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