The discussion revolves around finding the range of the function \( y = \sqrt{\ln(\cos(\sin(x)))} \). Participants explore both graphical and analytical methods to determine this range, questioning the implications of the function's domain and the behavior of the logarithmic and trigonometric components involved.
The discussion is active, with participants providing insights into the domain of the function and the implications of the square root and logarithm. There is an ongoing exploration of the range, with some participants suggesting specific values and others questioning the assumptions made in the reasoning.
Participants note the importance of understanding the domain of the function, particularly in relation to the behavior of \( \sin(x) \) and its zeros. The discussion includes considerations of the constraints imposed by the square root and logarithmic functions, as well as the need for clarity in mathematical notation.
All that you actually found here is that if ##\ x=0\,,\ ## then ##\ y=1\,.\ ## Therefore, 1 is in the range of your function.Buffu said:Homework Statement
Find the range ##y = \sqrt{\ln({\cos(\sin (x)}))}##
Homework Equations
The Attempt at a Solution
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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.
SammyS said: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.
PetSounds said:What is the range of ##y = cos (x)## ?
And how does that overlap with the domain of ##y = ln (x)##?Buffu said:[-1,1]
PetSounds said:And how does that overlap with the domain of ##y = ln (x)##?
And what is the range of ##ln (x)## for ##0 < x \leq 1## ?Buffu said:domain of ln x is (0, ##\infty##) .
So ##(0, 1]## part of cos x domain is only useful in this problem
PetSounds said:And what is the range of ##ln (x)## for ##0 < x \leq 1## ?
Bingo.Buffu said:less than 0 but we cannot have less than zero because of square root. So only 1 is left; Thus range is {0}.
Buffu said:less than 0 but we cannot have less than zero because of square root. So only 1 is left; Thus range is {0}.
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 ?Ray Vickson said:Yes. And the domain of ##f## is also very limited in the real line. What would it (the domain) be?
There are other solutions.Buffu said:when sin x is 0, that is 2pi
Oh yes sin x is also zero at π So the domain should be {x : x = π * n ∀n ∈ ℤ}haruspex said:There are other solutions.
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 }Buffu said:Oh yes sin x is also zero at π So the domain should be {x : x = π * n ∀n ∈ ℤ}
I did not get it. you just removed x.haruspex said: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 }
What you had posted said :Buffu said:I did not get it. you just removed x.
haruspex said:But why not omit x and write it my way?
haruspex 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.