Use Graph to Find Domain and Range

mathxyz · 2024-03-13 14:35:24

How does one use the graph of a function to find the domain and range?

Bob · 2024-03-13 21:22:43

The domain is all the x values for which the function is defined

eg the graph of y = 1/x doesn't have a point when x = 0

The range is the set of y values that result from the function.

eg y = x^2 has no negative values of y.

Usually looking at the graph is enough to answer these questions.

On a graph plotter you might need to zoom out to see all relavent values.

Bob

mathxyz · 2024-03-14 04:19:14

Bob wrote:

The domain is all the x values for which the function is defined
eg the graph of y = 1/x doesn't have a point when x = 0
The range is the set of y values that result from the function.
eg y = x^2 has no negative values of y.
Usually looking at the graph is enough to answer these questions.
On a graph plotter you might need to zoom out to see all relavent values.
Bob

Since I don't know how to upload photos here, can you show me with a graph or two?

Bob · 2024-03-14 21:54:37

This post shows how you can sketch a graph to help answering these questions. I tried using Imgur on my phone to upload the picture but I couldn't do it. The app for Android doesn't seems to allow this, so I've had to use my laptop. I'll try to find another way.

Let's take y = (x^2 + 2)/(x + 4) as an example. You can choose some x values, work out the corresponding y values and hence plot those points. When you see the actual graph you may realise that the chances of choosing the right x values to show the full details of the graph is low. As an alternative you can use a function plotter but I think there's a lot to be gained by learning how to work out the key details of a function and hence to sketch it (ie. show the main features without an accurate plot).

The above function has enough interesting features to allow me to show you the full sketching process.

(1) Where are the intercepts? When x = 0 y = 1/2 so we know this graph crosses the y axis at (0 , 1/2)

When is y = 0 ? We need to solve (x^2 + 2)/(x + 4) = 0

Multiply both sides by x + 4 to give x^2 + 2 = 0 which needs x^2 = -2

There are no real x values for that so the graph never crosses the x axis.

(2) Are there any values of x that make y zoom off to + or - infinity?

Put x + 4 = 0 ie x = -4

So x = -4 is excluded from the domain.

Just to the right of x = -4 (say x = -3) y = (9 + 2)/(-3 + 4) = 11/1 I'm not so interested in the actual value of y, but rather whether it is positive or negative. As it's positive this tells me that as x approaches -4 from the right y is positive and getting bigger. So I know the graph zooms off to + infinity as x approaches -4 from the right.

Just to the left of x = -4 (say x = -5) y = (25 + 2)/(-5 + 4) This is negative so I can say that y zooms off to minus infinity as x approaches -4 from the left.

The vertical line x = -4 is called an asymptote, a line that the curve approaches but never meets.

(3) How does the graph behave when x is large and positive. In the numerator the 2 is negligible compared to large x^2. The +4 will similarly be negligible compared to large x so we can say the function will behave like x^2/x = x when x is very large. This means that the line y = x is another asymptote. The curve approaches this line as x is very large and positive.

When x is large but negative we can make a similar approximation, so the curve approaches the asymptote for large negative x too.

(4) If you've done differential calculus you can also find when the gradient of the curve is zero. This allows you to identify any local maximum and minimum points. These are places where the curve bends round in an upside down U shape for a maximum or a U shape for a minimum. But I can figure these out just from what I've already done.

As x starts close to the line y = x in negative values but zooms off to negative infinity as x approaches -4, the curve must have a maximum somewhere between - infinity and -4.

After that the curve 'comes down' from plus infinity and crosses the y axis at 1/2. Somehow it doesn't cross the x axis but starts moving towards the line y = x again, so there must be a minimum after x = 0 to allow the curve to bend round towards the line.

All the above allows me to sketch the curve. I could do that on paper; take a picture; and upload that but I'll instead use the function plotter so it's neater. You can compare my graph with the above 4 steps to verify it has the right properties.

