Question 1197653
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Choices A and C are immediately ruled out because their domain is "set of all real numbers". 


The answer is one of the inverse trig functions either arcsine or arccosine.


Refer to this page
<a href = "https://www.alamo.edu/contentassets/35e1aad11a064ee2ae161ba2ae3b2559/trigonometric/math2412-inverse-trig-functions.pdf">https://www.alamo.edu/contentassets/35e1aad11a064ee2ae161ba2ae3b2559/trigonometric/math2412-inverse-trig-functions.pdf</a>
To see how the y = sin(x) function has its domain restricted to the interval [-pi/2, pi/2] so that it becomes one-to-one. This allows the inverse to be possible. 


On the domain [-pi/2, pi/2] the corresponding range is [-1, 1]
When computing the inverse, the domain and range swap.
Therefore, the domain of arcsine is [-1,1] and its range is [-pi/2, pi/2]
This rules out choice B.


You should find that y = cos(x) should have a domain restriction of [0, pi] so that we have a one-to-one portion.
This covers a range of [-1,1]
The domain and range swap leading arccos to have a domain of [-1,1] and range of [0,pi]
In short, this shows why choice D is the answer.


Side note: I don't know why your teacher has excluded the endpoints. The endpoints should be included.
Something like f(x) = 0 is possible for f(x) = arccos(x). It occurs when x = 1.


Answer: Choice D.  f(x) = arccos(x)
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