Hi. I have a problem. I don't know how to solve these kind of equations...and I don't know if the info is enough to solve them...could you help me?

Find f(x) given:

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- July 2nd 2013, 02:10 PMLeviathantheesperFinding functions.
Hi. I have a problem. I don't know how to solve these kind of equations...and I don't know if the info is enough to solve them...could you help me?

Find f(x) given:

- July 2nd 2013, 03:20 PMemakarovRe: Finding functions.
https://s3.amazonaws.com/grapher/exports/sjaerkthfj.png

Let and . We need to find f(x) such that f(g(x)) = h(x).

If g(x) were a bijection, then there would exist a unique such that for all . Then we could represent any y as and get . But in this case, g(x) is neither surjective (if the codomain is ) nor injective. More precisely, g(x) is not surjective because g(x) ≤ 3 for all x, so we have no information about f(y) when y is not in the image of g, i.e., when y > 3. And g(x) is not injective because there are two possible values of for each y < 3. Unfortunately, h(x) returns different results for these two values. This means we can't assign f(y) that would work for both possible values of . For example, g(-1) = g(-3) = 0, so f(0) has to equal both h(-1) = -2 and h(-3) = 18. If we set f(0) = -2, then f(g(x)) = h(x) is satisfied for x = -1 but not for x = -3, and if we set f(0) = 18, then f(g(x)) = h(x) works for x = -3, but not for x = -1.

The most we can do is to define, say, for y ≤ 3 using the right branch of g(x), i.e., for x ≥ -2. Then f(g(x)) = h(x) will hold for x ≥ -2. - July 2nd 2013, 06:41 PMProve ItRe: Finding functions.
- July 5th 2013, 02:08 PMLeviathantheesperRe: Finding functions.
Thanks. What I wanted to do is to convert into and convert the result into some functions, so I can plot it. The way you solve that is useful, so thanks. I hadn't thought it but converting it into parametric equations is useful too for the plotting software.

- July 6th 2013, 08:45 PMLeviathantheesperRe: Finding functions.
I have done something more general to transform:

into .

The result is:

Attachment 28748

And the procedure is here:

Attachment 28750

(Click on the pictures to see them and sorry for the spanish part).

Again, Thanks.

Sorry for the double post, for some reason I can't edit.