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How to show tanx is surjective?

  1. Nov 12, 2004 #1

    quasar987

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    How can I show the function [itex]f:]-\frac{\pi}{2},\frac{\pi}{2}[ \rightarrow R[/itex] defined by [itex]f(x)=tan(x)[/itex] is surjective?

    If the domain was a closed interval I could use the intermediate value theorem, but now?

    Thank you.
     
  2. jcsd
  3. Nov 12, 2004 #2

    Hurkyl

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    Try working directly with the definition. Can you find a preimage for every object in the range of f?
     
  4. Nov 12, 2004 #3

    AKG

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    The domain can't be the closed interval, so I don't see how that would help. tan(x) is not defined when x = [itex]\pi / 2[/itex]. Anyways, you know that it is continuous. You know that the limit as x approaches [itex]\pi /2[/itex] from the left is [itex]\infty[/itex] and by the definition of the limit, you can argue that "all the big numbers" are achieved by f. The definition of such a limit says that for any N, you can find a [itex]\delta[/itex] such that for all x in [itex](\pi /2 - \delta, \pi /2)[/itex], f(x) > N. Since you can choose any N, you can show that all the numbers greater than all N are achieved. Choose N to be something simple like 1. You can also show that the values 1 and -1 are achieved, and by intermediate value theorem, all numbers in (-1, 1) are thus achieved. Finally, the you know the limit as x approaches [itex]\pi /2[/itex] from the right and argue in a similar fashion that all numbers less than 1 are achieved. I'm not sure, but the last few steps might be superfluous. It might be sufficient to show that f is continuous and the left-hand limit as x approaches [itex]\pi / 2[/itex] is [itex]\infty[/itex]. I'm not sure. The above, I believe, is enough to prove it, if not too much.
     
  5. Nov 12, 2004 #4

    Hurkyl

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    Ah, right, I may have been answering the question at the wrong level.


    One thing you might consider is that if a and b are in the domain of f, then the closed interval [a, b] is a subset of the domain of f.


    If you're familiar with the extended real line, then you could extend f continuously to [-pi/2, pi/2] as a function whose range is the extended real line, rather than the reals.
     
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