Question

Analyze the function f(x) = - 2 cot 3x. Include:

Answer 1

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, $\sin3x\neq 0$

So, $\sin3x\neq \sin n\pi$

$3x\neq n\pi$

$x\neq \frac{n\pi}{3}$

The value of x is define as $x\neq n\pi+(-1)^n\frac{\pi}{3}$

The domain of the function is all real numbers except $x\neq n\pi+(-1)^n\frac{\pi}{3}$

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

Answer 2

The domain for the function is:
all real numbers except n π/3 where n is an an integer

The range for the function is:
all real numbers

The period is
π/3

Vertical asymptotes
x = n π/3 where n is an integer