Domain/range of inverse functions

AI Thread Summary
The discussion focuses on determining the domain and range of various inverse functions. For the function cos[arcsin(-2x)], the domain is found to be [-1/2, 1/2], with an expected range of [0, 1]. The second function, sin[arccos(x)] + 2cos(arcsin(x)], has a domain of [-1, 1] and an anticipated range of [0, 3]. The third function, sin[arccos(sin(arccos(x)))], also has a domain of [-1, 1], with the range expected to be [0, 180 degrees]. Participants suggest using rearrangement of equations to better understand how to derive the range from the given domain.
sk2nightfire
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Homework Statement

Find the domain/range:

1)cos[arcsin(-2x)]=f(x)
2)sin[arccos(x)] + 2cos(arcsin(x)]=f(x)
3)sin[arccos(sin(arccos(x)))]

Homework Equations


arccos
domain:{-1,1]
range[0,pi]
arcsin
domain:{-1,1]
range[-pi/2,pi/2]
tan
domain:{-infinity,infinity]
range[-pi/2,pi/2]

The Attempt at a Solution


1)d: -1 < -2x<1
so domain is [-1/2,1/2]

How do I find the range? Its supposed to be [0,1]
2)[-1,1] is the domain

the range is supposed to be [0,3],but why?3)for domain i got [-1,1],but range?
range Is supposed to be [0,180 degrees],but once again,I'm lost.
 
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There are several ways to approach this question, how did you find the domain?
by observation? or trial and error? or did you approach this a more formal way.
If you attempted by observation or trial and error simply rearrange the equation so it is in terms of x and do the same as what you did for domain (except using the 'f(x)' term) that is your range.
Showing the exact steps you used to solve the domain would be useful in showing you how to approach the range part
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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