4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do you know calculus? $\endgroup$ – vadim123 Feb 1 '15 at 17:54
  • 1
    $\begingroup$ 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. $\endgroup$ – Michael Hardy Feb 1 '15 at 17:54
  • 1
    $\begingroup$ The easiest way to prove injectivity is to show that $f$ is increasing on $\mathbb R$, if you know calculus... $\endgroup$ – 5xum Feb 1 '15 at 17:54
  • $\begingroup$ 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 $\mathbb R$ $\endgroup$ – Ant Feb 1 '15 at 19:46
17
$\begingroup$

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 $(-\infty, \infty).$

$\endgroup$
  • $\begingroup$ Yes, but this only proves that it is injective. You should also show that it takes on every possible value $\in \mathbb R$ :-) $\endgroup$ – Ant Feb 1 '15 at 18:40
  • $\begingroup$ Thank you - very helpful for proving injectiveness. Can you help with surjectiveness as well? $\endgroup$ – bkaiser Feb 1 '15 at 18:41
  • $\begingroup$ Your edit shows that the range is $(-\infty, \infty)$ but does that prove that every one of those elements is enumerated by some $f(x)$? $\endgroup$ – bkaiser Feb 2 '15 at 17:32
  • $\begingroup$ @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 $f(x) = y.$ $\endgroup$ – abel Feb 2 '15 at 17:45
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.