Answer and Explanation :
Given : Function \(f(x)=-2\cot 3x\)
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.
\(f(x)=-2\cot 3x\)
\(f(x)=-2(\frac{\cos 3x}{\sin 3x})\)
For domain, \(\sin3x
eq 0\)
So, \(\sin3x
eq \sin n\pi\)
\(3x
eq n\pi\)
\(x
eq \frac{n\pi}{3}\)
The value of x is define as \(x
eq n\pi+(-1)^n\frac{\pi}{3}\)
The domain of the function is all real numbers except \(x
eq n\pi+(-1)^n\frac{\pi}{3}\)
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 \(y=A\cot (Bx-C)+D\)
Where, Period is \(P=\frac{\pi}{|B|}\)
On comparing, B=3
So, The period of the given function is \(P=\frac{\pi}{|3|}\)
3) Vertical asymptote is defined as the line which approaches to infinity but never touches the line.
The vertical asymptote is at \(x= n\pi+(-1)^n\frac{\pi}{3}\) where function is not defined.
The two vertical asymptote is
Put n=0,
\(x= (0)\pi+(-1)^{0}\frac{\pi}{3}\)
\(x=\frac{\pi}{3}\)
Put n=1,
\(x= (1)\pi+(-1)^{1}\frac{\pi}{3}\)
\(x=\pi-\frac{\pi}{3}\)
\(x=\frac{2\pi}{3}\)
So, The two vertical asymptote are \(x=\frac{\pi}{3},\frac{2\pi}{3}\)
