My Math Forum Domain !

 Algebra Pre-Algebra and Basic Algebra Math Forum

 December 16th, 2010, 08:03 AM #1 Newbie   Joined: Nov 2010 Posts: 26 Thanks: 0 Domain ! find the domain 1\ root(4x^3-3x^2-x) 2\root(x^2+3x-4) × root ( 2-x-x^2) 3\ root(x^2+3x-4)\root(x-2)^2
December 16th, 2010, 08:16 AM   #2
Newbie

Joined: Dec 2010

Posts: 26
Thanks: 0

Re: Domain !

Quote:
 Originally Posted by ahmed-ar find the domain 1\ root(4x^3-3x^2-x) 2\root(x^2+3x-4) × root ( 2-x-x^2) 3\ root(x^2+3x-4)\root(x-2)^2
What's the condition for the function in the radical to be defined?

December 16th, 2010, 08:24 AM   #3
Newbie

Joined: Nov 2010

Posts: 26
Thanks: 0

Re: Domain !

Quote:
Originally Posted by ace
Quote:
 Originally Posted by ahmed-ar find the domain 1\ root(4x^3-3x^2-x) 2\root(x^2+3x-4) × root ( 2-x-x^2) 3\ root(x^2+3x-4)\root(x-2)^2
What's the condition for the function in the radical to be defined?
as i know x should be equal or bigger than 0 but i don't know how to find the domain!

December 16th, 2010, 08:26 AM   #4
Newbie

Joined: Dec 2010

Posts: 26
Thanks: 0

Re: Domain !

[quote=ahmed-ar]
Quote:
Originally Posted by ace
Quote:
 Originally Posted by "ahmed-ar":26i3ji5f find the domain 1\ root(4x^3-3x^2-x) 2\root(x^2+3x-4) × root ( 2-x-x^2) 3\ root(x^2+3x-4)\root(x-2)^2
What's the condition for the function in the radical to be defined?
as i know x should be equal or bigger than 0 but i don't know how to find the domain![/quote:26i3ji5f]

Yes.

(1) 4x^3-3x^2-x>=0

Can you solve this? The solutions are the domain of this function. Do the same to the rest and take the intersection of the solutions.

 December 16th, 2010, 08:32 AM #5 Newbie   Joined: Nov 2010 Posts: 26 Thanks: 0 Re: Domain ! i know how to solve but how to write the domain i will find more then one answer of x
December 16th, 2010, 08:41 AM   #6
Newbie

Joined: Dec 2010

Posts: 26
Thanks: 0

Re: Domain !

Quote:
 Originally Posted by ahmed-ar i know how to solve but how to write the domain i will find more then one answer of x
The same way you would write the solutions of an inequality.

If you were to write it in solution sets, {x: x>a or x<-b for x is all reals}. I am sure you can google this or refer to the examples in your textbook.

December 16th, 2010, 08:56 AM   #7
Newbie

Joined: Nov 2010

Posts: 26
Thanks: 0

Re: Domain !

Quote:
Originally Posted by ace
Quote:
 Originally Posted by ahmed-ar i know how to solve but how to write the domain i will find more then one answer of x
The same way you would write the solutions of an inequality.

If you were to write it in solution sets, {x: x>a or x<-b for x is all reals}. I am sure you can google this or refer to the examples in your textbook.
i haven't got it!

December 16th, 2010, 09:00 AM   #8
Newbie

Joined: Dec 2010

Posts: 26
Thanks: 0

Re: Domain !

[quote=ahmed-ar]
Quote:
Originally Posted by ace
Quote:
 Originally Posted by "ahmed-ar":2bm9pdw7 i know how to solve but how to write the domain i will find more then one answer of x
The same way you would write the solutions of an inequality.

If you were to write it in solution sets, {x: x>a or x<-b for x is all reals}. I am sure you can google this or refer to the examples in your textbook.
i haven't got it![/quote:2bm9pdw7]

Lets say the domain is x>3. You can write,

(1) The domain is (3 , infinity)

(2) The domain is x>3, $x\in R$

(3) {x>3 , $x\in R$}

These are the few ways of representing a domain. Is this what you are asking?

 December 16th, 2010, 09:14 AM #9 Newbie   Joined: Nov 2010 Posts: 26 Thanks: 0 Re: Domain ! thanks and what about if i got more than one value of x? for example in first question i will get x(4x+1)(x-1)>=0 so what will be the answer ?
December 16th, 2010, 09:19 AM   #10
Newbie

Joined: Dec 2010

Posts: 26
Thanks: 0

Re: Domain !

Quote:
 Originally Posted by ahmed-ar thanks and what about if i got more than one value of x? for example in first question i will get x(4x+1)(x-1)>=0 so what will be the answer ?
It's simply the domain of the inequality. The way of writing it is the same.

 Tags domain

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post sheran Algebra 6 July 17th, 2013 09:27 AM panky Calculus 1 October 5th, 2011 04:46 AM B521 Calculus 8 July 22nd, 2010 08:09 AM rhta Algebra 4 October 29th, 2009 07:30 PM good_phy Calculus 1 October 12th, 2008 10:26 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top