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Prove injectivity of function.

  1. Oct 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that f is injective.

    2. Relevant equations


    [itex]f:(- \infty, 3] \rightarrow [-7,\infty) \vert f(x) = x^2 -6x+2[/itex]

    3. The attempt at a solution

    I wish to prove this by calculus. I know that the maximum is three, and this is the only way the quadratic can be one-to-one. However; I'm having problems formalizing this into a proper proof.

    This is what I have so far:

    Definition of injective function: [itex]f(x_{2}) = f(x_{1}) \Rightarrow x_{2}=x_{1}[/itex]

    [itex]f(x)' = 2x-6[/itex]

    When [itex] f(x)' = 0, x=3 [/itex]

    This is true, only if [itex](3, f(3))[/itex] is a maximum or a minimum.

    In this case [itex](3,f(3))[/itex] is a minimum, and [itex]f(3) = -7.[/itex]

    Since 3 is the only critical point, f(x) is always increasing when x < 3.

    This implies injectivity, because ..... (how do I conclude this, it makes sense intuitively, but I don't know how to write it?)

    ------------------------------------------------------------------

    How can I explain that since 3 is a minimum and also the only critical number, the function is injective upon these domain constraints? Must I use images and pre-images? Also I know there is certainly a way to prove this without calculus, since calculus isn't a pre-requisite for the course this is for, but I'd like to use differentiation, since I have the tool and it's the first thing I had thought of upon noting that this function is quadratic and the only way for it to be injective is if f(x)' = 0 when x=3 and that it is the only place where f(x)'=0. Thank you!
     
    Last edited: Oct 27, 2012
  2. jcsd
  3. Oct 27, 2012 #2

    LCKurtz

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    If you want a calculus argument, you have f'(x) = 2(x-3) < 0 if x < 3. That tells you the function is decreasing on that interval, hence injective.
     
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