# Domain and Range Functions

• June 4th 2008, 03:12 PM
snakeman11689
Domain and Range Functions
I have this math problem that needs me to find the Domain:

F(x)=2x^3-x^2+3

How do you find the Domain or Range

I noticed some range or domains look something like this:

(2, infinity]
[2, infinity]
(infinity, 1) U (1, infinity)<------ this one confuses me a bit
• June 4th 2008, 03:34 PM
Mathstud28
Quote:

Originally Posted by snakeman11689
I have this math problem that needs me to find the Domain:

F(x)=2x^3-x^2+3

How do you find the Domain or Range

I noticed some range or domains look something like this:

(2, infinity]
[2, infinity]
(infinity, 1) U (1, infinity)<------ this one confuses me a bit

The first one is a function that is defined everywhere after two, but everywhere before and at two make the function undefined
consider
$\sqrt{x-2}\ln^2(x-2)$

For the second one it is just a function that is undefined everywhere less than two. consider
$\sqrt{x-2}$

For the hred one it woul be someting defined for all numbers except one

like

$\frac{1}{1-x}$
• June 4th 2008, 03:48 PM
snakeman11689
Quote:

Originally Posted by Mathstud28
The first one is a function that is defined everywhere after two, but everywhere before and at two make the function undefined
consider
$\sqrt{x-2}\ln(x-2)$

For the second one it is just a function that is undefined everywhere less than two. consider
$\sqrt{x-2}$

For the hred one it woul be someting defined for all numbers except one

like

$\frac{1}{1-x}$

I still don't quite understand that
• June 4th 2008, 03:55 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
The first one is a function that is defined everywhere after two, but everywhere before and at two make the function undefined
consider
$\sqrt{x-2}\ln(x-2)$

For the second one it is just a function that is undefined everywhere less than two. consider
$\sqrt{x-2}$

For the hred one it woul be someting defined for all numbers except one

like

$\frac{1}{1-x}$

Quote:

Originally Posted by snakeman11689
I still don't quite understand that

Ok for the first one we see that for any value less than two say x=1 if we put it in the function we get

$\sqrt{1-2}\ln^2(1-2)=\sqrt{-1}\cdot\ln^2(-1)=i(\pi{i})^2=-\pi^2i$

and this is obviously not a real number, so any numberless than two will cause a non-real result, therefore it is out of the domain of these numbers

For the second one the same thing applies

if lets say x=0 we have $\sqrt{0-2}=\sqrt{-2}=\sqrt{2}i$

once again not a real number, the same thing occurs with any other number less than two

For the third one everything on the reals makes the function defined except x=1

the reason being if you put one in we get

$\frac{1}{1-1}=\frac{1}{0}$

and as we both know we cannot divide by zero
• June 4th 2008, 04:35 PM
snakeman11689
Quote:

Originally Posted by Mathstud28
Ok for the first one we see that for any value less than two say x=1 if we put it in the function we get

$\sqrt{1-2}\ln^2(1-2)=\sqrt{-1}\cdot\ln^2(-1)=i(\pi{i})^2=-\pi^2i$

and this is obviously not a real number, so any numberless than two will cause a non-real result, therefore it is out of the domain of these numbers

For the second one the same thing applies

if lets say x=0 we have $\sqrt{0-2}=\sqrt{-2}=\sqrt{2}i$

once again not a real number, the same thing occurs with any other number less than two

For the third one everything on the reals makes the function defined except x=1

the reason being if you put one in we get

$\frac{1}{1-1}=\frac{1}{0}$

and as we both know we cannot divide by zero

gotcha!! thanks!!