Chapter 13.8, Problem 47PPSE

Solution

a.

To find: the domain, range and period of $y=\csc x$

Range of $y=\csc x$ is $\left\{y:y\in R,y\ge 1\text{ or }y\le -1\right\}$ .

Domain of $x$ in $y=\csc x$ is $\left\{x:x\in R,x\ne n\pi ,n\in Z\right\}$ .

Period of $y=\csc x$ is $2\pi$ .

Given:

Function $y=\csc x$

Concept used:

Reciprocal of $\csc x=\frac{1}{\sin x}$

Range of $y=\csc x$ will include the reciprocal of every value in the range of sine.

The domain of $y=\csc x$ is equal to domain of sine except where $\sin x=0$ .

Period of $\sec x$ is equal to period of $\sin x$ .

Calculation:

  $y=\csc x$ is the reciprocal of $y=\sin x$ . So, it’s domain and range are related to sine’s domain and range.

Range of $y=\sin x$ is $-1\le y\le 1$ .

Hence, the range of $y=\csc x$ is $\left\{y:y\in R,y\ge 1\text{ or }y\le -1\right\}$ here $R$ is any real number.

Domain of $x$ in $y=\csc x$ is $\left\{x:x\in R,x\ne n\pi ,n\in Z\right\}$ here $Z$ is any integer.

From the graph of $y=\sin x$ , period of $\sin x$ is $2\pi$

Period of $y=\csc x$ is $2\pi$ .

Conclusion:

Hence, the range of $y=\csc x$ is $\left\{y:y\in R,y\ge 1\text{ or }y\le -1\right\}$

Domain of $x$ in $y=\csc x$ is $\left\{x:x\in R,x\ne n\pi ,n\in Z\right\}$

Period of $y=\csc x$ is $2\pi$

b.

To find: the relative minimum in the interval $0\le x\le \pi$

  $\left(\frac{\pi }{2},1\right)$ is relative minimum of $y=\csc x$ in the interval $0\le x\le \pi$

Given:

Function $y=\csc x$

Concept used:

1. To find a max/min, find the first derivative and find values for which the derivative is zero.

2. Derivation of $y=\csc x$ by chain rule.

Calculation:

  $\begin{array}{c}y=\csc x\\ ={\left(\sin x\right)}^{-1}\\ \frac{dy}{dx}=\left(-1\right){\left(\sin x\right)}^{-2}\left(\cos x\right)\\ \frac{dy}{dx}=-\frac{\cos x}{{\sin }^{2}x}\end{array}$

At minima, $\frac{dy}{dx}=0$

  $\begin{array}{c}\Rightarrow -\frac{\cos x}{{\sin }^{2}x}=0\\ \therefore \cos x=0\\ x=\frac{-3\pi }{2},\frac{-\pi }{2},\frac{\pi }{2}\mathrm{.....}\end{array}$

At $x=\frac{\pi }{2}$

  $y=\frac{1}{\sin \frac{\pi }{2}}=1$

Conclusion:

Hence, $\left(\frac{\pi }{2},1\right)$ is relative minimum of $y=\csc x$ in the interval $0\le x\le \pi$

c.

To find: the relative maximum in the interval $\pi \le x\le 2\pi$

  $\left(\frac{3\pi }{2},-1\right)$ is relative maximum of $y=\csc x$ in the interval $\pi \le x\le 2\pi$

Given:

Function $y=\csc x$

Concept used:

1. To find a max/min, find the first derivative and find values for which the derivative is zero.

2. Derivation of $y=\csc x$ by chain rule.

Calculation:

  $\begin{array}{c}y=\csc x\\ ={\left(\sin x\right)}^{-1}\\ \frac{dy}{dx}=\left(-1\right){\left(\sin x\right)}^{-2}\left(\cos x\right)\\ \frac{dy}{dx}=-\frac{\cos x}{{\sin }^{2}x}\end{array}$

At maxima, $\frac{dy}{dx}=0$

  $\begin{array}{c}\Rightarrow -\frac{\cos x}{{\sin }^{2}x}=0\\ \therefore \cos x=0\\ x=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}\mathrm{.....}\end{array}$

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

  $y=\frac{1}{\sin \frac{3\pi }{2}}=-1$

Conclusion:

Hence, $\left(\frac{3\pi }{2},-1\right)$ is relative maximum of $y=\csc x$ in the interval $\pi \le x\le 2\pi$