prove $f(x)=x^3+x$ is one to one and onto

Question

The Problem

For my discrete structures course, I need to prove that $f(x)$ is one-to-one and onto, with $f:{\rm I\!R}\rightarrow{\rm I\!R}$ where $f(x)=x^3+x$. Based on the graph, this function is one-to-one and onto, and Wolfram confirms this, but I don't know how to approach the actual proof.

Proving it One-to-One

I understand a function $f(x)$ is one-to-one if for $x_1,x_2\in{\rm I\!R}$, if $f(x_1)=f(x_2)$ implies $x_1=x_2$. The problem is when I set $f(x_1)=f(x_2)$ this, I eventually get to $$\sqrt[3]{x_1^3+x_1}=\sqrt[3]{x_2^3+x_2}$$ It's at this point I'm stuck, and don't know how to progress any further.

Proving it Onto

So I understand to prove a function is onto, you solve $f(x)$ for $y$, and use the result as input to $f(x)$, and if $f(x)=y$, the function is onto. But I run into a similar problems where $$x^3+x=y$$ I could take a cubic root or rearrange the variables all I want but I can't think of a way to isolate $x$ and get $x=something$

Answer 1

It is one-to-one because it is strictly increasing ($x>y\implies f(x)>f(y)$) and it is surjective be cause $\lim_{x\to\pm\infty}f(x)=\pm\infty$ and by the intermediate value theorem.

Answer 2

Let $f:\Bbb R\to\Bbb R$ be the function $f(x)=x^3+x$. Then $f'(x)=3x^2+1\ge 1>0$, so $f$ is strictly monotone, thus injective (one-to-one). Then the limits of $f$ at $\pm\infty$ are respectively $\pm\infty$, and from the continuity of $f$ each value in between is taken.

Note: One can also show algebraically that $f$ is injective, so assume $f(a)=f(b)$, then $$ 0=f(a)-f(b)=(a-b)\underbrace{(a^2+ab+b^2+1)}_{\ge 0+1>0}\ , $$ so the factor $(a-b)$ must vanish, so $a=b$.

Answer 3

Surjectivity: For each $y$ polynomial equation $$x^3+x=y$$ is of odd degree, so it must have at least one real solution and thus $x^3+x$ is surjective.

Injectivety: Say there are $a$, $b$ such that $f(a)= f(b)$ and suppose $a\ne b$, then $$ (a-b)(a^2+ab+b^2+1)=0\implies a^2+ab+b^2+1=0$$

so multilpying this with 2 we get $$ (a+b)^2+a^2+b^2 +2 =0$$ which is clearly nosense. So $a=b$.

Answer 4

Partial answer:

1) Injective:

$x_1,x_2 \in \mathbb{R}$, and let

$f(x_1)=f(x_2).$

$x_1^3 +x_1=x_2^3+x_2$;

$x_1^3-x_2^3 +x_1-x_2=0;$

$(x_1-x_2)(x_1^2+x_1x_2+x_2) +(x_1-x_2)=0;$

$(x_1-x_2)(x_1^2+x_1x_2+x_2^2+1)=0;$

It follows $x_1=x_2$ , and we are done,

since $(x_1^2+x_1x_2+x_2^2+ 1) >0$.

Recall: $a^2+b^2 \ge 2|ab|$,

$a^2+ab+b^2 \ge 2|ab| +ab \ge$

$|ab| \ge 0.$

$a^2+ab+b^2 +1 \ge 1 >0.$

2) $y=x^3+x$ , a polynomial of degree $3$ has at least one real root.(Cf. Answer by Maria Mazur).

For given $y$, it has exactly one real root since $y=f(x)$ is injective.

Answer 5

Just to take a different approach, you can use Cardano's formula: write the equation $x^3+x=y$ as $x^3+x-y=0$. The discriminant is $$ \frac{y^2}{4}+\frac{1}{27}>0 $$ so the equation has a single real solution, for every $y$. This proves at once both injectivity and surjectivity.

You can even solve for the inverse function: $$ x=\sqrt[3]{\frac{y}{2}+\sqrt{\frac{y^2}{4}+\frac{1}{27}}}+ \sqrt[3]{\frac{y}{2}-\sqrt{\frac{y^2}{4}+\frac{1}{27}}} $$