Question

consider the inverse cosine function, defined by ( y = cos^{-1}x ) or ( y = arccos x ). complete parts (a) through (d).
(a) what is its domain?
( -1,1 )
(type your answer in interval notation. simplify your answer. type an exact answer, using ( pi ) as needed. use integers or fractions for a
(b) what is its range?
( 0,pi )
(type your answer in interval notation. simplify your answer. type an exact answer, using ( pi ) as needed. use integers or fractions for an
(c) is this function increasing or decreasing?
increasing
decreasing

Explanation:

Step1: Analyze the inverse cosine function

The inverse cosine function \(y = \cos^{-1}x\) or \(y=\arccos x\) is the inverse of the cosine function. For a function \(y = f(x)\) and its inverse \(y = f^{-1}(x)\), the domain of \(y = f^{-1}(x)\) is the range of \(y = f(x)\), and the range of \(y = f^{-1}(x)\) is the domain of \(y = f(x)\) (with appropriate restrictions for one - to - one nature).

The cosine function \(y=\cos x\) has a domain of \((-\infty,\infty)\) and a range of \([- 1,1]\). When we find the inverse of the cosine function (restricting the domain of \(y = \cos x\) to \([0,\pi]\) to make it one - to - one), the domain of \(y=\arccos x\) is the range of \(y = \cos x\) (i.e., \([-1,1]\)) and the range of \(y=\arccos x\) is the restricted domain of \(y=\cos x\) (i.e., \([0,\pi]\)).

To check if the function \(y = \arccos x\) is increasing or decreasing, we can use the derivative. The derivative of \(y=\arccos x\) is given by \(y'=-\frac{1}{\sqrt{1 - x^{2}}}\), for \(x\in(-1,1)\). Since \(y'<0\) for all \(x\in(-1,1)\), the function is decreasing.

Answer:

(a) The domain of \(y = \arccos x\) is \([-1,1]\)
(b) The range of \(y=\arccos x\) is \([0,\pi]\)
(c) The function \(y=\arccos x\) is decreasing.