Domain/range of inverse functions

In summary, the conversation is discussing finding the domain and range for three equations involving inverse trigonometric functions. For the first equation, the domain is found to be [-1/2,1/2] but the range is unclear. For the second equation, the domain is [-1,1] and the range is [0,3]. For the third equation, the domain is [-1,1] and the range is [0,180 degrees]. The person being asked for help is unsure of how to approach finding the range and is asking for guidance on how to do so.
  • #1
sk2nightfire
2
0

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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  • #2
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
 
  • #3


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].
 

Related to Domain/range of inverse functions

What is the domain of an inverse function?

The domain of an inverse function is the range of the original function. In other words, it is the set of all possible input values for the inverse function.

What is the range of an inverse function?

The range of an inverse function is the domain of the original function. It is the set of all possible output values for the inverse function.

Can the domain of an inverse function be different from the domain of the original function?

Yes, the domain of an inverse function can be different from the domain of the original function. This is because the inverse function is essentially a reflection of the original function over the line y=x, which can result in a different set of input values.

Why is it important to consider the domain and range of inverse functions?

Understanding the domain and range of inverse functions is important because it helps us determine the validity of the inverse function. If the domain and range are not carefully considered, the inverse function may not be a valid function.

How can we find the domain and range of an inverse function?

To find the domain and range of an inverse function, we can use algebraic methods such as solving for the independent variable in terms of the dependent variable. We can also use graphical methods by plotting the original function and its inverse on a graph and identifying the domain and range from the graph.

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