# Proving that $f(x)=x^3+x+2$ is bijective without calculus

I want to prove that $f(x)=x^3+x+2$, $f: \mathbb R \rightarrow \mathbb R$ is bijective without calculus. My attempts at showing to prove that it' injective and surjective are written below:

$1)$ Injectivity:

I want to show that $\forall a,b \in \mathbb R$ $f(a)=f(b) \implies a=b$.

I started like this: $$f(a)=f(b) \implies a^3+a+2=b^3+b+2$$ $$\implies a(a^2+1)=b(b^2+1)$$ $$\implies \frac{a}{b}=\frac{b^2+1}{a^2+1}$$ Then I said since $\frac{b^2+1}{a^2+1}>0$ $\forall a,b \in \mathbb R$ then either $a \land b < 0$ or $a \land b > 0$. (For the case when $b=0 \land a \in \mathbb R$ it would be easy to prove that $a=b$.) From there it seemed pretty obvious that $\frac{a}{b}=\frac{b^2+1}{a^2+1} \implies a=b$ so I couldn't really draw a logical argument to show that $a=b$.

$2)$ Surjectivity:

I want to show that $\forall b \in \mathbb R$ $\exists a \in \mathbb R$ s.t. $f(a)=b$.

I started like this:

Let $b \in \mathbb R$ and set $f(a)=b$ then we have:

$$a^3+a+2=b$$ $$\implies a^3+a=b-2$$ $$\implies a(a^2+1)=b-2$$

But then I couldn't find an expression for $a \in \mathbb R$ in terms of $b$. So I'm wondering if anyone can tell me how I can proceed with my surjectivity and injectivity proofs.

• You have to assume $b\neq 0$ when you want to devide through it. Therefore you have to view the case b=0 seperate. Oct 8 '17 at 11:59
• @Cornman I edited my question.
– iza
Oct 8 '17 at 12:03
• Suffices to show $f(x)$ is strictly increasing. if $x >y$ then $x^3>y^3$ so $x^3+x+2>y^3+y+2$. Oct 8 '17 at 12:04
• All function $f(x)=x^3+ax+b$ with $a$ positive is a bijection. Oct 8 '17 at 12:24
• @Bernard No, we haven't covered it yet or else I reckon it'd be pretty easy to prove surjectivity.
– iza
Oct 8 '17 at 12:27

As for the injectivity, if $a>b$, then $a^3>b^3$, so $f(a)=a^3+a+2>b^3+b+2=f(b)$.

As for the surjectivity, there is a formula to find a real solution of an equation of degree 3; it is not very nice, but you can use it. You can find it here

• Is there any easier methods not involving calculus for the surjectivity because I'm pretty sure I'm not supposed to use the formula for an equation of degree $3$ for this question.
– iza
Oct 8 '17 at 12:05
• @iza: Note that surjectivity also follows from the continuity of $f$ together with the fact that $f(x)\to\pm\infty$ as $x\to\pm\infty$.
– MPW
Oct 8 '17 at 12:10
• @MPW In the comments OP says that IVT is not allowed (they've said it after you've posted your comment). Oct 8 '17 at 13:06
• @Wojowu : Ok, thanks. It doesn't require that, but OP may not have basic topology at his disposal. You can easily show from first principles that the continuous image of a connected set is connected, the only connected sets in $\mathbb R$ are intervals, and $[f(-n),f(n)]\subseteq f([-n,n])$. Take union over all $n$ to get the desired result.
– MPW
Oct 8 '17 at 13:23

Injectivity is clear and so is for the fact that there is not a positive root.

Applying now Descartes's Rule we have $$f(x)=x^3+x+2 \text{ have no change signs }\\f(-x)=-x^3-x+2 \text{ has one change sign }$$ Consequently the maximum number of negative roots is $0+1=1$ and $f^{-1}$ exists.

1) Injective:

$f(x) = x^3 +x+2$, $x \in \mathbb{R}$ is strictly monotonically increasing.

Let $x_1< x_2$, then

$x_1^3 + (x_1+2) \lt x_2^3 + (x_2 +2)$,

is strictly monotonically increasing,

sum of $y_1= x^3$ and a linear function $y_2= x+2$, both strictly monotonically increasing.

2) Surjective.

A)$\lim_{x \rightarrow \infty} f(x) = \infty.$

B)$\lim_{x \rightarrow -\infty} f(x) = - \infty$.

Let $c \in \mathbb{R}$.

Choose $a$ with $f(a) < c$, and $b$ with $f(b) >c$,

(property A,B).

Consider the closed interval $[a,b] .$

Since $f$ is continuos, and $f(a) < c < f(b)$,

there is a $p \in (a,b)$ such that $f(p) =c$.

• In the comments OP says that you can't use intermediate value theorem. Oct 8 '17 at 13:07