# Maximal domain and range

• Apr 12th 2010, 03:30 PM
matlabnoob
Maximal domain and range
im very vERY confused..

first of all,can anyone confirm to me if 'maximal' domain and 'domain' are the same?(Headbang)

i am trying to find maximal domain & range of these on the real line:

F(x) = x² + 2x – 3
F(x) = √ x² + 2x – 3
F(x) = 1/√ x² + 2x – 3
F(x) = (x² + 2x – 3) -⅓

aren't they all going to be the same?

factorising you get = (x+3)(x-1)
so domain is the whole real line except -3, 1
and range will be whole real line...! where am i going wrong.....

thankyou!
• Apr 12th 2010, 03:31 PM
stapel
How does your book define "maximal domain"?

Thank you! (Wink)
• Apr 12th 2010, 03:34 PM
matlabnoob
my notes is very vague in the definition of maximal domain, which is why im confused. it says this:
'The maximal domain is the set of points at which the rule f could legitimately be used'

which sounds like domain to me(Thinking)
• Apr 12th 2010, 03:40 PM
stapel
That's sure what it sounds like to me, too. (Wondering)

All I can find online are references to "maximal domain" being what people usually mean by "domain", with any difference maybe being between the mathematical domain (being anything allowable mathematically) and the practical domain (being, say, only non-negative values of the radius for an "area" function).
• Apr 12th 2010, 03:45 PM
matlabnoob
Quote:

Originally Posted by stapel
That's sure what it sounds like to me, too. (Wondering)

All I can find online are references to "maximal domain" being what people usually mean by "domain", with any difference maybe being between the mathematical domain (being anything allowable mathematically) and the practical domain (being, say, only non-negative values of the radius for an "area" function).

thanks

so maximal is just an extra word to make it sound complicated. uggh!
does that mean i am right?? so the domain for all of the functions are the whole real line (Worried) except -3 and 1. the only thing that is confusing me now is that these functions are separate problems... so they cant be the same answer! or can they(Wondering)
• Apr 13th 2010, 12:18 AM
HallsofIvy
No, "domain" and "maximal domain" are NOT the same.

For example, I can find $\frac{x-1}{x-2}$ for any value of x except 2 so its "maximal domain" is all "all real numbers except 2".

But If I specifically define " $f(x)= \frac{x-1}{x-2}$ with domain 1< x< 2 I now have a completely different function with "domain" [1, 2).

The "maximal domain" for a formula is the set of all numbers for which that formula can be evaluated. The "domain" of a function can be any subset of the maximal domain for whatever formula is used in defining the function. It often is the maximal domain, but any subset can be given as the domain for that specific function.

And, no, while the maximal domain is the same for these functions, the maximal domain is NOT "all real numbers except -3 and 1" for any of those functions.

Yes, $x^2+ 2x- 3= (x+ 3)(x- 1)$ but whatever x is, x+ 3 and x- 1 are just numbers and you can multiply any two numbers together. The "maximal domain" for this formula is "all real numbers".

Also, completing the square, $x^2+ 2x+ 1- 1+ 3= (x+ 1)^2+ 2$. Since a square is never negative, when x= -1, this give a value of 2 and for x anything except 1, it gives numbers larger than 2. It's range is "all real numbers larger than or equal to 2".

The only "difficulty" with $\sqrt{x^2+ 2x+ 3}$ would be that you cannot take the squareroot of a negative number. Since $x^2+ 2x+ 3$ is never smaller than 2, it is never negative. The maximal domain is still all real numbers and now the range is all real numbers larger than or equal to $\sqrt{2}$.

$\frac{1}{\sqrt{x^2+ 2x+ 3}$ and $(x^2+ 2x+ 3)^{-1/2}$ are exactly the same thing. Now the only possible "difficulty" is that we cannot take the square root of a negative number and we cannot divide by 0. But, again, $x^2+ 2x+ 3$ is never less than 2 so is never either negative or 0. Again, the maximal domain is "all real numbers". And, now, the range is $0< x\le \frac{1}{\sqrt{2}}$ since $\frac{1}{x}$ goes to 0 as x goes to infinity.