I've already mentioned the domain limit. What about the range? With calculus you can determine the actual coordinates of the maximum and minimum points. Let's say max is (a,b) and the minimum (c,d) Then the range is (-infinity , b] and [d , +infinity)

ob

mathxyz · 2024-03-15 04:16:40

Bob wrote:

This post shows how you can sketch a graph to help answering these questions. I tried using Imgur on my phone to upload the picture but I couldn't do it. The app for Android doesn't seems to allow this, so I've had to use my laptop. I'll try to find another way.
Let's take y = (x^2 + 2)/(x + 4) as an example. You can choose some x values, work out the corresponding y values and hence plot those points. When you see the actual graph you may realise that the chances of choosing the right x values to show the full details of the graph is low. As an alternative you can use a function plotter but I think there's a lot to be gained by learning how to work out the key details of a function and hence to sketch it (ie. show the main features without an accurate plot).
The above function has enough interesting features to allow me to show you the full sketching process.
(1) Where are the intercepts? When x = 0 y = 1/2 so we know this graph crosses the y axis at (0 , 1/2)
When is y = 0 ? We need to solve (x^2 + 2)/(x + 4) = 0
Multiply both sides by x + 4 to give x^2 + 2 = 0 which needs x^2 = -2
There are no real x values for that so the graph never crosses the x axis.
(2) Are there any values of x that make y zoom off to + or - infinity?
Put x + 4 = 0 ie x = -4
So x = -4 is excluded from the domain.
Just to the right of x = -4 (say x = -3) y = (9 + 2)/(-3 + 4) = 11/1 I'm not so interested in the actual value of y, but rather whether it is positive or negative. As it's positive this tells me that as x approaches -4 from the right y is positive and getting bigger. So I know the graph zooms off to + infinity as x approaches -4 from the right.
Just to the left of x = -4 (say x = -5) y = (25 + 2)/(-5 + 4) This is negative so I can say that y zooms off to minus infinity as x approaches -4 from the left.
The vertical line x = -4 is called an asymptote, a line that the curve approaches but never meets.
(3) How does the graph behave when x is large and positive. In the numerator the 2 is negligible compared to large x^2. The +4 will similarly be negligible compared to large x so we can say the function will behave like x^2/x = x when x is very large. This means that the line y = x is another asymptote. The curve approaches this line as x is very large and positive.
When x is large but negative we can make a similar approximation, so the curve approaches the asymptote for large negative x too.
(4) If you've done differential calculus you can also find when the gradient of the curve is zero. This allows you to identify any local maximum and minimum points. These are places where the curve bends round in an upside down U shape for a maximum or a U shape for a minimum. But I can figure these out just from what I've already done.
As x starts close to the line y = x in negative values but zooms off to negative infinity as x approaches -4, the curve must have a maximum somewhere between - infinity and -4.
After that the curve 'comes down' from plus infinity and crosses the y axis at 1/2. Somehow it doesn't cross the x axis but starts moving towards the line y = x again, so there must be a minimum after x = 0 to allow the curve to bend round towards the line.
All the above allows me to sketch the curve. I could do that on paper; take a picture; and upload that but I'll instead use the function plotter so it's neater. You can compare my graph with the above 4 steps to verify it has the right properties.
https://i.imgur.com/ZWA9YyB.gif
I've already mentioned the domain limit. What about the range? With calculus you can determine the actual coordinates of the maximum and minimum points. Let's say max is (a,b) and the minimum (c,d) Then the range is (-infinity , b] and [d , +infinity)
ob

A fabulous reply. Thank you so much.

Here is my imgur image url:

https://i.imgur.com/0KUuvCI.jpeg

How do I post this image on this site?

Bob · 2024-03-15 21:40:27

hi,

I've tried putting (in square brackets) url and /url tags at the start and end of the address. That doesn't embed the image but rather gives the reader the option to click the link to get to your image. Not much better than just being told the link but it saves having to copy it.

My graph link is in bbcode format and, minus the tags, https://i.imgur.com/0KUuvCI.jpeg looks like this. That leading "i" makes all the difference. The MIF server can interpret that as insert this image here so you get to see the image rather than the link to it.

The laptop version of imgur allows this. At the moment the Android version doesn't. I don't know whether they have plans to include it; there's certainly some clammer for it by imgur users.

I tried photobucket. It seems that would work but they charge for the service. Also tried flickr, tapatalk and imgbb but none of these lived up to the promise. For the moment what you have done will have to do.

If it's really important for members to see the image I can put it through my imgur account and insert it but it's a bit of a task to do this so keep this for important cases only please.

