16

Here they are
\[
\left\{\left\{x\to \frac{1}{3} \left(3-\sqrt{42}\right)\right\},\left\{x\to
\frac{1}{3} \left(3+\sqrt{42}\right)\right\}\right\}
\]

Answer 17

The course I am in doesn't really actually explain how to do anything for the most part, which is what is so frustrating
thanks for trying to help out!

Answer 18

Also there are two values for x such that x=-1 here they are
https://www.wolframalpha.com/input/?i=solve+2%2F(x%5E2+-+2+x+-+3)%3D-1

Answer 19

Yeah, but I'd like to understand HOW to do it by hand, instead of using a calculator

Answer 20

I am tying to tell you that the answer giving above and saying that 3 and -1 are not part of the range is false

Answer 21

Oh okay

Answer 22

@eliesaab it's clear from the graph the range is not all real numbers

Answer 23

In fact the graph tells you that the range is all real numbers except 0

Answer 24

You don't see that gap between -0.5 and 0?

Answer 25

The answer is
\[
(-\infty ,-1/2) \cup (0,\infty)
\]

Answer 26

I never said anything about 3 or -1, since they're clearly in the range. I was referring to your range of all reals.

\[\large y=\frac{ 2 }{ x^2-2x-3 }\] \[\large x=\frac{ 2 }{ y^2-2y-3 }\] \[\large y^2-2y-3 =\frac{ 2 }{ x }\]
You can probably use the quad formula or complete the square from here. I am not going to, because this should be done using calculus in the first place.

Answer 27

Sorry, the original poster said it.

Answer 28

https://www.wolframalpha.com/input/?i=solve+for+y+x%3D+2%2F(y%5E2+-+2y+-+3)
You could find the domain from there. But i mean, why? Just either use the graph, or do things like this in calculus.

Answer 29

Actually I found a way first take a any number and solve
f(x)=a
you find two roots, using the quadratic formula
Here they are
\[
\left\{\left\{x\to \frac{a-\sqrt{2} \sqrt{2 a^2+a}}{a}\right\},\left\{x\to
\frac{\sqrt{2} \sqrt{2 a^2+a}+a}{a}\right\}\right\}
\]
Now
\[
2 a^2 + a\ge 0
\]
for a >0 or a <-.5
That is the proof

Answer 30

You do not need Calculus here. You only need the quadratic formula

Answer 31

It's not that you need calculus, but that it makes it far easier.

Answer 32

Actually -.5 is in the range since f(1)=-.5
So finally the answer is
\[
(-\infty, -.5]\cup (0,\infty)\]

Answer 33

Did I read above that the student is in pre-calc

Answer 34

Yeah, but that doesn't change what I said. Finding range of rational functions is much easier with calculus. You won't be able to do much with a more complex rational function (eg. a cubic or higher degree in the denominator)

Answer 35

wait find the domain?
you know that the denominator cannot be equal to zero

factor the denominator find the roots and those are the points in the domain that cant exist

every other point on the number line is fair game

Answer 36

oh wait find the range,
just look at the extreme points

or the value of y as you reach extreme points like as you reach negative infinity, as you reach positive infinity, limit of asymptotes from the left and from the right

Answer 37

chances are, the y=limit as x-> asymptote might be infinity
and the limit as x-> asymptote from the other side might be negative infinity

then your range would be (-infinity, infinity)

though if all your limits go to 0 and maybe infinity, then you might have to use calculus in order to determine the local minima