Finding an "inverse function" symmetrical to y=2x not y=x

Question

Hello I'm very inexperienced in math (I know a little about derivation/integrals etc but nothing on university level) so my terminology will not be on point (as well due to english not being my native language). I have a function

$$ y=\sqrt2*\sqrt x $$ graph 1 and 2

and I want to find the symmetrical function by the function $y=2x$ like this:

[in previous file on the right]

as far as I know the inverse function is always symmetrical by the function $y=x$. so I dont know how to call this. Expected result:

graph 3 and 4 I'm doing this because I wanna to "connect" these function so I can "draw" this

[in previous file on the right]

I can only post 2 links so I have combined graph 1 and 2 into one image and 3 and 4 into another one. Thank you in advance for responding! <3

Answer 1

There is actually quite a trivial way to do this. So as you know the inverse swaps the $x$ and $y$ values and is therefore symmetric to the line $y=x$, and note we get the inverse by plugging in $y$ for $x$ and vise versa.
Now if we want to make it symmetric to the line $y=2x$, we just have to exchange all the $y$s for $2x$, and all the $2x$s for $y$. So we have $$y=\sqrt{2x} \Rightarrow 2x=\sqrt{y}$$ or $$y=4x^2, x\ge 0$$ Note: We constrict the domain because in the original function, defined implicitly by $2x=\sqrt{y}$, $2x$ must be greater than $0$ for there to ever be a corresponding $y$ value. And if you graph it it will make even more sense.

Answer 2

The reflection in a line $y=x\tan \theta$ is represented by the matrix

\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}

In your case $\tan \theta =2$. Now using so called t-substitution, letting $t=\tan\theta$ we have

$$ \cos 2\theta = \frac{1-t^2}{1+t^2} \; \; \text{and} \;\; \sin 2\theta = \frac{2t}{1+t^2} $$

So the matrix is

\begin{pmatrix} -3/5 & 4/5 \\ 4/5 & 3/5 \end{pmatrix}

Hence position vector of reflected points are

$$ \begin{pmatrix} -3/5 & 4/5 \\ 4/5 & 3/5 \end{pmatrix} \begin{pmatrix} x\\ \sqrt{2x} \end{pmatrix} = \begin{pmatrix} -\frac{3}{5}x + \frac{4}{5} \sqrt{2x}\\ \frac{4}{5} x + \frac{3}{5} \sqrt{2x} \end{pmatrix} $$

The plot of this on Wolfram alpha shows that it is not a function: http://www.wolframalpha.com/input/?i=plot+%28x%2Cy%29%3D+%28-3t%2F5+%2B4sqrt%282t%29+%2F5%2C+4t%2F5+%2B3sqrt%282t%29+%2F5%29

Edit: Also this parametric curve interset $y=\sqrt{2x}$ at two different points, also computed by Wolfram alpha - http://www.wolframalpha.com/input/?i=plot+%28x%2Cy%29%3D+%28-3t%2F5+%2B4sqrt%282t%29+%2F5%2C+4t%2F5+%2B3sqrt%282t%29+%2F5%29

So the part of graph which you want is parameterized by paramters lying between these two solutions.

Answer 3

In order to reflect about this line, we should put this equation in the context of the line:

$$y = \sqrt{2x} \to y^2-2x=0\\ y = 2x\to y-2x=0$$

Line $u$ would be the axis of reflection, for $y-2x = 0$, line $v$ would be the perpendicular, with $2y+x = 0$. In terms of $u$ and $v$, we have

$$\left(\frac 15u+\frac25v\right)^2-2\left(\frac15v-\frac25u\right)=y^2-2x$$

Now a reflection about the line $v$ is simply a matter of negating $u$, giving us

$$\left(\frac 15u-\frac25v\right)^2-\frac25v-\frac45u=\left(\frac35y+\frac45x\right)^2-\frac85y+\frac65x$$

This can be algebraically expanded further as desired.

By comparison, here is the graph of the suggested function $y=4x^2$ vs. $y^2=2x$:

(Graph testing done using https://www.desmos.com/calculator.)