1
$\begingroup$

Given the following two functions

$$ f: \mathbb R \to \mathbb R, \quad f(x)=5-x$$ $$ g: [3,\infty[ \to [0,\infty[, \quad g(x)=\sqrt{x-3}$$

determine whether $f \circ g^{-1} $ can be formed. If it can be formed, then find its composite function, and write down the domain and codomain of your composite function.

If it cannot be formed, give a counter-example to support your answer.


I'm not sure whether $g(x)$ is injective and / or surjective.

I guess $g(x)$ is injective because because $g(x)$ can only be non-negative for any $x \ge 3$. And $g(x)$ is surjective because every $g(x)$ has a corresponding element $x$. Therefore $g(x)$ is bijective and has an inverse function.

But I can't figure out $g^{-1}$.

  1. Grateful if you can help me confirm whether $g(x)$ is bijective.
  2. Can you give me some hints how to fund $g^{-1}$.

Then I can decide whether $ f\circ g^{-1}$ can be formed.

Thank you.

$\endgroup$
2
$\begingroup$

You are correct, $g$ is bijective:

  • Injectivity: let $\sqrt{x_1-3}=\sqrt{x_2-3} \implies x_1 - 3 = x_2 -3 \implies x_1 = x_2$
  • Surjectivity: let $y \in [0, +\infty)$ and let us find $x$ such that $\sqrt{x-3} = y$. Then we get $x = y^2+3 \in [3, +\infty)$

In the last calculation we have also determined that the inverse function is $g^{-1}(x) = x^2+3$.

$\endgroup$
  • $\begingroup$ Thank you! So, the final answer for the function is 5 - (x^2 + 3), right? $\endgroup$ – ronzenith Dec 24 '16 at 14:13
  • $\begingroup$ Exactly that ;) $\endgroup$ – Harnak Dec 24 '16 at 14:15
  • $\begingroup$ Also, Domain = [0, infinity[ Codomain = R Correct? $\endgroup$ – ronzenith Dec 24 '16 at 14:15
  • $\begingroup$ Yes, that is also correct, since $f \circ g^{-1}$ domain is $g^{-1}$ domain and its codomain is $f$ codomain. $\endgroup$ – Harnak Dec 24 '16 at 14:17
  • $\begingroup$ 1000000 thanks! You've been so helpful! $\endgroup$ – ronzenith Dec 24 '16 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.