Domain/range of inverse functions

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SUMMARY

The discussion focuses on determining the domain and range of inverse trigonometric functions, specifically for the functions: cos[arcsin(-2x)], sin[arccos(x)] + 2cos(arcsin(x)], and sin[arccos(sin(arccos(x)))]. The domain for cos[arcsin(-2x)] is established as [-1/2, 1/2], while the expected range is [0, 1]. For sin[arccos(x)] + 2cos(arcsin(x)], the domain is [-1, 1] with a range of [0, 3]. Lastly, the domain for sin[arccos(sin(arccos(x)))] is also [-1, 1], with a range of [0, 180 degrees]. The discussion emphasizes the importance of rearranging equations to find ranges effectively.

PREREQUISITES
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  • Knowledge of function composition and transformations
  • Familiarity with domain and range concepts in mathematics
  • Ability to manipulate inequalities for solving domains
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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.
 
Last edited:
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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
 

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