Okay, so I know there are plenty of injective/surjective (and thus, bijective) questions out there but I'm still not happy with the rigor of what I have done. So I'd really appreciate some help!
So the question actually asks me to do two things:
(a) give an example of a cubic function that is bijective. Explain why it is bijective.
(b) give an example of a cubic function that is not bijective. Explain why it is not bijective.
So for (a) I'm fairly happy with what I've done (I think):
$$ f: \mathbb R \rightarrow \mathbb R , f(x) = x^3$$
So we know that to prove if a function is bijective, we must prove it is both injective and surjective.
Proof: $f$ is injective
Let: $$x,y \in \mathbb R : f(x) = f(y)$$
$$x^3 = y^3$$ (take cube root of both sides)
$$x=y$$
Proof: $f$ is surjective
Let: $$y \in \mathbb R$$
$$x = \sqrt[3]{y}$$
$$f(x) = (\sqrt[3]{y})^3 = y$$
So I believe that is enough to prove bijectivity for $f(x) = x^3$. Keep in mind I have cut out some of the formalities i.e. invoking definitions and sentences explaining steps to save readers time. This is just 'bare essentials'.
So for (b)
$$f: \mathbb R \rightarrow \mathbb R , f(x) = x^3 – x$$
Now I'm just going to try and prove it is NOT injective, as that should be sufficient to prove it is NOT bijective.
Let:
$$x,y \in \mathbb R : f(x) = f(y)$$
$$x^3 – x = y^3 – y$$
This is about as far as I get. Send help. Thanks everyone.
Best Answer
Injectivity:
A function $f$ from $X\to Y$ is said to be injective iff the following statement holds true:
The negation of this then yields:
A function $f$ from $X\to Y$ is not injective iff there exists $x_1,x_2\in X$ such that $x_1\neq x_2$ but $f(x_1)=f(x_2)$
In the case of the cubic in question, it is an easily factorable polynomial and we can find multiple distinct roots. $f(x)=x^3-x=x(x^2-1)=x(x+1)(x-1)$
We know that a root of a polynomial is a number $\alpha$ such that $f(\alpha)=0$
As an aside, one can prove that any odd degree polynomial from $\Bbb R\to \Bbb R$ must be surjective by the fact that polynomials are continuous and the intermediate value theorem.