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Range of a weird function

  1. Jan 28, 2017 #1
    1. The problem statement, all variables and given/known data

    Find the range ##y = \sqrt{\ln({\cos(\sin (x)}))}##

    2. Relevant equations


    3. The attempt at a solution

    https://www.desmos.com/calculator

    I used a graphing calculator to find the intersection between ##y = e^{x^2}## and ##y = \cos(\sin(x))##.
    Which I get as ##(0,1)##. So the range is ##\{0\}##.

    But I want to find the range without graphs and by analytical methods.
    Thanks for help.
     
  2. jcsd
  3. Jan 28, 2017 #2
    What is the range of ##y = cos (x)## ?
     
  4. Jan 28, 2017 #3

    SammyS

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    All that you actually found here is that if ##\ x=0\,,\ ## then ##\ y=1\,.\ ## Therefore, 1 is in the range of your function.

    I suggest the first thing to do is to determine the (implied) domain of your function.
     
  5. Jan 28, 2017 #4
    Putting x = 0 ##y = \sqrt{\ln(\cos(\sin(0)))} =\sqrt{\ln(\cos 0))} = \sqrt{\ln(1)} = 0##, So y = 0 is also in range.
    So the range is {0,1}.

    Domain of function is (0 to pi/2) + 2n*pi.
    [-1,1]
     
    Last edited: Jan 28, 2017
  6. Jan 28, 2017 #5
    And how does that overlap with the domain of ##y = ln (x)##?
     
  7. Jan 28, 2017 #6
    domain of ln x is (0, ##\infty##) .

    So ##(0, 1]## part of cos x domain is only useful in this problem
     
  8. Jan 28, 2017 #7
    And what is the range of ##ln (x)## for ##0 < x \leq 1## ?
     
  9. Jan 29, 2017 #8
    less than 0 but we cannot have less than zero because of square root. So only 1 is left; Thus range is {0}.
     
    Last edited: Jan 29, 2017
  10. Jan 29, 2017 #9
    Bingo.
     
  11. Jan 29, 2017 #10

    Ray Vickson

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    Yes. And the domain of ##f## is also very limited in the real line. What would it (the domain) be?
     
  12. Jan 29, 2017 #11
    Domain of my original function would be when sin x is 0, that is 2pi or for general solution 2* pi *n. So my domain would be {x : x = 2pi * n ##\forall n \in \mathbb Z##}. Right ?
     
  13. Feb 3, 2017 #12

    haruspex

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    There are other solutions.
     
  14. Feb 3, 2017 #13
    Oh yes sin x is also zero at π So the domain should be {x : x = π * n ∀n ∈ ℤ}
     
  15. Feb 3, 2017 #14

    haruspex

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    Looks right. Your use of the predicates is a little inaccurate. There does not exist an x such that it equals π * n for all integers n. You mean {π * n : n∈ ℤ }
     
  16. Feb 3, 2017 #15
    I did not get it. you just removed x.
     
  17. Feb 3, 2017 #16

    haruspex

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    What you had posted said :
    "the set of things x such that x equals πn for all integers n".
    There is no number that can equal πn for two different integers n, let alone all infinity of them.
    If you want to use x and n then I suggest using ∃n. Maybe {x:∃n∈ℕ:x=πn}. But why not omit x and write it my way?
     
  18. Feb 4, 2017 #17
    Your way is better.

    Oh I understand what you mean.
     
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