# Finding inverse of functions[methods of]

I am now trying to understand functions, inverses and composites. I must admit am not getting a thing. But following some leads, I managed to work one as below. Is this a good understanding on hows and whys?

## Find the inverse of function $y=\sqrt{6+x}$

$y=\sqrt{6+x}$ can be expanded as $y^2=6+x$

solving for $x$ we get $x=y^2-6$

swapping the position of $x$ and $y$ we get $y=x^2-6$

$\therefore$ the inverse of $y=\sqrt{6+x}$ is $$y=x^2-6$$

-

Careful here, your work is actually correct (the swapping method is in fact standard), except that you must keep track of your domain. The inverse of $y=\sqrt{6+x}$ is $y=x^2-6$, and your work is absolutely correct. However the domain on $\sqrt{6+x}$ is $x \geq -6$, hence on the inverse, we have that $y \geq -6$.

Now for the inverse $y=x^2-6$, since $y \geq -6$, it follows that $x \geq 0$ as, $$x^2 -6 \geq -6 \Rightarrow x^2 \geq 0 \Rightarrow x \geq 0.$$ Now if you plot the two solutions you will have something that looks like this.

If you do not keep track of your domains and ranges like this, then your work is wrong as the inverse must be 1-1, and the inverse of the inverse must also be 1-1. The inverse of the inverse must give the function you started with on the correct domain.

-
Thank you @J.W.Perry –  Sylvester Oct 6 '13 at 14:48
@Sylvester You are quite welcome. –  J. W. Perry Oct 6 '13 at 14:48

That's almost it:

\begin{align} y&=\sqrt{6+x}\\ y^2&=6+x\\ x&=y^2-6 \end{align}

And that's your answer. You don't "swap" x and y positions, you just solve for x or solve for y.

Many inverses are rather simple... Take for example:

$$y=x\iff x=y$$

When you have more complicated things like $$y=x^2$$ then you have $$x=\sqrt{y}$$

More complicated functions have defined inverses like $$y=10^x\iff x=\log_{10} y$$

and

$$y=e^x\iff x=\ln y$$

Then there are the trig functions:

$$y=\sin x\iff x=\arcsin y$$ $$y=\cos x\iff x=\arccos y$$ $$y=\tan x\iff x=\arctan y$$

At an elementary level, this is pretty much all you need to know.

-
so you imply my workings look pretty correct and follows the correct method? Given that I picked this method from some site, I'd thought it belongs to some cook book. –  Sylvester Oct 6 '13 at 13:49
You were correct up until you decided to swap $x$ for $y$. You just need to solve for $x$ or $y$ and then you're done. –  agent154 Oct 6 '13 at 13:50
I don't know what level you're at, but be wary that the inverse of a function is not always a function due to the limitations in the definition of a function. It may not pass the "vertical line test" when drawn. Therefore, many inverse functions are only defined over specific intervals. –  agent154 Oct 6 '13 at 13:52
am still at precalculus level. Just started on this topic. I am trying to get myself going, meticulously but correctly. But am not understanding your comment about " not swapping" –  Sylvester Oct 6 '13 at 13:56
I thought the swapping method is a "standard" because I used it to solve for $y^2=6=x$ and getting my solution as $y=x^2-6$ and solving for $x$ and swapping. –  Sylvester Oct 6 '13 at 14:00