Domain of f(x) = sqrt(1-sin(x)): Understanding and Calculating

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The domain of the function f(x) = sqrt(1-sin(x)) includes all real numbers except for values where sin(x) equals 1, which occurs at increments of π/2 plus multiples of π. The function is undefined when 1-sin(x) becomes negative, specifically when sin(x) is greater than 1, which is impossible since sin(x) can only reach a maximum of 1. The derivative f'(x) was calculated as (1/2 - sin(x))(-cos(x)), but its correctness was questioned. Understanding the behavior of the sine function is crucial for determining the domain accurately. The discussion emphasizes the importance of recognizing when sin(x) reaches its maximum to define the domain of f(x).
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Homework Statement


let f(x) = sqrt(1-sin(x))


Homework Equations


What is the domain of f?
What is the domain of f'(x)?


The Attempt at a Solution


I understand that the domain of f is all real numbers not including every increment of 90 degrees, but I am not sure how to state that.
I also found that f'(x) = (1/2 -sin(x))(-cos(x)) I am not sure if that's correct though
 
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A point is not in the domain if it does not have a function value. The only possible situation that f(x)=\sqrt{1-sin(x)} does not exist, is when you take the root of a negative number. So you will have to search for what x we have that 1-sin(x)<0.
 
go on google and type

wolfrom alpha

look at the graphs and try understand
 
Micromass is spot on here, for what values of x satisfy \sin x\leqslant 1? Do you know the graph of \sin x?
 
so the domain is all real numbers except when sin(x)<1?
 
Yes! But can you say explicitly when sin(x)<1?
 
No that's not what we're saying the domain if all numbers which satisfy sin(x)<1, not the other way around.
 
thats where i got lost
 

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