another controversial kilbaha thread :P (help required)

[Whatlol]

i came across this question in kilbaha 2006 exam. Working out the inverse function is not a problem, but determining whether to take the positive and negative root has worked up some dilema formyself. any clarification will be much appreciated.

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/]

[brightsky]

Since the domain of is , therefore hence positive root. (Is this what you're refering to?)

[8039]

Isn't that just asking you to make it fall within the limits of the specified domain? (as brigthsky suggested)

[Nomvalt]

lol, funny signature 8039
are you really gonna repeat those subjects?

[Whatlol]

both the positive and negative solutions have positive and negative y values.... so how can you eliminate them using this logic. I think i am misunderstanding it.

[kenhung123]

Since the domain of is , therefore hence positive root. (Is this what you're refering to?)

Yea, 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?

[Duck]

The range of f(x) is from 0 to infinity hence the domain of the inverse is from 0 to infinity.

[kenhung123]

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

[brightsky]

The range of is (0, 1/2]. So the domain of must 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.

EDIT: Thanks Martoman.

[HERculina]

BRIGHTSKY IS A YEAR NINE.

[Martoman]

The range of is (0, 1/2]. So the domain of must 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]

[Whatlol]

The range of is (0, 1/2]. So the domain of must 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]

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.

[brightsky]

The range of is (0, 1/2]. So the domain of must 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]

Gah, silly mistake. Positive 1 yes so in that case the inverse is ACTUALLY defined to be consisting of both roots, i.e.

[Whatlol]

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 

[brightsky]

It seems so. :p