Results 1 to 3 of 3

Thread: How to prove that this function is bijective...

  1. #1
    Junior Member gusztav's Avatar
    Joined
    Jan 2008
    Posts
    48
    Awards
    1

    How to prove that this function is bijective...

    Hi, how can I prove that $\displaystyle f:\mathbb{R} \to \mathbb{R}$, where

    $\displaystyle f(x)=x^3+4x-\cos(\pi x)$

    is a bijective function?

    If the function were just $\displaystyle g(x)=x^3+4x$, the problem could be easily solved by noting that $\displaystyle x^3$ and $\displaystyle x$ are strictly increasing functions, and their linear combination is therefore also strictly increasing. As the image of $\displaystyle g(x)$ is whole $\displaystyle \mathbb{R}$, $\displaystyle g(x)$ is bijective.

    But in $\displaystyle f(x)$, we also have $\displaystyle \cos(\pi x)$, which is a periodic, and not strictly monotonous function. Do we have to show that $\displaystyle f(x)$ is simultaneously injective and surjective to be a bijection or is there some other way?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,742
    Thanks
    2814
    Awards
    1
    $\displaystyle f'(x) = 3x^2 + 4 + \pi \sin (\pi x)$.
    What does that tell you?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member gusztav's Avatar
    Joined
    Jan 2008
    Posts
    48
    Awards
    1
    Quote Originally Posted by Plato View Post
    $\displaystyle f'(x) = 3x^2 + 4 + \pi \sin (\pi x)$.
    What does that tell you?
    Aha! Plato, thank you so much!

    Of course,

    $\displaystyle \sin (\pi x) \in [-1,1], \forall x \in \mathbb{R}$,

    $\displaystyle \pi \sin (\pi x) \in [-\pi,\pi], \forall x \in \mathbb{R}$

    and naturally $\displaystyle 4 + \pi \sin (\pi x) > 0, \forall x \in \mathbb{R}$

    And because $\displaystyle x^2$ is also positive, f$\displaystyle '(x)>0, \forall x \in \mathbb{R}$, so our $\displaystyle f$ must be a strictly increasing function.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Is my work correct? Prove the following is bijective.
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: Nov 28th 2010, 07:48 PM
  2. Prove comnplement function on [n] is bijective
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jun 24th 2010, 08:54 AM
  3. How to prove this function is bijective
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: May 8th 2010, 07:30 AM
  4. Prove about no bijective fuction exist
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Nov 8th 2009, 06:31 AM
  5. Replies: 1
    Last Post: Apr 21st 2009, 10:45 AM

Search Tags


/mathhelpforum @mathhelpforum