# Finding functions.

• Jul 2nd 2013, 01:10 PM
Leviathantheesper
Finding 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:
$f(-3x^2-12x-9)=4x^2+6x$
• Jul 2nd 2013, 02:20 PM
emakarov
Re: Finding functions.
https://s3.amazonaws.com/grapher/exports/sjaerkthfj.png

Let $g(x) = -3x^2 - 12x - 9$ and $h(x) = 4x^2 + 6x$. 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 $g^{-1}(y)$ such that $g(g^{-1}(y))=y$ for all $y\in\mathbb{R}$. Then we could represent any y as $g(g^{-1}(y))$ and get $f(y)=f(g(g^{-1}(y)))=h(g^{-1}(y))$. But in this case, g(x) is neither surjective (if the codomain is $\mathbb{R}$) 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 $g^{-1}(y)$ 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 $g^{-1}(y)$. 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, $f(y) = h(g^{-1}(y))$ 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.
• Jul 2nd 2013, 05:41 PM
Prove It
Re: Finding functions.
Quote:

Originally Posted by Leviathantheesper
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:
$f(-3x^2-12x-9)=4x^2+6x$

Let \displaystyle \begin{align*} u = -3x^2 - 12x - 9 \end{align*}, then we have

\displaystyle \begin{align*} -\frac{u}{3} &= x^2 + 4x + 3 \\ \frac{-u-9}{3} &= x^2 + 4x \\ \frac{-u-9}{3} + 2^2 &= x^2 + 4x + 2^2 \\ \frac{3 - u}{3} &= (x + 2)^2 \\ \pm \sqrt{ \frac{ 3 - u}{3} } &= x + 2 \\ \pm \frac{ \sqrt{ 9 - 3u } }{ 3 } &= x + 2 \\ \frac{ -6 \pm \sqrt{ 9 - 3u } }{3} &= x \end{align*}

So that means if you have \displaystyle \begin{align*} f \left( -3x^2 - 12x - 9 \right) &= 4x^2 + 6x \end{align*}, then Case 1:

\displaystyle \begin{align*} f \left( -3x^2 - 12x - 9 \right) &= 4x^2 + 6x \\ &= 4 \left( \frac{ -6 - \sqrt{ 9 - 3u }}{3} \right) ^2 + 6 \left( \frac{ -6 - \sqrt{ 9 - 3u } }{3} \right) \\ &= 4 \left( \frac{36 + 12 \sqrt{9 - 3u} + 9 - 3u }{9} \right) - 12 - 2\sqrt{ 9 -3u} \\ &= \frac{60 + 16\sqrt{9 - 3u} - 4u}{3} + \frac{-36 - 6\sqrt{9 - 3u}}{3} \\ &= \frac{24 + 10\sqrt{9 - 3u} - 4u}{3} \\ &= \frac{ 24 + 10 \sqrt{9 - 3 \left( -3x^2 - 12x - 9 \right) } - 4 \left( -3x^2 - 12x - 9 \right) }{3} \end{align*}

And from there we can now see that \displaystyle \begin{align*} f(x) = \frac{24 + 10\sqrt{ 9 - 3x} - 4x }{3} \end{align*}. Now see how you go with Case 2.
• Jul 5th 2013, 01:08 PM
Leviathantheesper
Re: Finding functions.
Thanks. What I wanted to do is to convert $(a,f(a))$ into $(a^2-f(a)^2,2af(a))$ 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.
• Jul 6th 2013, 07:45 PM
Leviathantheesper
Re: Finding functions.
I have done something more general to transform:

$(x,ax+b)$ into $(x^2-(ax+b)^2,2x(ax+b))$.

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.