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.


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 $g^{-1}(x) = x^2+3$.

  • $\begingroup$ 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

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