Question

What is the range of this function?

Answer 1

\[y = (4x^2)/(x^2 + 2)\]

Answer 2

it is asking what y values you can "reach"

Answer 3

How do I figure that out?

Answer 4

one hint is we only use x^2 (which means x*x)
and that means we can never get a negative number.

Answer 5

But that's for the domain, right? Since we can't get a negative number, the domain is all real numbers.

Answer 6

you could check what number do you get for x=0 ?

Answer 7

You would get 0 for x = 0

Answer 8

the domain are the allowed x's , and here we can use any x value.

ok, so 0 is the smallest this thing can get .

Answer 9

4 is a horizontal asymptote, right?

Answer 10

Why is 0 the smallest thing we can get if the domain is all real numbers?

Answer 11

yes. I was going to say divide top and bottom by x^2
we get
\[ \frac{4}{1+\frac{2}{x^2} }\]

as x gets big the 2/huge number gets close to 0
and we get something close to 4/1 or 4
(but never reach 4)
so that is the range
[0, 4)

Answer 12

Oh. For future problems, how would I find the range though? Get all the variables on the bottom?

Answer 13

I don't really understand why 0 is the lowest y value we can get.

Answer 14

I would check -infinity (i.e. a "big negative" number) , 0 and + infinity
to see if I can figure out what y values I get

But it helps to do a bunch of problems. (It is annoying to learn how to do something that is so useless)

Answer 15

Again, looking at a graph is always a good aid.

Also notice that x^2 on top can never be negative, and the denom. can never be negative.

Answer 16

Oh - I see why 0 is the lowest value now.

Answer 17

**I don't really understand why 0 is the lowest y value we can get.***

do you agree that x^2 is *never* negative?
and adding or dividing as in this case will never get us a negative number.
so the smallest we can expect is 0. When you test with x=0, we get 0
so that must be the min value in the range.

Answer 18

Yes :) thanks for explaining that. For the maximum y value, do I just move all the variables to the denominator?

Answer 19

that is one strategy (trick, if you like)
it may not always be needed, which is why you have to do a few problems to "get a feel" for this stuff.

Answer 20

What about something like: (x^2 + 36)/(x^2 - 36) ?

The range is all real numbers except +/- 6 but I'm stuck on the range again :/

Answer 21

*domain

Answer 22

Right.

Answer 23

To get the asymptote (note that it's not necessarily the max value)

you can just use what you used earlier - use the highest degree on top and bottom, remember

Answer 24

The asymptote would be 1, correct? How does that help if it isn't necessarily the max value?

Answer 25

Asymptote is not 1.

Divide the highest degree term on top by highest degree term on bottom

Answer 26

x^2/x^2 = 1 ? Are we talking about the original problem or the second one?

Answer 27

Original of course

Answer 28

Oh, then the asymptote is 4

Answer 29

Why wouldn't the range of the orig. equation be all real numbers except 4?

Answer 30

Correct. But note that isn't always the max value. You should get used to looking at graphs
Look:
https://www.google.com/search?q=y%3D4x%5E2%2F(x%5E2%2B2)&oq=y%3D4x%5E2%2F(x%5E2%2B2)&aqs=chrome..69i57.5658j0j1&sourceid=chrome&ie=UTF-8

Answer 31

Hmmm okay. Do most math classes prefer you to look at a graph or find the domain/range solely from the equation?

Answer 32

I would use the graph as a very helpful hint, and verify with the equation if I had it.

Answer 33

Is the range all real numbers except 1? for the second equation.

Answer 34

You should do both. The more graphs you see, the better you can picture some in your head.

Answer 35

The second equation? That was equivalent to the first...