You have plotted the points; tick; but I don't thing you're justified in joining them up like this. If you have results from some experiment and the points follow a clear curve or line, then joining up is usual and ok. But these points are all over the place with no clear curve so it's not really justified to join them.

That's the problem with the question that asks you do interpolate a point between two known points. How do we know that's valid? I think the question shouldn't be asked like that. Better to say "Can you predict whether .....?" And then maybe "Why might that be wrong?"

Bob

mathxyz · 2024-03-15 23:30:12

Bob wrote:

hi,
I've tried putting (in square brackets) url and /url tags at the start and end of the address. That doesn't embed the image but rather gives the reader the option to click the link to get to your image. Not much better than just being told the link but it saves having to copy it.
My graph link is in bbcode format and, minus the tags, https://i.imgur.com/0KUuvCI.jpeg looks like this. That leading "i" makes all the difference. The MIF server can interpret that as insert this image here so you get to see the image rather than the link to it.
The laptop version of imgur allows this. At the moment the Android version doesn't. I don't know whether they have plans to include it; there's certainly some clammer for it by imgur users.
I tried photobucket. It seems that would work but they charge for the service. Also tried flickr, tapatalk and imgbb but none of these lived up to the promise. For the moment what you have done will have to do.
If it's really important for members to see the image I can put it through my imgur account and insert it but it's a bit of a task to do this so keep this for important cases only please.
You have plotted the points; tick; but I don't thing you're justified in joining them up like this. If you have results from some experiment and the points follow a clear curve or line, then joining up is usual and ok. But these points are all over the place with no clear curve so it's not really justified to join them.
That's the problem with the question that asks you do interpolate a point between two known points. How do we know that's valid? I think the question shouldn't be asked like that. Better to say "Can you predict whether .....?" And then maybe "Why might that be wrong?"
Bob

Thank you for the information.

You said:

"If it's really important for members to see the image I can put it through my imgur account and insert it but it's a bit of a task to do this so keep this for important cases only please."

This means a lot to me. I don't have a computer or laptop or iPhone. Android is limited in service. I will not post too many questions with photos. This means limited geometry, trigonometry, vector graphs, applications with pictures, etc. It's a shame because I really like this site.

As you know, I am doing a simultaneous self-study of mathematics racing toward my James Stewart Calculus textbook. I will post from my College Algebra textbook and Precalculus textbook keeping track of chapters and sections. As a full-time employee, it is really hard to find time for mathematics and my classical guitar arrangements and sight reading exercises. I will do the best I can. Now, some math questions.

Last edited by mathxyz (2024-03-16 00:41:52)

phrontister · 2024-03-16 01:34:44

FelizNYC wrote:

Here is my imgur image url:
https://i.imgur.com/0KUuvCI.jpeg
How do I post this image on this site?

Bob wrote:

I've tried putting (in square brackets) url and /url tags at the start and end of the address. That doesn't embed the image but rather gives the reader the option to click the link to get to your image. Not much better than just being told the link but it saves having to copy it.

This morning I copy/pasted FelizNYC's link into Google Chrome, which then opened Imgur's web page displaying the image and other content...as shown in this link: https://imgur.com/0KUuvCI

Tonight, FelizNYC's link displays just the image!

Anyway, this is the code that now works (see Examples below):

[img]https://i.imgur.com/0KUuvCI.jpeg[/img]

FWIW, I've also recently discovered that by appending a 'size modifier' (the letter t, m or l) to the url immediately preceding the full stop before the image extension (".jpeg", in this case), a smaller image than the default one displays. That can be useful at times (eg, if your image is too huge).

Examples:

mathxyz · 2024-03-16 01:47:34

phrontister wrote:

FelizNYC wrote:
Here is my imgur image url:
https://i.imgur.com/0KUuvCI.jpeg
How do I post this image on this site?
Bob wrote:
I've tried putting (in square brackets) url and /url tags at the start and end of the address. That doesn't embed the image but rather gives the reader the option to click the link to get to your image. Not much better than just being told the link but it saves having to copy it.
This morning I copy/pasted FelizNYC's link into Google Chrome, which then opened Imgur's web page displaying the image and other content...as shown in this link: https://imgur.com/0KUuvCI
Tonight, FelizNYC's link displays just the image!
Anyway, this is the code that now works (see Examples below):
[img]https://i.imgur.com/0KUuvCI.jpeg[/img]
FWIW, I've also recently discovered that by appending a 'size modifier' (the letter t, m or l) to the url immediately preceding the full stop before the image extension (".jpeg", in this case), a smaller image than the default one displays. That can be useful at times (eg, if your image is too huge).
Examples:

