Prove that $(x+3)^2$ is not one-to-one So the domain given is $f\colon\mathbb{R}\to\mathbb{R}^+\cup\{0\}$.
Does this mean the set of all negative numbers and $0$ but no positive numbers? I am asking because if it does include positive and negative numbers I believe I can prove this by making $x_1= -8$ and $x_2=2$ giving different $x$ values but the same $y$ value. 
At least I think that is how this is done, any resources, pointers and information is much appreciated! 
 A: So the domain is ALL of $\mathbb R$, whereas the codomain is limited to all reals greater than or equal to $0$.
$$f\colon\mathbb{R}\to\mathbb{R}^+\cup\{0\}$$
Note that the codomain consists of values greater than or equal to zero, because whatever the $x\in \mathbb R$, $f(x) = (x+3)^2 \geq 0$.
HINT: Consider $x= -7$, with $x=1$
$f(-7) = (-4)^2 = 16.$
$f(1) = (4)^2 = 16.$
Can you see why it fails to be one-to-one?  
A: As the problem is stated, $f$ is a function with domain the set of all real numbers and codomain the set of positive real numbers and zero.
The action is $f(x)=(x+3)^2$, and the function is well defined, because for each $x\in\mathbb{R}$, $f(x)\in\mathbb{R}^+\cup\{0\}$.
Your strategy is good: since $f(-8)=(-8+3)^2=25$ and $f(2)=(2+3)^2=25$, the function is not one-to-one.

Just one example is sufficient. In order to prove that $f$ is one-to-one you have to see that for each choice of $x_1\ne x_2$ you have $f(x_1)\ne f(x_2)$ (or, which is the same, that $f(x_1)=f(x_2)$ implies $x_1=x_2$). Thus an example where this does not happen immediately proves the function is not one-to-one.
Were the function defined over $\mathbb{R}^+\cup\{0\}$ it would be one-to-one. Indeed, if $(x_1+3)^2=(x_2+3)^2$, we get either
$$
x_1+3=x_2+3
$$
(and so $x_1=x_2$), or
$$
x_1+3=-(x_2+3)
$$
or
$$
x_1+x_2=-6
$$
which is impossible if $x_1\ge0$ and $x_2\ge0$.
Of course the function would be also one-to-one if the domain was taken as the interval $[-3,\infty)$ (try it).
