Question 1197653: Which if the following functions has domain (-1,1)and range (0,pi)
A. f(x)=sin((pi/2)x)
B. f(x)=arcsin(x)
C. f(x)=cos(x)
D. f(x)=arccos(x)
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
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
https://www.alamo.edu/contentassets/35e1aad11a064ee2ae161ba2ae3b2559/trigonometric/math2412-inverse-trig-functions.pdf
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)
Answer by ikleyn(52788)  (Show Source):
You can put this solution on YOUR website! .
If to consider your request formally, as it is literally written,
then the correct answer is " No one of the given functions ".
Why ? - Because the domain and the range are pointed INCORRECTLY in your question.
The correct writing is " the domain [-1,1] and range [0,pi] ".
Do you see the difference ?
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