# Show that $f(x) = \frac{x^3}{1+x^2}$ is bijective

This seems like a simple question, but I'm stuck: how do I show that $f: \mathbb{R} \rightarrow \mathbb{R}$ defined as $f(x) = \frac{x^3}{1+x^2}$ is bijective?

I want to demonstrate that it is both injective and surjective. To show that it's injective, I need to show that $f(x) = f(y)$ implies $x = y$. However, I can't see a way to reduce $\frac{x^3}{1+x^2} = \frac{y^3}{1+y^2}$ to $x = y$ (since there are no like terms to combine). I'm also unsure of how to prove surjectiveness.

• Do you know calculus? – vadim123 Feb 1 '15 at 17:54
• If you know calculus, you can do it by showing that the derivative is everywhere defined and positive. Or if there are isolated points where the derivative is $0$, you're still OK. – Michael Hardy Feb 1 '15 at 17:54
• The easiest way to prove injectivity is to show that $f$ is increasing on $\mathbb R$, if you know calculus... – 5xum Feb 1 '15 at 17:54
• For surjectiveness, note that $f(x)$ has limit $+\infty$ if $x \to +\infty$ and has limit $-\infty$ if $x \to -\infty$. Also $f(x)$ is continuos, so it "takes on" all the values between $-\infty$ and $+\infty$, that is $\mathbb R$ – Ant Feb 1 '15 at 19:46

you don't need calculus to show that $f(x) = \dfrac{x^3}{1+x^2}$ is 1-1. suppose $$\dfrac{a^3}{1+a^2} = \dfrac{b^3}{1+b^2} \tag 1$$ we will show that this implies $a = b$ proving $f$ is 1-1. cleaning up $(1)$ gives $$(a-b)(a^2 + ab + b^2 + a^2b^2) = 0$$ now use the fact that $a^2 + ab + b^2 > 0$ for $a \neq 0, b \neq 0$ to conclude $a = b.$

$\bf Edit:$ To show that $f$ is onto note that $f(x) = \dfrac{x^3}{1+x^2}$ is odd and $\lim_{x \to \infty} \dfrac{x^3}{1+x^2} = \infty$. That is, the range of $f$ is $(-\infty, \infty).$

• Yes, but this only proves that it is injective. You should also show that it takes on every possible value $\in \mathbb R$ :-) – Ant Feb 1 '15 at 18:40
• Thank you - very helpful for proving injectiveness. Can you help with surjectiveness as well? – bkaiser Feb 1 '15 at 18:41
• Your edit shows that the range is $(-\infty, \infty)$ but does that prove that every one of those elements is enumerated by some $f(x)$? – bkaiser Feb 2 '15 at 17:32
• @caesar, what range of $f$ is $(-infty, \infty)$ shows is that for every number $y,$ there is an $x$ such that $f(x) = y.$ this is what $f$ is onto means. we have already shown that there cannot be more than one. what we have now is there must be at least one. putting two together, you have exactly $x$ for every $y$ so that $f(x) = y.$ – abel Feb 2 '15 at 17:45

Hint: $f(x) = x - \dfrac{x}{1+x^2}$, and show $f'(x) > 0$

Using the quotient rule $\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v\dfrac{du}{dx}-u\dfrac{d}{dx}}{v^2}$, we get

$$\frac{df}{dx}=\frac{x^4+3x^2}{1+x^2}$$

which is continuous, and strictly positive except at $x=0$. Therefore $f$ is strictly increasing, hence injective.