Thanks. I will try using the code that now works best.

mathxyz · 2024-03-16 02:26:21

Why does the image show?

https://imgur.com/a/gpB0TL4

mathxyz · 2024-03-16 02:29:17

Why does the image not show?

https://imgur.com/a/gpB0TL4.jpeg

phrontister · 2024-03-16 11:39:13

FelizNYC wrote:

Why does the image show?
https://imgur.com/a/gpB0TL4

FelizNYC wrote:

Why does the image not show?
https://imgur.com/a/gpB0TL4.jpeg

An image will only display on a forum post if the actual image address (ie, the image url) is enclosed in 'image tags'...as shown in my Post #9 above. Also see instructions about that here (about halfway down the page): BBCode).

The two addresses you gave are to the webpage containing the image, not to the image itself.

The image url will have an image extension (eg, ".jpeg"), but merely adding that extension to a webpage address, as you did, doesn't work.

In fact, the image address looks entirely different from the webpage address you gave, and it's this:

[img]https://i.imgur.com/7NtStGD.jpeg[/img]

Btw, I obtained that address with my laptop, on which it's easily done with a couple of mouse clicks. I don't have a smartphone.

In other posts, amnkb and Bob have tried to help you obtain an image address via your smartphone, but the process is a bit tedious and you said here "Thanks but I am not going through all those steps".

HOWEVER, on your post #5, the link you gave is an image address, not just one to the webpage containing the image!!! How did you get that?

If you can repeat that process...

Last edited by phrontister (2024-03-16 12:06:55)

mathxyz · 2024-03-16 11:45:15

phrontister wrote:

FelizNYC wrote:
Why does the image show?
https://imgur.com/a/gpB0TL4
FelizNYC wrote:
Why does the image not show?
https://imgur.com/a/gpB0TL4.jpeg
An image will only display on a forum post if the actual image address (ie, the image url) is enclosed in 'image tags'...as shown in my Post #9 above. Also see instructions about that here (about halfway down the page): BBCode).
The two addresses you gave are to the webpage containing the image, not to the image itself.
The image url will have an image extension (eg, ".jpeg"), but merely adding that extension to a webpage address, as you did, doesn't work.
In fact, the image address looks entirely different from the webpage address you gave, and it's this:
[img]https://i.imgur.com/7NtStGD.jpeg[/img
Btw, I obtained that address with my laptop, on which it's easily done with a couple of mouse clicks. I don't have a smartphone.
In other posts, amnkb and Bob have tried to help you obtain an image address via your smartphone, but the process is a bit tedious and you said here "Thanks but I am not going through all those steps".
HOWEVER, on your post #5, the link you gave is an image address, not just one to the webpage containing the image!!! How did you get that?
If you can repeat that process...

I don't have time to repeat the process now. Going to work. Remind me on Wednesday and Thursday (my days off) to go over whatever it is that I did.

phrontister · 2024-03-16 12:30:44

FelizNYC wrote:

I don't have time to repeat the process now. Going to work. Remind me on Wednesday and Thursday (my days off) to go over whatever it is that I did.

I'll try to remember.

Btw, my link in post #12 has an error (the closing square bracket is missing).

I've fixed it, and this is the correct link:

[img]https://i.imgur.com/7NtStGD.jpeg[/img]

Just copy that link and paste it into a post to display the image.

The reason the image doesn't display in my post is that I've enclosed the link in 'Code' tags, which deactivates the img tags (see the section headed 'Code' in BBCode, about 2/3rds down the page).

Last edited by phrontister (2024-03-16 12:31:57)

mathxyz · 2024-03-17 01:04:31

phrontister wrote:

FelizNYC wrote:
I don't have time to repeat the process now. Going to work. Remind me on Wednesday and Thursday (my days off) to go over whatever it is that I did.
I'll try to remember.
Btw, my link in post #12 has an error (the closing square bracket is missing).
I've fixed it, and this is the correct link:
[img]https://i.imgur.com/7NtStGD.jpeg[/img]
Just copy that link and paste it into a post to display the image.
The reason the image doesn't display in my post is that I've enclosed the link in 'Code' tags, which deactivates the img tags (see the section headed 'Code' in BBCode, about 2/3rds down the page).

