consider the inverse sine function, defined by y = sin⁻¹x or y = arcsin x. complete parts (a) through (d)…

Explanation:

Step1: Recall domain of inverse - sine

The domain of the inverse - sine function $y = \sin^{-1}(x)$ or $y=\arcsin(x)$ is the set of values of $x$ for which the function is defined. Since the sine function $y = \sin(t)$ has a range of $[- 1,1]$, the domain of $y=\arcsin(x)$ is $[-1,1]$.

Step2: Recall range of inverse - sine

The range of the inverse - sine function $y=\arcsin(x)$ is the set of angles $y$ such that $y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. This is because the sine function is one - to - one on the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Step3: Analyze the monotonicity

The derivative of $y = \arcsin(x)$ is $y'=\frac{1}{\sqrt{1 - x^{2}}}$, and for $x\in(-1,1)$, $y'>0$. So the function $y = \arcsin(x)$ is increasing on its domain $[-1,1]$.

Step4: Check the value for non - definition

Since the domain of $y=\arcsin(x)$ is $[-1,1]$, and $-2\notin[-1,1]$, $\arcsin(-2)$ is not defined.

Answer:

(a) $[-1,1]$ (b) $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (c) increasing (d) $-2$ is not in the domain