Chapter 4, Problem 118CR

To determine

To graph: the function manually.

Explanation of Solution

Given information:

Need to create graph of function $f\left(x\right)=\frac{1}{2}\csc \left(2x-\pi \right)$ .

Graph:

Give function is $f\left(x\right)=\frac{1}{2}\csc \left(2x-\pi \right)$ .

Lets first sketch the graph of $y=\frac{1}{2}\sin \left(2x-\pi \right)$

Amplitude of the function is $\frac{1}{2}$ .

Period of the sine function is $2\pi .$

So evaluate the value of $x$ at interval $0$ and $2\pi$ by solving follows two equation

  $2x-\pi =0$ and $2x-\pi =\pi$

  $x=\frac{\pi }{2}$ , $x=\frac{3\pi }{2}$

As value of sine function is zero at midpoint and endpoint of the interval $x=\frac{\pi }{2}$ , $x=\frac{3\pi }{2}$ ,

Corresponding cosecant function $f\left(x\right)=\frac{1}{2}\csc \left(2x-\pi \right)=\frac{1}{2}\left(\frac{1}{\sin \left(2x-\pi \right)}\right)$ will have vertical asymptotes at all the points where $\frac{1}{2}\sin \left(2x-\pi \right)$ is zero that is $f\left(x\right)$ have vertical asymptote at $-\frac{3\pi }{2},-\pi ,-\frac{\pi }{2},0,\frac{\pi }{2},\pi ,\frac{3\pi }{2}$ .

Based on above calculation, Graph of $f\left(x\right)=\frac{1}{2}\csc \left(2x-\pi \right)$ represented by blue curve is as follows.