Hi i have a question to find the inverse, and i am kinda stuck. Can anyone please help me.

Y = Srt of X + 2X

Find the inverse.

Thanks in advance =]

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- Oct 24th 2008, 01:24 PMferkenHelp Finding the Inverse!!!
Hi i have a question to find the inverse, and i am kinda stuck. Can anyone please help me.

Y = Srt of X + 2X

Find the inverse.

Thanks in advance =] - Oct 24th 2008, 01:36 PMicemanfan
Question: Find the inverse of $\displaystyle y = \sqrt{x} + 2x$.

The first step is to switch x and y:

$\displaystyle x = \sqrt{y} + 2y$

Then it's time for a little equation rearrangement:

$\displaystyle x - 2y = \sqrt{y}$

$\displaystyle (x - 2y)^2 = y$

$\displaystyle x^2 - 4xy + 4y^2 = y$

$\displaystyle 4y^2 - 4xy - y + x^2 = 0$

$\displaystyle 4y^2 - (4x + 1)y + x^2 = 0$

$\displaystyle y = \frac{4x + 1 \pm \sqrt{(4x + 1)^2 - 4(4)(x^2)}}{8}$

$\displaystyle y = \frac{4x + 1 \pm \sqrt{(4x + 1)^2 - 16x^2}}{8}$

$\displaystyle y = \frac{4x + 1 \pm \sqrt{16x^2 + 8x + 1 - 16x^2}}{8}$

$\displaystyle y = \frac{4x + 1 \pm \sqrt{8x + 1}}{8}$

Now I'm not sure how to proceed with this, but perhaps someone else can give a little insight. - Oct 24th 2008, 01:39 PMferken
Thanks!

However, i actually figured it out all the way up to that part. I know the inverse is also a function, but i do not know how to figure out which answer i reject and which answer to accept. - Oct 24th 2008, 01:48 PMicemanfan
I think, after thinking about it, that the correct solution is $\displaystyle y = \frac{4x + 1 - \sqrt{8x + 1}}{8}$. My reasoning is that the original function $\displaystyle y = \sqrt{x} + 2x$ is always above the line $\displaystyle y = 2x$, so the inverse function must always be below the line $\displaystyle y = 0.5x$. Note that the inverse is only defined for positive values of x, because the original function, which is also defined for only positive values of x, always produces positive outputs.

- Oct 24th 2008, 01:57 PM11rdc11
- Oct 24th 2008, 02:01 PMKrizalid
Haven't you guys considered inyection and surjection? Does make sense to find the inverse?

- Oct 24th 2008, 02:25 PMicemanfan
This method of checking confirms that my intuition was correct. For example: (4, 10) is a point on the original function, and (10, 4) is a point on the inverse. (1, 3) is a point on the original function, and (3, 1) is a point on the inverse. Edit: 11rdc11, your proof is comprehensive and thank you.

- Oct 24th 2008, 02:26 PM11rdc11
Also remember that

$\displaystyle f(f^{-1}(x))=x$

So

$\displaystyle y = \frac{4x + 1 - \sqrt{8x + 1}}{8}$

is correct because

$\displaystyle y = \frac{4(\sqrt{x} + 2x) + 1 - \sqrt{8( \sqrt{x} + 2x)+ 1}}{8}$

$\displaystyle y = \frac{4\sqrt{x} + 8x + 1 - \sqrt{8\sqrt{x} + 16x+ 1}}{8}$

let

$\displaystyle u = \sqrt{x}$

$\displaystyle \frac{8u^2+4u+1 - \sqrt{16u^2+8u+1}}{8}$

$\displaystyle \frac{8u^2+4u+1 - \sqrt{(4u+1)^2}}{8}$

$\displaystyle \frac{8u^2+4u+1 - (4u+1)}{8}$

$\displaystyle \frac{8u^2}{8}$

$\displaystyle u^2$

backsub in for u

$\displaystyle y = (\sqrt{x})^2$

$\displaystyle y=x$