Question

Inverse trigonometric function domains and ranges: For each of the following functions ${\sin }^{-1}\left(x\right)$,${\tan }^{-1}\left(x\right)$, ${\sec }^{-1}\left(x\right)$, ${\sin }^{-1}\left(\frac{x}{3}\right)$(a) list the domain and range, (b) draw a labeled graph, and (c) talk about the domains and ranges in relation to the unit circle.

Short answer

1. The domain of this function $f\left(x\right)={\sin }^{-1}\left(x\right)$is [-$1$,$1]$ and the range of the function is $\left(-\frac{\pi }{2},\frac{\pi }{2}\right).$
2. The domain of this function $f\left(x\right)={\tan }^{-1}\left(x\right)$is $(-\infty ,\infty )$and the range of the function $f\left(x\right)={\tan }^{-1}\left(x\right)$is $\left(-\frac{\pi }{2},\frac{\pi }{2}\right).$
3. The domain of this function $f\left(x\right)={\sec }^{-1}\left(x\right)$is $(-\infty ,-1]\cup [1,\infty )$and the range of this function $f\left(x\right)={\sec }^{-1}\left(x\right)$is $\left[0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right].$
4. The domain of this function $f\left(x\right)={\sin }^{-1}\left(\frac{x}{3}\right)$is [-$3$,$3$] and the range of this function $f\left(x\right)={\sin }^{-1}\left(\frac{x}{3}\right)$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right].$

Step-by-step solution

Part(a) Step 1: Given information.

The given function is$f\left(x\right)={\sin }^{-1}\left(x\right)$

Part (a) Step 2: Calculation.

The trigonometric functions' domain and range are well known. All we have to do is adjust them in accordance with the function's argument.

The given function is $f\left(x\right)={\sin }^{-1}\left(x\right)$.

The domain of this function is localid="1661320810672" $\left[-1,1\right]$and the range of this function is localid="1660460048803" $\left(-\frac{\pi }{2},\frac{\pi }{2}\right).$

The domain and range of this function, when applied to the unit circle, are the first and fourth quadrants of the circle, respectively.

Part(b) Step 1: Given information

The given function is$f\left(x\right)={\tan }^{-1}\left(x\right)$.

Part(b) Step 2: Calculation.

The trigonometric functions' domain and range are well known. All we have to do is adjust them in accordance with the function's argument.

The given function is $f\left(x\right)={\tan }^{-1}\left(x\right).$

The domain of this function is $(-\infty ,\infty )$and the range of this function is $\left(-\frac{\pi }{2},\frac{\pi }{2}\right).$

The entire real line is this function's domain in the setting of the unit circle, and its range is the first and fourth quadrants of the circle.

The domain and range of this function, when applied to the unit circle, are the first and fourth quadrants of the circle, respectively.

Part(c) Step 1: Given information

The function $f\left(x\right)={\sec }^{-1}\left(x\right).$

Part(c) Step 2: Calculation

The trigonometric functions' domain and range are well known. All we have to do is adjust them in accordance with the function's argument.

The given function is $f\left(x\right)={\sec }^{-1}\left(x\right).$

The domain of this function is $(-\infty ,-1]\cup [1,\infty )$and the range of this function is $\left[0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right].$

The first and second quadrants of the unit circle, with the exception of the portion inside the circle, are the domain and range of this function in the context of the unit circle$\frac{\pi }{2}$respectively.

Part(d) Step 1: Given information

The function $f\left(x\right)={\sin }^{-1}\left(\frac{x}{3}\right).$

Part(d) Step 2: Calculation

The trigonometric functions' domain and range are well known. All we have to do is adjust them in accordance with the function's argument.

The given function is $f\left(x\right)={\sin }^{-1}\left(\frac{x}{3}\right).$

As was mentioned in the part before, the domain of ${\sin }^{-1}\left(x\right)$is $[-1,1]$and the range of ${\sin }^{-1}\left(x\right)$is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right].$

Then, the domain of the given function;$f\left(x\right)={\sin }^{-1}\left(\frac{x}{3}\right)$is $[-3,3]$and the range of the given function;$f\left(x\right)={\sin }^{-1}\left(\frac{x}{3}\right)$is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right].$

This function's domain in the context of the unit circle is three times the diameter, expanded on both sides, and its range is the first and fourth quadrants of the unit circle.