Injection / Surjection Here's the question

(source: gyazo.com)
I got that it is injective. By saying:
$f(y) = 3 - y^2$
Suppose;
$f(x) = f(y)$
$3 - x^2 = 3 - y^2$
$x^2 = y^2$
$sqrt(x) = sqrt(y)$
$±x = ±y $
I conclude the it's not injective because $ -x =!$ $y$ Is this the right way to come to this conclusion?
I'm not sure how to find out if it's surjective or not.
 A: That is the right way of testing whether or not it is injective, although personally I would prefer to use $a$ and $b$ instead of $x$ and $y$. As for surjectivity, try to solve $f(x) = 4$.
A: First, drawing a picture for something like this in invaluable:

Injective means that if $f(x)=f(x')$ then you must have $x=x'$. You can see that $f(1)=f(-1)$ for example, so it cannot be injective.
Surjective means that for any $y$ in the range, there is some $x$ in the domain such that $f(x) = y$. Since $x^2 \ge 0$ for any $x$, you can see that $f(x) \le 3$ for all $x$. Hence any point $y>3$ cannot be in the range and so it is not surjective.
A: This is a good calculation, that will help you find an answer.  To prove a function is NOT injective you need a single counterexample.  Using your calculation, let's pick some $x,y$ that we expect will map to the same place.  For example, $x=4, y=-4$.  Then $f(4)=3-4^2=-13$ and $f(y)=3-(-4)^2=3-16=-13$.
To determine if it's surjective, let $a\in \mathbb{R}$ and try to find some $x$ such that $f(x)=a$. (i.e. solve for $x$).  If you succeed regardless of $a$, you've proved it's surjective.  If you fail for some $a$, you will have a roadmap to find a counterexample.
A: Easiest method 
1.Draw Graph  2.Draw Horizontal lines. If any horizontal line intersects the graph more than once,then the graph is not injective. 
 I hope I have illustrated that and surjectivity in the image :)
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A: A function is injective if each element in the codomain is mapped to by at most one element in the domain.$$\forall x_1\in \Bbb R~~\forall x_2\in\Bbb R:(f(x_1){=}f(x_2)\to x_1{=}x_2)$$
As you have shown, there exists some elements in the codomain which are mapped to by more than one element in the domain.  That does indeed allow you to declare that the function is not injective.    Well done.

A function is surjective if every element in the codomain is mapped to by at least one element in the domain.  Ie: The entire codomain is the image of the domain.
$$\forall y\in\Bbb R~~\exists x\in\Bbb R:y=f(x)\\[2ex]\Bbb R=f(\Bbb R)$$
So are there any elements in the codomain ($\Bbb R$) which are not in the image of the domain?
