# Functions..

• Jun 12th 2008, 04:31 PM
eh501
Functions..
I have no idea what to do here, can anyone help?:

{$\displaystyle {(x,y)|y=2x+1}$}

I have to find the Domain, Range and Inverse.
• Jun 12th 2008, 04:36 PM
topsquark
Quote:

Originally Posted by eh501
I have no idea what to do here, can anyone help?:

{$\displaystyle {(x,y)|y=2x+1}$}

I have to find the Domain, Range and Inverse.

$\displaystyle y = 2x + 1$

Domain: Are there any restrictions for x in this expression?
Range: Given the domain, are there any values that y cannot take on?

To find the inverse we must first determine that this is 1 to 1 (which it is.) Then:
$\displaystyle y = 2x + 1$

Reverse the roles of x and y. (This effectively reflects the function over the line y = x.)
$\displaystyle x = 2y + 1$

Solve for y:
$\displaystyle y = \frac{x - 1}{2}$

-Dan
• Jun 12th 2008, 04:39 PM
eh501
Quote:

Originally Posted by topsquark
$\displaystyle y = 2x + 1$

Domain: Are there any restrictions for x in this expression?
Range: Given the domain, are there any values that y cannot take on?

I still dont get it sorry.. I know the domain and range are all real numbers i just need to know how to find it.
• Jun 12th 2008, 04:44 PM
topsquark
Quote:

Originally Posted by eh501
I still dont get it sorry.. I know the domain and range are all real numbers i just need to know how to find it.

Is there any real number that makes $\displaystyle 2x + 1$ undefined or otherwise impossible to calculate? No. Thus the domain is all real numbers.

The situations you need to worry about are functions with vertical asymptotes or that are undefined. For example
$\displaystyle f(x) = \frac{g(x)}{h(x)}$
is undefined for any values of x that make h(x) = 0. Similarly we can't have any negative numbers or zero in $\displaystyle ln(x)$. We also have the case for any "even" radicals that we forbid negative numbers: $\displaystyle x^{1/2} = \sqrt{x}$, or $\displaystyle x^{1/4} = \sqrt[4]{x}$, etc. Since your function contains none of these expressions there is nothing to worry about.

-Dan
• Jun 12th 2008, 04:48 PM
eh501
Ok so does that mean that the answer for both the Domain and Range for this question is just "all real #'s"?
• Jun 12th 2008, 04:55 PM
eh501
And i have another question on functions:

"Given the function, $\displaystyle g(x)=4x-5$, and it's domain, {$\displaystyle x|0<x<4$}, find the range.

What do i do?
• Jun 12th 2008, 04:56 PM
topsquark
Quote:

Originally Posted by eh501
Ok so does that mean that the answer for both the Domain and Range for this question is just "all real #'s"?

In a word, yes. :)

-Dan
• Jun 12th 2008, 04:58 PM
topsquark
Quote:

Originally Posted by eh501
And i have another question on functions:

"Given the function, $\displaystyle g(x)=4x-5$, and it's domain, {$\displaystyle x|0<x<4$}, find the range.

What do i do?

The domain is all values of x from 0 to 4. As this is a linear function the range is easy: just calculate g(x) for the end points of the interval and that is your range.

This is slightly more complicated for a non-linear function. For example, using the same domain,
$\displaystyle h(x) = \frac{1}{x - 2}$
has a range of all real numbers. The best strategy that I have is to graph the function and visually determine the range.

-Dan
• Jun 12th 2008, 05:06 PM
eh501
Im still having trouble finding the range. Do i sub in $\displaystyle x$ with anything?
• Jun 12th 2008, 05:11 PM
topsquark
Quote:

Originally Posted by eh501
Im still having trouble finding the range. Do i sub in $\displaystyle x$ with anything?

Quote:

Originally Posted by topsquark
The domain is all values of x from 0 to 4. As this is a linear function the range is easy: just calculate g(x) for the end points of the interval and that is your range.

Your endpoints are 0 and 4. So g(0) = ? g(4) = ?

And, of course, note that your domain is $\displaystyle (0, 4)$ which does not include either x = 0 or x = 4 so you have to make sure you write your range interval to not include those endpoints.

-Dan
• Jun 12th 2008, 05:19 PM
eh501
Ohh so it would be {$\displaystyle g|-5<x<11$} right?
• Jun 13th 2008, 02:59 AM
topsquark
Quote:

Originally Posted by eh501
Ohh so it would be {$\displaystyle g|-5<x<11$} right?

Yes. :)

-Dan