Answer 36

\[ \frac{x^2 + 36}{x^2 - 36}= \frac{1+\frac{36}{x^2}}{1-\frac{36}{x^2}} \]
it looks like it asymptotes to 1 but
the top is bigger than 1 and the bottom less than 1
(example 1.1/0.9 > 1) so it gets bigger than 1 somewhere before infinity

Answer 37

The second one I asked about was (x^2 + 36)/(x^2 - 36) - different problem :)

Answer 38

That's why it's better to post new questions for a new question ;P

Answer 39

What would the range be then? @phi You're confusing me lol

Answer 40

I put in x=0 and got -1
and at x= infinity it's +1

Answer 41

What does that mean?

Answer 42

That y can be equal to 1?

Answer 43

Again a graph helps :P

https://www.google.com/search?q=(x%5E2+%2B+36)%2F(x%5E2+-+36)&oq=(x%5E2+%2B+36)%2F(x%5E2+-+36)&aqs=chrome..69i57&sourceid=chrome&ie=UTF-8

graphs help see what to look for when you do it algebraically

Answer 44

it does not mean too much.

if we factor the bottom we get
\[ \frac{x^2+36}{(x-6)(x+6)}\]
they don't allow x to be 6 (or -6)
but as x gets close to 6 (say 5.9999) (x-6) becomes a tiny negative number
so we zoom down to negative infinity

Answer 45

@agent0smith How do you do it algebraically? Find the inverse of the function?

Answer 46

No, i mean what phi is doing. like "I put in x=0 and got -1"

Answer 47

Go nuts if you want to try to find the inverse of this function, but don't expect me to help with that haha

Answer 48

So I just plug in different values of x and see what works?

Answer 49

Haha

Answer 50

I don't understand what phi is doing :P no offense :)

Answer 51

No i mean use the graph to see key points... like at x=0 is the only real key point. And as x approaches + or - infinity

Answer 52

Are you allowed to use graphs to do this problem ?

Answer 53

Ah okay. So plug in the key points?

Answer 54

I don't know if I'm allowed to use graphs, the lesson doesn't specify. That is also what I was confused about..

Answer 55

in that case, if you can, factor the top and the bottom
in this case you get
\[ \frac{x^3+36}{(x-6)(x+6) }\]

look for values that make the bottom 0. those will give you vertical asymptotes
you will have to figure out if the go up or down.

Answer 56

Whether you're allowed to or not, doesn't mean you can't use the graph as an aid. You can still show evidence for the range in other ways.

Answer 57

How do the vertical asymptotes help with finding the range?

Answer 58

Thank you both so much for explaining this to me

Answer 59

The behaviour of the graph very near the asymptote tells you if the graph approaches + or - infinity... that tells you something about the range.

Answer 60

if you go to negative infinity then the range starts a -infinity
if you go to positive infinity, the upper bound of the range is + infinity

Answer 61

In this case it goes to both negative and positive infinity, correct?

Answer 62

Yes, so that'll be part of the range.

Answer 63

yes. If you can't use a graph, you should try to make a rough sketch

we should check numbers near -6: say -6.1 and -5.9 (this is close to the vertical asymptotes)

mostly we are interested in the sign of y: is the graph zooming up or down on each side of the asymptote ?

Answer 64

Okay. You guys have been really helpful, I think I have a good foundation now. Thank you both.

Answer 65

Yeah even if you can't use things like a google graph, you can still make sketches by hand

Answer 66

If you look at the graph, you see there will be some y values that the curve never reaches.
specifically -1 < y < = 1 are excluded. the range would be
\[ (\infty, -1] \cup (1, +\infty) \]
or
\[ y \le -1 \text{ or } y>1\]

Answer 67

you don't know calculus , right ? Because people use calculus to figure this stuff out.

Answer 68

Nope, currently in precalc. So the problems I have should be relatively simple.

Answer 69

I think I got it from here :)

Answer 70

Here are the steps I did
1) divided top and bottom by x^2 to get a fraction
\[ \frac{1+\frac{36}{x^2}}{1-\frac{36}{x^2}}\]
at either plus or minus infinity, we approach 1 (but are above it)

2) look at
\[ \frac{x^2+36}{x^2-36} \]
at x= -6.1 (left side of the asymptote). top is +, the bottom is +
we are zooming up.

that gives this picture