# Thread: Help Finding the Inverse!!!

1. ## Help 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.

2. 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.

3. 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.

4. 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.

5. Originally Posted by icemanfan
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.
Another way I think you can find out which is the inverse is that the domain and range just switch. f(x) = Domain A and range B and the inverse function of x equals Domain B and Range A. I'm not sure though.

6. Haven't you guys considered inyection and surjection? Does make sense to find the inverse?

7. Originally Posted by 11rdc11
Another way I think you can find out which is the inverse is that the domain and range just switch. f(x) = Domain A and range B and the inverse function of x equals Domain B and Range A. I'm not sure though.
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.

8. 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

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

$\displaystyle y=x$