You said:

"Just copy that link and paste it into a post to display the image."

What do you mean by that?

mathxyz · 2024-03-17 01:12:34

phrontister wrote:

FelizNYC wrote:
I don't have time to repeat the process now. Going to work. Remind me on Wednesday and Thursday (my days off) to go over whatever it is that I did.
I'll try to remember.
Btw, my link in post #12 has an error (the closing square bracket is missing).
I've fixed it, and this is the correct link:
[img]https://i.imgur.com/7NtStGD.jpeg[/img]
Just copy that link and paste it into a post to display the image.
The reason the image doesn't display in my post is that I've enclosed the link in 'Code' tags, which deactivates the img tags (see the section headed 'Code' in BBCode, about 2/3rds down the page).

Last edited by mathxyz (2024-03-17 01:34:21)

KerimF · 2024-03-17 01:33:03

phrontister wrote:

FelizNYC wrote:
I don't have time to repeat the process now. Going to work. Remind me on Wednesday and Thursday (my days off) to go over whatever it is that I did.
I'll try to remember.
Btw, my link in post #12 has an error (the closing square bracket is missing).
I've fixed it, and this is the correct link:
[img]https://i.imgur.com/7NtStGD.jpeg[/img]
Just copy that link and paste it into a post to display the image.
The reason the image doesn't display in my post is that I've enclosed the link in 'Code' tags, which deactivates the img tags (see the section headed 'Code' in BBCode, about 2/3rds down the page).

Thank you for clarifying this point.

I will try it here too before our friend Feliz will do.

I also appended the lettelr 'l' before ".jpeg" since the image is somehow big.

mathxyz · 2024-03-17 01:35:35

KerimF wrote:

phrontister wrote:
FelizNYC wrote:
I don't have time to repeat the process now. Going to work. Remind me on Wednesday and Thursday (my days off) to go over whatever it is that I did.
I'll try to remember.
Btw, my link in post #12 has an error (the closing square bracket is missing).
I've fixed it, and this is the correct link:
[img]https://i.imgur.com/7NtStGD.jpeg[/img]
Just copy that link and paste it into a post to display the image.
The reason the image doesn't display in my post is that I've enclosed the link in 'Code' tags, which deactivates the img tags (see the section headed 'Code' in BBCode, about 2/3rds down the page).
Thank you for clarifying this point.
I will try it here too before our friend Feliz will do.
https://i.imgur.com/7NtStGDl.jpeg
I also appended the lettelr 'l' before ".jpeg" since the image is somehow big.

How come it's not working for me?

KerimF · 2024-03-17 02:00:08

FelizNYC wrote:

How come it's not working for me?

I suggest, till you will succeed doing it on your side, that you may like to include the URL of your image of interest with a posted question and the first member who will read your post will attach it to a subsequent post for you (as I did here).

What do you think?

mathxyz · 2024-03-17 11:50:02

KerimF wrote:

FelizNYC wrote:
How come it's not working for me?
I suggest, till you will succeed doing it on your side, that you may like to include the URL of your image of interest with a posted question and the first member who will read your post will attach it to a subsequent post for you (as I did here).
What do you think?

I will try the URL thing at my end to see what happens.

mathxyz · 2024-03-17 19:50:47

phrontister wrote:

FelizNYC wrote:
I don't have time to repeat the process now. Going to work. Remind me on Wednesday and Thursday (my days off) to go over whatever it is that I did.
I'll try to remember.
Btw, my link in post #12 has an error (the closing square bracket is missing).
I've fixed it, and this is the correct link:
[img]https://i.imgur.com/7NtStGD.jpeg[/img]
Just copy that link and paste it into a post to display the image.
The reason the image doesn't display in my post is that I've enclosed the link in 'Code' tags, which deactivates the img tags (see the section headed 'Code' in BBCode, about 2/3rds down the page).

