Chapter 4.5, Problem 59E

Solution

To graph: The function $f\left(x\right)=-\cot x$ on the interval $\left(-\pi ,\pi \right)$ . Explain why is it correct to say $f$ is increasing on the interval $\left(0,\pi \right)$ but not correct to say $f$ is increasing on the interval $\left(-\pi ,\pi \right)$

Given information:

The function $f\left(x\right)=-\cot x$ on the interval $\left(-\pi ,\pi \right)$ .

Graph:

Consider the functions $f\left(x\right)=-\cot x$ on the interval $\left(-\pi ,\pi \right)$ ,

Sketch the graph of the functions on the interval $\left[-\pi ,\pi \right]$ .

  

From the graph, it can be seen that the function is not defined at $x=0$ .

As it is not defined everywhere at $\left(-\pi ,\pi \right)$ so it is not increasing on $\left(-\pi ,\pi \right)$ .

Also, $\frac{\pi }{4}>\frac{\pi }{6}$ where $a=\frac{\pi }{6}$ and $b=\frac{\pi }{4}$ .

But,

  $f\left(a\right)=-\sqrt{3}$

  $f\left(b\right)=-1$

So,

  $f\left(b\right)>f\left(a\right)$ for all $a,b\in \left(0,\pi \right)$

Hence, function is increasing on $\left(0,\pi \right)$ .

Also, $\frac{\pi }{4}>-\frac{\pi }{4}$ where $a=-\frac{\pi }{4}$ and $b=\frac{\pi }{4}$ .

But,

  $f\left(a\right)=1$

  $f\left(b\right)=-1$

So,

  $f\left(a\right)>f\left(b\right)$ for all $a,b\in \left(0,\pi \right)$

Hence, function is not increasing on $\left(-\pi ,\pi \right)$ .

Also, $f>g$ from $x=-\frac{\pi }{2}$ to $x=0.7$ .

Again, intersecting at $x=2.2$ and $f>g$ from $x=2.2$ to $x=\frac{\pi }{2}$ .

Interpretation:

The function $f$ is increasing on the interval $\left(0,\pi \right)$ but $f$ is not increasing on the interval $\left(-\pi ,\pi \right)$ .