Find the domain of and range of the function f(x) ,and express your answer in interval form $f(x)=\sqrt{\csc(3x)}$

I got the domain, like how its restricted and $\sin(x)$ can't equal to zero. However, I'm stuck on the range part of the question. I'm not sure how to do it. I got 0 ≤ $\sqrt{csc(3x)}$ ≤ 1, which is wrong when I checked online graphing calculator

This is a review question for University Calculus.


To determine the domain of $\sqrt{\csc(3x)}$, you have to first look at where $\sin(3x) = 0$ and $\csc(3x) < 0$. Notice that this is determined in the following manner:

$$\sin(3x) =0 \implies 3x = \sin^{-1}(0) \implies 3x = k\pi, k \in \mathbb{Z}, \implies x = \frac{k\pi}{3}, k \in \mathbb{Z}.$$ We also note that $\csc(3x) < 0$ when $\sin(3x) < 0$, so we see that $$\sin(3x) < 0 \implies \pi < 3x< 2\pi \implies \frac{\pi}{3} + 2k\pi < x < \frac{2\pi}{3} + 2k\pi, k \in \mathbb{Z}.$$ Therefore the domain of $\sqrt{\csc(3x)}$ is the set of all $x \in \mathbb{R}$ such that $x \neq \frac{k\pi}{3}$ and $x \not \in \left( \frac{\pi}{3} + 2k\pi, \frac{2\pi}{3} + 2k\pi \right)$.

To determine the range of $\sqrt{\csc(3x)}$ note that $$\sqrt{\csc(3x)} = \frac{1}{\sqrt{\sin(3x)}}. $$ Since $\sin(3x) \to 0$ as $x \to 0$, the function is clearly unbounded (above).

It is also worth noting that the range of the function $f(g(x))$ is the range of $f$ when restricted to the domain of $g(x)$ such that the function is well defined. Notice that $\sqrt{\csc(3x)}$ is the composite of $\sin(3x)$ and $\frac{1}{\sqrt{x}}$. Therefore, the range of $\sqrt{\csc(3x)}$ is the range of $\frac{1}{\sqrt{x}}$ when restricted to the domain determined above. It so follows that the range of $\sqrt{\csc(3x)}$ is $[1, \infty)$, since the largest value of $\sin(3x)$ on the restricted domain is $1$. Hence the minimum of $\sqrt{\csc(3x)}$ is $\sqrt{1}=1$.

Hope this helps.

I disagree with the comments made by Dr. Sonnhard Graubner. One should not be forced to rely on graphical techniques. While the student should be able to graph this type of functions, a purely analytic solutions should be able to be provided.


Assuming the function is real-valued:


For the expression $\sqrt{a}$ to be defined, we need $a\ge0$. Thus, $\csc(3x)\ge0$.

Since $\csc(x)=\frac{1}{\sin(x)}$, we also need $\sin(3x)\ne0\Leftrightarrow3x\ne n\cdot\pi\Leftrightarrow x\ne n\cdot\frac{\pi}{3}, n\in\mathbb{Z}$

Now $\csc(3x)\ge0$ iff $\sin(3x)\ge0$

We know $\sin(x)\ge0$ iff $x\in\left[2n\pi;(2n+1)\pi\right], n\in\mathbb{Z}$

Then $\sin(3x)\ge0$ iff $x\in\left[\frac{2n}{3}\pi;\frac{2n+1}{3}\pi\right], n\in\mathbb{Z}$

Combining these two restrictions yields the Domain $D=\{x\in\mathbb{R}\colon \exists n\in\mathbb{Z}\left(x\in\left(\frac{2n}{3}\pi;\frac{2n+1}{3}\pi\right)\right)\}$

As seen above, for this domain the expression $\sin(3x)$ takes on the values in the interval $(0;1]$

Then $\csc(3x)=\frac{1}{\sin(3x)}$ takes on values in the interval $[1;+\infty)$

Finally, this means that $\sqrt{\csc(3x)}$ will take on values in the interval $[1;+\infty)$, meaning that the range is $R=\{x\in\mathbb{R}\colon x\ge1\}$

Note: Knowing the sines periodic behavior we can see that $f(x)$ behaves the same in any interval $\left(\frac{2n}{3}\pi;\frac{2n+1}{3}\pi\right)$ for any $n\in\mathbb{Z}$ and will take on all the values in its range in any of these intervals.


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