Proving Bijections

Question

i'm working on this question here but I am having some trouble:

$f: ℝ ⇒ ℝ$, $f(x) = x^3 - 6x$.

A) Is $f(x)$ injective?

B) Is $f(x)$ surjective?

C) Is $f(x)$ bijective?

My attempted solution:

A) $f(x)$ is injective if and only if $f(x) = f(y)$ $⇒$ $x = y$

$⇔$ $x^3 - 6x = y^3 - 6y$

$⇔$ $x(x^2 - 6) = y(y^2 - 6)$

$⇔$ $x(x - √6)(x + √6) = y(y - √6)(y + √6)$ I saw this step in a similar example but don't understand exactly what is going on here, an explanation would be great

At this point I imagine from that last line above these $2$ values for x are derived from it?($0$ and $√6$)

$f(0) = x^3 - 6x$

$=$ $0^3 - 6(0) $

$= 0$

$f(√6) = x^3 - 6x$

$= √6^3 - 6(√6)$

$= 0$

$∴ f(x)$ is not injective since $f(0) = 0 = f(√6)$

B) To show f(x) is surjective, we can assume $y = x^3 - 6x$(a bit stuck on this step, or maybe there is nothing that can be done ∴ I know it's not surjective right away?)

$y = x^3 - 6x $

what more can I do from here?

C) Regardless of figuring out part B or not, I still know that $f(x)$ is not a bijection since it's not injective.

Thanks!

Answer 1

For proving surjectivity, note that $\lim\limits_{x\to\infty}f(x)=\infty$ and $\lim\limits_{x\to -\infty}f(x)=-\infty$. Further note that $f(x)$ is a polynomial and it is known that polynomials are continuous. An application of the Intermediate-Value-Theorem will show that for each possible value of $y$ it can be achieved by some $x$.

More specifically, we know that cubics (being polynomials of odd degree) will always have at least one real root.

Given a specific $y$, we try to find what value of $x$ will give that value of $y$.

$y=f(x)=x^3-6x\Rightarrow 0=x^3-6x-y$ where $y$ is known. We know it will have a real root for $x$. We do not necessarily need to find what exactly it is to be satisfied, but via cardano's method we find that it would be something along the lines of $\frac{\sqrt[3]{\sqrt{y^2-32}+y}}{\sqrt[3]{2}}+\frac{2\sqrt[3]{2}}{\sqrt[3]{\sqrt{y^2-32}+y}}$. You are not expected to know this.

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For proving/disproving injectivity, all you need to do is find an instance where $f(x)=f(y)$ despite $x\neq y$ (to prove it is not injective) or you need to show that $f(x)=f(y)$ implies that $x=y$.

With $f$ a polynomial of low degree, we attempt to factor it. A polynomial is in factored form if it is of the form $f(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)\cdots (x-\alpha_n)$. Each of these terms $\alpha_i$ are what are called "roots" and are the points for which $f(x)=0$.

In your case $f(x)=x^3-x=(x-0)(x-\sqrt{6})(x-(-\sqrt{6}))$ so there are three roots. $f(0)=f(\sqrt{6})=f(-\sqrt{6})=0$. In particular, these roots are distinct, so we have found an occurrence of an $x$ and a $y$ where $f(x)=f(y)$ but $x\neq y$.

Answer 2

Your argument for (A) is correct, but longer than necessary. All that you have to say is that $f(0) = f(\sqrt{6})$.

(C) is right too, for the reason you give.

For (B) - sketch the graph of the function. Can you see what happens as $x$ becomes large, or large (in absolute value) and negative?

(For future reference: sketching the graph would have been a good place to start thinking about all three questions.)

PS Please use mathjax to format the mathematics in any other questions or answers you want to post here.

Answer 3

$$\begin{align} & f({{x}_{1}})=f({{x}_{2}})\Rightarrow {{x}_{1}}^{3}-6{{x}_{1}}={{x}_{2}}^{3}-6{{x}_{2}}\Rightarrow (\,{{x}_{1}}^{3}-{{x}_{2}}^{3})+(-6{{x}_{1}}+6{{x}_{2}})=0 \\ & ({{x}_{1}}-{{x}_{2}})({{x}_{1}}^{2}+{{x}_{1}}{{x}_{2}}+x_{2}^{2})-6({{x}_{1}}-{{x}_{2}})=({{x}_{1}}-{{x}_{2}})({{x}_{1}}^{2}+{{x}_{1}}{{x}_{2}}+x_{2}^{2}-6)=0\, \\ & \large\left\{ \begin{matrix} {{x}_{1}}={{x}_{2}} \\ {{x}_{1}}^{2}+{{x}_{1}}{{x}_{2}}+x_{2}^{2}-6=0\,\,\,\,\,\,\Rightarrow \,\,\,\,{{x}_{1}}=\frac{-{{x}_{2}}\pm \sqrt{-3{{x}_{2}}^{2}+24}}{2} \\ \end{matrix} \right. \\ \end{align}$$

$f$ is surjective:

let $t\in\mathbb{R}$. We have $x^3-6x=t$. This equation has a real solution, because it's degree is odd.