Is the Tangent Function Surjective on Its Defined Interval?

[quasar987]

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

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

Thank you.

 

[Hurkyl]

Try working directly with the definition. Can you find a preimage for every object in the range of f?

 

[AKG]

The domain can't be the closed interval, so I don't see how that would help. tan(x) is not defined when x = \pi / 2. Anyways, you know that it is continuous. You know that the limit as x approaches \pi /2 from the left is \infty 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 \delta such that for all x in (\pi /2 - \delta, \pi /2), 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 \pi /2 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 \pi / 2 is \infty. I'm not sure. The above, I believe, is enough to prove it, if not too much.

 

[Hurkyl]

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.