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Domain and Range of A Function

  1. Oct 23, 2012 #1
    I have a question about domain and range.What are the domain and range of [itex]cos(e^{-×})[/itex] and [itex]\frac{\left|2×-1\right|}{sin(\frac{1}{2}\pi×-\pi)}[/itex] ??

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  3. Oct 23, 2012 #2


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    The domain of the first functions is pretty simple, isn't it? You can take the exponential of any numberr and can take cosine of any number. As for the range, well, cosine is always between -1 and 1 isn't it?

    For the second function, the only difficulty with the domain is that you can't divide by 0. For what x is [itex]sin(\pi((1/2)x- 1))[/itex] equal to 0? And, I don't see any reason why the range cannot be all real numbers.
  4. Oct 23, 2012 #3


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    Stated like this, the question does not have a unique answer.

    You can consider ##f: [0,1] \to [-10,10]##, ##f(x)=\cos(e^{-x})##. It is a valid function, it has the correct images for all allowed x, and it has a (nearly) arbitrary domain and codomain. And it has an image different from ##g: [-10,10] \to [-10,10]##, ##g(x)=\cos(e^{-x})##.

    Oh, and we can extend the function to complex numbers! Or even matrices with complex entries, or anything else where the exponential function and the cosine function have a clear definition.

    Something like that would have a unique answer: "What is the maximal* domain as subset of R where the function definition is meaningful, and what is the image of the implied function?"
    *with the obvious meaning

    @HallsofIvy: Small, positive values in the image might be tricky with the second expression.
  5. Oct 23, 2012 #4
    The first question's solution:
    If we take [itex]lim_{x\rightarrow+∞}cos(e^{-x})\approx1[/itex]. However if we take [itex]lim_{x\rightarrow-∞}cos(e^{-x})[/itex] the function takes values on closed interval -1 and 1 [itex]\left[-1,1\right][/itex]. That's why we can easily say that [itex]f:ℝ\rightarrow\left[-1,1\right][/itex].
    Second one:
    For this function denominator can not be zero [itex]sin(\frac{\pi}{2}x-\pi)\neq0[/itex] So that, we have to except the such x values [itex]x=\left\{x:x=2k,k\in Z\right\}[/itex] that make it zero. That was easy part. The hard one is to find the set of range of this g(x) function. I'm actually stuck in that. Help for [itex]g:ℝ-\left\{x:x=2k,k\in Z\right\}\rightarrow?[/itex] ??
  6. Oct 23, 2012 #5


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    Limits? Why would you care about limits? The question is about the function and its values on its defined domain, not about the values of some hypothetical extension onto some hypothetical extended domain that includes plus or minus infinity.

    What is an inverse of f(x) = [itex]cos(e^{-x})[/itex]?

    It's a fairly easy inverse to define -- at least symbolically.

    The inverse of cos is arccos.
    The inverse of e^x is log
    The inverse of - is -

    Now you only have the problem of taking the log of a number that may not be strictly positive. But that can be dealt with.

    HINT: The inverse of cos is multi-valued. You can pick which inverse you want.

    mfb has given a hint about a problem with the second one. The numerator is non-negative. The denominator is bounded by [-1,1]. The only way to get a small result is for the numerator to be small.

    If the numerator is small, what can be said about the denominator? In particular, what is its sign? What does that imply about small values of f(x).

    On the bright side, the function is continuous and piecewise differentiable. You are searching for an extremum. If you can enumerate the local extrema and pick out the right ones then perhaps you can find a global extremum in a particular range?

    I have not tried to solve the rest of the problem myself. If it's an exercise in a textbook then it ought to be tractable.
  7. Oct 23, 2012 #6


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    Would you mind confirming the form of the second? It seems unnecessarily complicated, but that would make more sense if it were [itex]\frac{\left|2x-1\right|}{sin(\frac{1}{2}\pi-\pi{x})}[/itex]
    Anyway, assuming [itex]\frac{\left|2x-1\right|}{sin(\frac{1}{2}\pi{x}-\pi)}[/itex] is right...
    Consider the continuous pieces of the function, (0, 2) etc. For each you can find a local max/min. What will each piece do outside that local extremum? I think you'll find there's just one gap in the range, of the form (0, y).
  8. Oct 24, 2012 #7
    Why would not? Of course, with my self-study I realized that it's useful to find the range of a function is given. We can find the boundaries of a function through limit then make out what its range is.

    I have no idea why you told me the inverse of the function? We don't need to use the inverse of the function to find the function's range and domain values.
  9. Oct 24, 2012 #8


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    Taking a limit will not tell you what the boundaries of a function are. A limit, if it exists, is the value that the function f() approaches as its argument approaches x.

    If the limit exists, it tells you only one thing about the domain of f() -- that the domain of f includes values that are arbitrarily close to (and not equal to) x.

    If the limit exists, it tells you only one thing about the range of f() -- that the range of f includes values that are arbitrarily close to (and possibly equal to) the limit.

    If you are asking yourself whether a particular value of y is in the range of a function f() then one way to proceed is to try to explicitly find an x such that f(x) = y. Ideally you can come up with a formula that takes a value y and produces such an x. The range of f() is then equal to the domain of this formula.

    Such a formula is an inverse of f().

    There are other approaches -- such as the mean value theorem. That might be a way to formalize what you had in mind with the "limit" argument.
  10. Oct 24, 2012 #9
    What if I take [itex]\stackrel{lim}{x\rightarrow\mp∞}f(x)[/itex] then find the values which the function has boundaries between them. So I can say the limit of the function x goes to infinity either sides in other words that would determine the range of the function. Did you catch my point?

    I know the meaning and rules of an inverse function. I meant it will not all the time be useful to try to find the domain and range of a function by solving from its inverse. If you try what you said on the functions I've posted here, you'll get what I meant. It just gets complicated.
  11. Oct 24, 2012 #10


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    ##f(x)=\frac{1}{x^2+1}##, f:R->R

    ##f(x)=\frac{\sin(x)}{x^2+1}##, f:R->R
    Both have limits of 0 for both sides. How does that help?
  12. Oct 24, 2012 #11
    Okay, I got it now. I'm so glad that you disproved me. I was confused and not certainly sure anyway before you told me this, now I learnt something new. Thanks..

    But I couldn't still find the range of the [itex]f(x)=\frac{\left|2x-1\right|}{sin(\frac{\pi}{2}x-\pi)}[/itex] Can you help me?
  13. Oct 24, 2012 #12


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    haruspex gave a good hint how you can approach that.
  14. Oct 24, 2012 #13


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    I am not sure that I understand your point.

    It sounds like you may be talking about a function like arctan(x). Its limit as x increases without bound is pi/2. Its limit as x decreases without bound is -pi/2.

    One could note that this function is continuous and invoke the mean value theorem to demonstrate that the range of arctan() must include all values in the open interval (-pi/2, pi/2)

    There are at least four problems with such an approach.

    1. mfb identified this one. The values of the function in the limit may not be global maximums or minimums.

    2. The function may or may not be continuous.

    3. As with arctan, your argument does not establish whether the limiting values are part of the range. [This is really just a restatement of problem 1]

    4. One or both limits may not even exist.

    On the first function it does not get complicated. It gets easy. That's why I selected that approach.

    On the second function you are correct. The inverse is difficult and perhaps even impossible to obtain as a simple formula. For that function, finding a local extremum for f(x) for 0 < x < 2 would be a useful exercise. That has been suggested multiple times already. Have you attempted it yet?
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