# Determine the Range

• October 23rd 2006, 06:01 AM
qbkr21
Determine the Range
Yeh I know I posted quite a bit over last several hours, however I have one last question. My teacher wants me to look at the graphing calculator to find the range. Can you all please help me. What is the easiest way to go about doing this and what am I too look for? Are there any rules, ect.? Thanks for all you help.
• October 23rd 2006, 06:58 AM
earboth
Quote:

Originally Posted by qbkr21
Yeh I know I posted quite a bit over last several hours, however I have one last question. My teacher wants me to look at the graphing calculator to find the range. Can you all please help me. What is the easiest way to go about doing this and what am I too look for? Are there any rules, ect.? Thanks for all you help.

Hi,

first you have to find the domain of a function which contains all x-values.

Then you can determine the range which contains all y-values.

I've attached a diagram to show you where you can find domain and range.

EB
• October 23rd 2006, 07:13 AM
qbkr21
Re:
Ok but what exactly are you looking at to l you if its all real numbers, infinate, or specific boundaries?

Thanks,
• October 23rd 2006, 07:17 AM
qbkr21
Re:
Take for example f(x)= (1)/(2-x)... What in this problem are looking at to determine the range. Getting the domain isn't a problem for me, in this case its x=2. Now where do I go from here?
• October 23rd 2006, 07:35 AM
ThePerfectHacker
Quote:

Originally Posted by qbkr21
Take for example f(x)= (1)/(2-x)... What in this problem are looking at to determine the range. Getting the domain isn't a problem for me, in this case its x=2. Now where do I go from here?

You are right about the domain, anything but 2, hooah!.

Now the range:
You can use the inequality argument as before.
Or you can do it the primitive way (your teachers way) and find the graph:
Shown below, thus the range is $y=(-\infty,0)\cup (0,+\infty)$
• October 23rd 2006, 07:46 AM
qbkr21
Re:
Ok so correct me if I'm wrong ,but from what I can tell in looking at the picture you want to see how the range correlates to domain. When looking up and over on the picture if you see a juncture where there is no correlation and a space then the range has stopped. Then as you progress even further upward on the graph the range picks back up from the end of the juncture and keeps going untill it makes another evasive stop. Is this correct?

Thanks
• October 23rd 2006, 07:58 AM
ThePerfectHacker
Quote:

Originally Posted by qbkr21
Ok so correct me if I'm wrong ,but from what I can tell in looking at the picture you want to see how the range correlates to domain. When looking up and over on the picture if you see a juncture where there is no correlation and a space then the range has stopped. Then as you progress even further upward on the graph the range picks back up from the end of the juncture and keeps going untill it makes another evasive stop. Is this correct?

Thanks

Take a ruler and move it horizontally. The lines for which it does not intersect the curve are not in the range. For example for any point $y>0$ the ruler will intersect at least once. For any point $y<0$ the ruler shall intersect at least once. But for $y=0$ it does not intersect (though it gets close to it).

The answer is simple, because for $y=0$ we have,
$0=\frac{1}{x-2}$
The only way this can be zero is when the numerator is zero which it is not, so no value of $x$ in the domain makes this statement true.
• October 23rd 2006, 08:02 AM
qbkr21
Re:
Ok great thanks this is exactly what I needed