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Math Help - How to prove that this function is bijective...

  1. #1
    Junior Member gusztav's Avatar
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    How to prove that this function is bijective...

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

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

    is a bijective function?

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

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

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

    Of course,

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

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

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

    And because x^2 is also positive, f '(x)>0, \forall x \in \mathbb{R}, so our f must be a strictly increasing function.
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