Q&A #6118

Teachers' Lounge Discussion: Domain and range of linear functions

T2T || FAQ || Ask T2T || Teachers' Lounge || Browse || Search || Thanks || About T2T

View entire discussion
[<< prev]

From: Loyd <loydlin@aol.com>
To: Teacher2Teacher Public Discussion
Date: 2002020309:10:13
Subject: Re: functions

On 2002020114:20:00, heather wrote:
>Why is the domain of the function 3*x^1/2 divided by 2*x^1/6, why is
>the domain all positive real number instead of all real numbers
>I would appreciate your help, thank you.
I thought your question was interesting.  YOur problem:
can be rewritten as:

If you the domain (x values) in the numerator are positive then the y
values (range) are all positive real numbers.  The same is true for
the denominator.  However, if the domain or x values are negative, the
numerator and denominator are imaginary values.  

However, If you re-write the equation as:

y= 3/2[x^(3/6-1/6)= 3/2[x^(1/3)

The last expression has a negative output in the range if the domain
is negative.  I hope someone at this site can explain why this
happens.  Try graphing the above on a TI-83 and you get two different
graphs; one is only in the first quadrant and the second is in 1st and
3rd quadrant.  

Maybe someone who has calculus limits fresh on their mind can explain
why this happens.

Post a reply to this message
Post a related public discussion message
Ask Teacher2Teacher a new question

[Privacy Policy] [Terms of Use]

Math Forum Home || The Math Library || Quick Reference || Math Forum Search

Teacher2Teacher - T2T ®
© 1994- Drexel University. All rights reserved.
The Math Forum is a research and educational enterprise of the Drexel School of Education.The Math Forum is a research and educational enterprise of the Drexel University School of Education.