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The given domain and codomain are: $$\mathbb{R} \rightarrow \{x \in \mathbb{R} | x \ge - 3\}$$

I'm suppossed to create a function that is surjective but not injective.

Now, does the codomain imply that in order for the function to be surjectice all real y values have to be covered by that function for all $x \ge - 3$ or is the more (or less?) to it? If that's the case, I supposse that the function has to approach a certain value in one way or another while also not being hyperbolic (i.e. $x^2$ wouldn't be surjectice, but $x^3$ would be injective), but I'm not really sure how to approach the problem mathematically. Could someone please give me a tip?

Thanks!

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  • $\begingroup$ Not functional analysis....or set theory (please read the tags first) $\endgroup$ Nov 8, 2016 at 21:38
  • $\begingroup$ Sorry, thought that the amount of analysis required to solve this problem would be enough to justify the tag. $\endgroup$
    – Skydiver
    Nov 8, 2016 at 21:40
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    $\begingroup$ What about $x^2-3$? $\endgroup$ Nov 8, 2016 at 21:43
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    $\begingroup$ Just surjective and not injective, no other conditions required? Why not $f(x)=x$ for $x\ge -3$ and $f(x)=-3$ otherwise $\endgroup$
    – user160738
    Nov 8, 2016 at 21:44
  • $\begingroup$ @arbervdullahu x^2 is not surjective... $\endgroup$
    – Skydiver
    Nov 8, 2016 at 21:50

1 Answer 1

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The codomain of the function is $\{ x \in \mathbb R | x \geq -3\}$. Hence, surjective means that you should cover all $y \in \mathbb R$ so that $y \geq -3$ (do not let the notation fool you).

Consider the function $f(x) = x^2 - 3$.

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  • $\begingroup$ You are completly right, I got fooled here. I kept wondering what that x was doing in my codomain, but I just got tricked as I'm used to the standard x-y coordinate system. I guess that my proffessor aimed for exactly that. Thank you! $\endgroup$
    – Skydiver
    Nov 8, 2016 at 21:55

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