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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!

Scientifica
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whatamii
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2 Answers2

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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?

amWhy
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    Ohhhh, so since they both equal 16, but one comes from -4 and the other from 4 they are not one-to-one

    I think I get it thank you!!

     – whatamii Sep 11 '16 at 14:44
  • Yes, you got it, whatamii! – amWhy Sep 11 '16 at 14:47
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    Not a problem @Scientifica. I'm glad you pointed out the deficit, since corrected. I appreciate your response! 8) – amWhy Sep 11 '16 at 15:12
  • I visited your profile page and scanned the "about me" section after my comment 24 mins ago. You seem to be a very considerate person and I TOTALLY agree with you when you say "Be welcoming, be patient, and assume good intentions.", leaving (sincerely speaking) an admiration. I'd be sad if this misunderstanding would leave me considerated as an obnoxious person by a person like you :/ – Scientifica Sep 11 '16 at 15:14
  • :o Thank you very much for your comprehension!! :D – Scientifica Sep 11 '16 at 15:16
  • No worries, @Scietifica. I will be smiling when I encounter your username! – amWhy Sep 11 '16 at 15:19
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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).

egreg
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