To save me some time writing everything out in latex, I'm adding a picture of the question and Ill try to explain what I understand for the problem. Just a heads up, I'm really not sure how to do this at all :/

enter image description here

So the definition is saying for a function G, A implies B and elements of subset C are also elements of subset B. The inverse function (given) is elements a for which g(a) is in the set, C?

Other than that, I'm fairly lost on how to do any of the a,b,c parts. I'm not asking for straight up answers for all three, but to actually understand/make sense as to what the question is asking/how I can find the solutions.


a.) f$^-1$({a,-2,16}) = {1,-1,2,-2}


c.) [0,$\infty$) ?



First, the notation $g:A\longrightarrow B$ does not mean $A$ implies $B$ but "$g$ is a function with domain $A$ and codomain $B$", so that $g(x)$ only makes sense if $s\in A$, and in that case, $g(x)\in B$.

for instance, there is a function from the set of people to the set of integers which gives the age in years of a person. So you can write $age: \mathrm{People} \to \mathbb N$ and $age(5)$ doe not make sense, but $age(John Doe)$ makes sense and it has a value, say $45\in\mathbb N$.

Now the inverse image $g^{-1}$ for subsets is defined as follows: for a subset of the codomain $C\subset B$, it gives the subset $g^{-1}(C)\subset A$ which consists of the elements whose image is in $C$. For instance, $age^{-1}(\{13, ..., 19\})$ is the set of teenagers which is a subset of the set of people.

Concretely, how to answer the questions in the exercise ? For each value in the subset of $B$ given, look in $A$ for all the elements which are mapped to this value. Regrouping all those elements in a subset gives you the inverse image subset of $A$.

  • $\begingroup$ Ok that example of the relation definitely made more sense, thanks! It's all the symbols that confuse me. I still don't fully understand that 3rd paragraph though. Like how the answer above yours, says for part a) it'd be {1,2,-2} $\endgroup$ – Cozen Feb 3 '14 at 21:37
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    $\begingroup$ From all the real numbers (in your domain $A=\mathbb R$), which ones give $1,-2$ or $16$ when put to the $4$th power ? $\endgroup$ – zozoens Feb 3 '14 at 21:40
  • $\begingroup$ Excellent. That makes much more sense. Thank you! $\endgroup$ – Cozen Feb 3 '14 at 21:41
  • $\begingroup$ If you don't mind me asking really quick, For part b, do I just plug 5 through 16 in as x.. to the 4th power? And would c.) be: [0,$\infty$)? $\endgroup$ – Cozen Feb 3 '14 at 22:00
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    $\begingroup$ The set $C=(4;16]$ is already an infinite set of values, so it makes sense that its inverse image is also infinite. You may find the set of the solution of the same form. $\endgroup$ – zozoens Feb 3 '14 at 22:14

Hint: for a) you need to find all such numbers x for which f(x) = 1, -2 or 16

  • $\begingroup$ x for which f(x) = 1,-2,16 or the inverse of f(x)? $\endgroup$ – Cozen Feb 3 '14 at 21:16
  • $\begingroup$ We are not talking about inverse FUNCTIONS at all - the f in a) doesn't even have an inverse - but the inverse IMAGE OF A SET under the function in question. $\endgroup$ – Pifagor Feb 3 '14 at 21:18
  • $\begingroup$ Ok that kinda makes sense lol. Thanks $\endgroup$ – Cozen Feb 3 '14 at 21:20
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    $\begingroup$ Ok. The answer to a) is the set {1, 2, -2} because f(1) = 1, f(2)=16 and also f(-2)=16. $\endgroup$ – Pifagor Feb 3 '14 at 21:27
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    $\begingroup$ This is the exact meaning of this inverse image application. $\endgroup$ – zozoens Feb 3 '14 at 21:41

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