I'm using this more as a method of verifying if I'm correct on a question I am having difficulty with. Keep in mind, I'm a complete beginner, so.. yeah.


Assume the function $g$ is defined as $g: ℤ \rightarrow ℕ$ $by$ $g(x) = x^2 + 6$. Use proper notation.

a. What is the domain of $g$?

Would this be $\left\{{0,1,2,3,4,5,...}\right\}$?

b. What is the codomain of $g$?

Would this be $\left\{{6,7,8,9,10...}\right\}$?

c. What is the range (full image) of $g$?


d. What is the image of $\left\{{-2,4,7}\right\}$?

Is this just the set of all natural numbers? So $ℕ$?

e. What is the pre-image of $\left\{{6,15,87}\right\}$?

Is this just the inverse? So $\left\{{-6,-15,-87}\right\}$?

f. Is the function injective (one-to-one)?

Yes. No two answers the same.

g. Is the function surjective (onto)?


h. Is the function bijective (one-to-one correspondence)?

No. Since it isn't onto.

Thanks for taking a look. I don't think I did very well on these.. What did I miss/get correct? Thanks a lot $:)$.

  • 2
    $\begingroup$ You are given the domain and codomain. $f:A \rightarrow B$ says the domain is $A$ and the codomain is $B$ (This is about question a and b) $\endgroup$
    – randomgirl
    Apr 30, 2015 at 3:54
  • 1
    $\begingroup$ Also if $f(a)=b$ then the image of $a$ is $b$. So rethink your image set question. $\endgroup$
    – randomgirl
    Apr 30, 2015 at 3:59
  • 1
    $\begingroup$ Oh another thing sorry. $f(a)=b$ also says a preimage of $b$ is $a$. Though you know there could be more than one preimage of $b$ if you have for some other $x$ we have $f(x)=b$ $\endgroup$
    – randomgirl
    Apr 30, 2015 at 4:00
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    $\begingroup$ The function is not injective, for $g(-10)=g(10)$. For the image of $\{-2,4,7\}$ calculate $g$ at these values of $x$. For the preimage question, who all gets mapped to these things? You should get a $5$-element set. $\endgroup$ Apr 30, 2015 at 4:01

1 Answer 1


The domain is what values you are allowed to put into the function.

(a) And so the domain is $\mathbb{Z}$

(b) The codomain is what you expect to get but not the exact values ! and so the codomain is $\mathbb{N}$

(c) The range is what values you are actually going to get which is $\{6,7,10,15,22,31,......... \}$ as you said

(d) The image of $-2$ is $g(-2) = (-2)^2 + 6 = 10$ The image of $4$ is $g(4) = 4^2 +6 = 22$ The image of $7$ is $g(7) = 7^2 + 6 = 55$ and so the image of $\{-2,4,7\}$ is $$\{10,22,55\}$$

(e) The preimage of $6$ is what values in the domain would give $6$. Notice that I said $\textbf{values}$ not a value !

The only value that gives $6$ is $0$ because $0^2 + 6 = 6$ and so the preimage of $6$ is $0$

The preimage of $15$ is both $3$ and $-3$ because $3^2 + 6 = (-3)^2 + 6 = 15$

The preimage of $87$ is both $-9$ and $9$ because $9^2 + 6 = (-9)^2 + 6 = 87$

And so the preimage of $\{6,15,87\}$ is $$ \{0,-3,3,9,9\}$$

(f) The function is clearly not one-to-one. Take for instance $g(-3)=g(3) = 15$

(g) Is the function surjective ?

No, It's not. A function is said to be surjective if it maps to (or cover) the whole codomain. There is no value in the domain that maps to $1,2,3,4$ or $5$ even though they are in the codomain. And so $g$ is not surjective

(h) Well the function is obviously not bijective. It's not injective or surjective.

And you are done my friend

  • $\begingroup$ 8 isn't in the range because $x^2+6 \neq8$ for any integer x. $\endgroup$
    – randomgirl
    Apr 30, 2015 at 4:18
  • $\begingroup$ But he has the range correct $\left\{{6,7,10,15,22,31,...}\right\}$ $\endgroup$
    – randomgirl
    Apr 30, 2015 at 4:20
  • $\begingroup$ Yes yes :D thanks :) $\endgroup$
    – alkabary
    Apr 30, 2015 at 4:20
  • $\begingroup$ I kinda thought he said 8 was in the range earlier but I think I also looked up on accident at first at his codomain answer. :p $\endgroup$
    – randomgirl
    Apr 30, 2015 at 4:22
  • $\begingroup$ mathematics always gets us ! $\endgroup$
    – alkabary
    Apr 30, 2015 at 4:24

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