Chapter 1.6, Problem 13E

(a)

To determine

To find: The period of the function $y=-3\tan \left(3x+\pi \right)+2$ .

Answer to Problem 13E

The period of the function is $\frac{\pi }{3}$ .

Explanation of Solution

Given information:

The given function is $y=-3\tan \left(3x+\pi \right)+2$ .

Calculation:

The period of a function is given by $\frac{2\pi }{B}$ .

Consider the given function.

  $y=-3\tan \left(3x+\pi \right)+2$

So, the period of the given function is:

  $\begin{array}{c}\frac{\pi }{B}=3\\ B=\frac{\pi }{3}\end{array}$

Therefore, the period of the function is $\frac{\pi }{3}$ .

(b)

To determine

To find: The domain of the function $y=-3\tan \left(3x+\pi \right)+2$ .

Answer to Problem 13E

The domain of the function is all real $x\ne \frac{k\pi }{6}$ for odd integers $k$ .

Explanation of Solution

Given information:

The given function is $y=-3\tan \left(3x+\pi \right)+2$ .

Calculation:

The domain of a function is the set of all input values such that the function is defined.

The domain included all value of $x$ except the value of $x$ that are vertical asymptotes.

The vertical asymptotes $y=\tan x$ has vertical asymptotes at $x=\frac{k\pi }{2}$ , where $k$ is an odd integer.

Calculate the domain of the function.

  $\begin{array}{c}3+\pi =\frac{k\pi }{2}\\ 3x=\frac{k\pi }{2}-\pi \\ x=\frac{k\pi -2\pi }{6}\end{array}$

So domain of the function is $x\ne \frac{k\pi -2\pi }{6}$ for all integers $k$ . Since $x\ne \frac{k\pi -2\pi }{66}=\frac{k\pi }{6}-\frac{\pi }{3}$ and the period of the function is $\frac{\pi }{3}$ , $x\ne \frac{k\pi -2\pi }{6}$ is equivalent to $x\ne \frac{k\pi }{6}$ for odd integer $k$ .

Therefore, domain of the function is all real $x\ne \frac{k\pi }{6}$ for odd integers $k$ .

(c)

To determine

To find: The range of the function $y=-3\tan \left(3x+\pi \right)+2$ .

Answer to Problem 13E

The range of the function is $\left(-\infty ,\infty \right)$ .

Explanation of Solution

Given information:

The given function is $y=-3\tan \left(3x+\pi \right)+2$ .

Calculation:

The range of a function is the set of all output values.

It is known that the range of the tangent function is the set of all real numbers.So, the function $y=-3\tan \left(3x+\pi \right)+2$ is also all real numbers.

Therefore, the range of the function is $\left(-\infty ,\infty \right)$ .

(d)

To determine

To graph:The function $y=-3\tan \left(3x+\pi \right)+2$ .

Explanation of Solution

Given information:

The given function is $y=-3\tan \left(3x+\pi \right)+2$ .

Graph:

To graph a function $y=-3\tan \left(3x+\pi \right)+2$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=-3\tan \left(3x+\pi \right)+2$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(-)}\boxed{3}\boxed{\text{TAN}}\boxed{3}\boxed{\text{X}}\boxed{+}\boxed{2\text{nd}}\boxed{^}\boxed{)}\boxed{+}\boxed{2}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=-3\tan \left(3\text{X}+\pi \right)+2$

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of both the equations as shown below.

  

Figure (1)

Interpretation: From the graph it can be noticed that three periods of the function are shown in the window.