Post by: Whatlol on October 06, 2010, 11:32:12 pm
In particular i am curious to know why you can eliminate the negative solution by saying y > 0.
Thanks.
[URL=http://img59.imageshack.us/i/kilbaha2006q1e.png/](http://img59.imageshack.us/img59/317/kilbaha2006q1e.png)
Post by: brightsky on October 07, 2010, 12:00:11 am
Post by: 8039 on October 07, 2010, 12:29:21 am
Post by: Nomvalt on October 07, 2010, 03:31:44 am
are you really gonna repeat those subjects?
Post by: Whatlol on October 07, 2010, 05:12:56 pm
Post by: kenhung123 on October 08, 2010, 05:21:10 pm
Since the domain ofYea, that is true but both inverse equations give sections y>0 and y<0. Is it just a convention that y>0 therefore positive root?is
, therefore
hence positive root. :) (Is this what you're refering to?)
Post by: Duck on October 08, 2010, 06:30:32 pm
Post by: kenhung123 on October 08, 2010, 06:33:37 pm
The range of f(x) is from 0 to infinity hence the domain of the inverse is from 0 to infinity.Lol, yes we know that. But this particular inverse contains a positive and negative root that produces y>0 and y<0 in both equations when looking at the graph
Post by: brightsky on October 08, 2010, 08:02:51 pm
We know that y > 0 so with the given domain, only the positive root is valid because 1) x cannot be less than 0; 2) The square root can't be negative; 3) For y to be positive, the numerator and denominator must be either both positive or both negative. If we go with the negative root, we can visualise the equation as:
This clearly can't be positive as the numerator is negative and there is no negative in the denominator to compensate for it.
EDIT: Thanks Martoman.
Post by: HERculina on October 08, 2010, 08:09:16 pm
Post by: Martoman on October 08, 2010, 08:42:05 pm
The range ofmust be (0,1/2]. With some algebra we find that the inverse function is given by:
We know that y > 0 so with the given domain, only the positive root is valid because 1) x cannot be less than 0; 2) The square root can't be negative; 3) For y to be positive, the numerator and denominator must be either both positive or both negative. If we go with the negative root, we can visualise the equation as:
This clearly can't be positive as the numerator is negative and there is no negative in the denominator to compensate for it.
I don't agree for starters its POSITIVE 1 +- blah blah
Restricting the domain to (0,1]
we have range (0,0.5]
This means on our inverse we have a domain (0,0.5] and a range of (0,1]
If you graph the -ve root you find that it is defined for the range (0,1] but the +ve root only is at the point [0.5,1]
Post by: Whatlol on October 08, 2010, 09:28:05 pm
The range of
We know that y > 0 so with the given domain, only the positive root is valid because 1) x cannot be less than 0; 2) The square root can't be negative; 3) For y to be positive, the numerator and denominator must be either both positive or both negative. If we go with the negative root, we can visualise the equation as:
This clearly can't be positive as the numerator is negative and there is no negative in the denominator to compensate for it.
I don't agree for starters its POSITIVE 1 +- blah blah
Restricting the domain to (0,1]
we have range (0,0.5]
This means on our inverse we have a domain (0,0.5] and a range of (0,1]
If you graph the -ve root you find that it is defined for the range (0,1] but the +ve root only is at the point [0.5,1]
yes the problem is you can restrict the domain so you have a one to one function on either side
i.e either [0, 1] or [1 , infinity]
so if you take the range as (0,0.5] and domain to be (1, infinity] (for the original function)
then positive solution can exist?
but as you just said, restricting domain from (0,1]
the negative solution can exist.
Post by: brightsky on October 08, 2010, 09:42:11 pm
The range of
We know that y > 0 so with the given domain, only the positive root is valid because 1) x cannot be less than 0; 2) The square root can't be negative; 3) For y to be positive, the numerator and denominator must be either both positive or both negative. If we go with the negative root, we can visualise the equation as:
This clearly can't be positive as the numerator is negative and there is no negative in the denominator to compensate for it.
I don't agree for starters its POSITIVE 1 +- blah blah
Restricting the domain to (0,1]
we have range (0,0.5]
This means on our inverse we have a domain (0,0.5] and a range of (0,1]
If you graph the -ve root you find that it is defined for the range (0,1] but the +ve root only is at the point [0.5,1]
Gah, silly mistake. Positive 1 yes so in that case the inverse is ACTUALLY defined to be consisting of both roots, i.e.
Post by: Whatlol on October 08, 2010, 09:43:23 pm
Gah, silly mistake. Positive 1 yes so in that case the inverse is ACTUALLY defined to be consisting of both roots, i.e.
so now what happens... do we contact kilbaha :D
Post by: brightsky on October 08, 2010, 09:50:59 pm
Post by: Martoman on October 08, 2010, 10:16:24 pm