Answer and Explanation :
Given : Function [tex]f(x)=-2\cot 3x[/tex]
To find :
1) Domain and range
2) Period
3) Two Vertical Asymptotes
Solution :
1) Domain is defined as the set of possible values of x where function is defined.
[tex]f(x)=-2\cot 3x[/tex]
[tex]f(x)=-2(\frac{\cos 3x}{\sin 3x})[/tex]
For domain, [tex]\sin3x\neq 0[/tex]
So, [tex]\sin3x\neq \sin n\pi[/tex]
[tex]3x\neq n\pi[/tex]
[tex]x\neq \frac{n\pi}{3}[/tex]
The value of x is define as [tex]x\neq n\pi+(-1)^n\frac{\pi}{3}[/tex]
The domain of the function is all real numbers except [tex]x\neq n\pi+(-1)^n\frac{\pi}{3}[/tex]
The range is defined as all the y values for every x.
So, The range of the function is all real numbers.
2) The general form of the cot function is [tex]y=A\cot (Bx-C)+D[/tex]
Where, Period is [tex]P=\frac{\pi}{|B|}[/tex]
On comparing, B=3
So, The period of the given function is [tex]P=\frac{\pi}{|3|}[/tex]
3) Vertical asymptote is defined as the line which approaches to infinity but never touches the line.
The vertical asymptote is at [tex]x= n\pi+(-1)^n\frac{\pi}{3}[/tex] where function is not defined.
The two vertical asymptote is
Put n=0,
[tex]x= (0)\pi+(-1)^{0}\frac{\pi}{3}[/tex]
[tex]x=\frac{\pi}{3}[/tex]
Put n=1,
[tex]x= (1)\pi+(-1)^{1}\frac{\pi}{3}[/tex]
[tex]x=\pi-\frac{\pi}{3}[/tex]
[tex]x=\frac{2\pi}{3}[/tex]
So, The two vertical asymptote are [tex]x=\frac{\pi}{3},\frac{2\pi}{3}[/tex]