[img]https://i.imgur.com/a/IbFoc46.jpeg[/img]

KerimF · 2024-03-17 22:19:13

[img]https://i.imgur.com/a/IbFoc46.jpeg[/img]

This time it didn't work on my side.

But the following one is ok.

[img]https://i.imgur.com/7NtStGDl.jpeg[/img]

Last edited by KerimF (2024-03-17 22:22:19)

mathxyz · 2024-03-18 01:42:15

KerimF wrote:

[img]https://i.imgur.com/a/IbFoc46.jpeg[/img]
https://i.imgur.com/a/IbFoc46.jpeg
This time it didn't work on my side.
But the following one is ok.
[img]https://i.imgur.com/7NtStGDl.jpeg[/img]
https://i.imgur.com/7NtStGDl.jpeg

What a pain in the neck it is to upload photos here!!!

amnkb · 2024-03-18 09:57:28

FelizNYC wrote:

How does one use the graph of a function to find the domain and range?

domain: Look at where the graph goes from side to side
Does it quit (like with square roots) or have gaps (like with vert. asymps), or does it go everywhere?
Domain is wherever graph goes along x-axis
(not necessarily right next to the axis but you can trace up or down from the axis to the graph)

range: Look at where the graph goes from bottom to top
Does it quit (like with parabolas) or have gaps (like with hor. asymps), or does it go everywhere?
Range is wherever graph goes along y-axis
(not necessarily right next to the axis but you can trace right or left from the axis to the graph)

mathxyz · 2024-03-18 19:56:14

amnkb wrote:

FelizNYC wrote:
How does one use the graph of a function to find the domain and range?
domain: Look at where the graph goes from side to side
Does it quit (like with square roots) or have gaps (like with vert. asymps), or does it go everywhere?
Domain is wherever graph goes along x-axis
(not necessarily right next to the axis but you can trace up or down from the axis to the graph)
range: Look at where the graph goes from bottom to top
Does it quit (like with parabolas) or have gaps (like with hor. asymps), or does it go everywhere?
Range is wherever graph goes along y-axis
(not necessarily right next to the axis but you can trace right or left from the axis to the graph)

I think searching YouTube for math video clips is also very helpful.

Math Is Fun Forum

#1 2024-03-13 14:35:24

Use Graph to Find Domain and Range

#2 2024-03-13 21:22:43

Re: Use Graph to Find Domain and Range

#3 2024-03-14 04:19:14

Re: Use Graph to Find Domain and Range

#4 2024-03-14 21:54:37

Re: Use Graph to Find Domain and Range

#5 2024-03-15 04:16:40

Re: Use Graph to Find Domain and Range

#6 2024-03-15 21:40:27

Re: Use Graph to Find Domain and Range

#7 2024-03-15 23:30:12

Re: Use Graph to Find Domain and Range

#8 2024-03-16 01:34:44

Re: Use Graph to Find Domain and Range

#9 2024-03-16 01:47:34

Re: Use Graph to Find Domain and Range

#10 2024-03-16 02:26:21

Re: Use Graph to Find Domain and Range

#11 2024-03-16 02:29:17

Re: Use Graph to Find Domain and Range

#12 2024-03-16 11:39:13

Re: Use Graph to Find Domain and Range

#13 2024-03-16 11:45:15

Re: Use Graph to Find Domain and Range

#14 2024-03-16 12:30:44

Re: Use Graph to Find Domain and Range

#15 2024-03-17 01:04:31

Re: Use Graph to Find Domain and Range

#16 2024-03-17 01:12:34

Re: Use Graph to Find Domain and Range

#17 2024-03-17 01:33:03

Re: Use Graph to Find Domain and Range

#18 2024-03-17 01:35:35

Re: Use Graph to Find Domain and Range

#19 2024-03-17 02:00:08

Re: Use Graph to Find Domain and Range

#20 2024-03-17 11:50:02

Re: Use Graph to Find Domain and Range

#21 2024-03-17 19:50:47

Re: Use Graph to Find Domain and Range

#22 2024-03-17 22:19:13

Re: Use Graph to Find Domain and Range

#23 2024-03-18 01:42:15

Re: Use Graph to Find Domain and Range

#24 2024-03-18 09:57:28

Re: Use Graph to Find Domain and Range

#25 2024-03-18 19:56:14

Re: Use Graph to Find Domain and Range

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