# Explain how to verify inverse function

Let $f(x) = 3x -1$

Can someone explain how to verify $f[f^{-1}(x)] = x$ and $f^{-1}[f(x)] = x$, each for $x$ in the appropriate domain?

I was able to determine that the inverse function of $f(x) = 3x - 1$ is $\frac {x + 1}{3}$.

Do I just substitute in the known equations and solve?

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Substitute and simplify. If your determination was carefully done, there is no formal need for a verification, though it can be useful as a check. –  André Nicolas Oct 25 '11 at 14:48
"0% accept rate" - have none of your previous questions been resolved to your desire? –  Ｊ. Ｍ. Oct 25 '11 at 14:49
Please note that the equations you are trying to verify automatically hold, if you found the correct inverse (and you did). So I would change notation slightly, and write as follows. Let $g(x)=(x+1)/3$. We show that $f(g(x))=x$ for all $x\in \mathbb{R}$, and $g(f(x))=x$ for all $x\in\mathbb{R}$. I am assuming $f$ is defined on the reals, things will be quite different if our function $f$ is from the integers to the integers. –  André Nicolas Oct 25 '11 at 15:02
@J.M. Didn't realize I could do that, thanks for pointing it out. –  erimar77 Oct 25 '11 at 15:57
No problem; it's kinda important here as feedback, you see. The check mark is our way of knowing that you were satisfied by the answer you have marked. –  Ｊ. Ｍ. Oct 25 '11 at 16:00

You just write the "instructions" $f(f^{-1}(x))$ and $f^{-1}(f(x))$. That is,
$$f(f^{-1}(x)) = f\left( \frac{x+1}{3} \right) = 3\frac{x+1}{3} -1 = x + 1 -1 = x \ .$$
And similarly for $f^{-1}(f(x))$.