0
$\begingroup$

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?

  • 5
    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.2011-10-25
  • 2
    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.2011-10-25
  • 0
    @J.M. Didn't realize I could do that, thanks for pointing it out.2011-10-25
  • 2
    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.2011-10-25

1 Answers 1

3

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 $f^{-1}(f(x))$.