Domain/range of inverse functions

[sk2nightfire]

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.

 

[cheff3r]

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

 

[Architeuthis Dux]

1) The domain of the inverse function, arcsin(-2x), is -1/2 ≤ -2x ≤ 1/2. This can be rewritten as -1/4 ≤ x ≤ 1/4. The range of the inverse function, cos(arcsin(-2x)), is [0,1]. This means that the domain of f(x) is -1/4 ≤ x ≤ 1/4 and the range is [0,1].

2) The domain of the inverse function, arccos(x), is -1 ≤ x ≤ 1. The domain of arcsin(x) is also -1 ≤ x ≤ 1. Therefore, the domain of the inverse function, 2cos(arcsin(x)), is -1 ≤ x ≤ 1. The range of the inverse function, sin(arccos(x)), is [0,1]. This means that the domain of f(x) is -1 ≤ x ≤ 1 and the range is [0,3].

3) The domain of the inverse function, sin(arccos(sin(arccos(x)))), is -1 ≤ x ≤ 1. The range of the inverse function, arccos(sin(arccos(x))), is [0,π]. The range of the inverse function, sin[arccos(sin(arccos(x)))], is [-1,1]. This means that the domain of f(x) is -1 ≤ x ≤ 1 and the range is [-1,1].