Question

For question 44 is f(2pi/4) a positive or a negative number

Please help ​

Answer 1

Amplitude = 3

Period = π

Explanation:

`\boxed{\begin{minipage}{8.7cm}\underline{Standard form of a cosecant function}\\\\$y=A\csc(Bx)$\\\\where:\\\\\phantom{ww}$\bullet$ $|A|$ is the stretch factor. \\ \\\phantom{ww}$\bullet$ $(2\pi)/(|B|)$ is the period.\\\\\phantom{ww}$\bullet$ $x=(\pi)/(|B|)$ is the interval of vertical asymptotes.\\\end{minipage}}`

Given function:

`f(x)=3 \csc 2x`

Therefore:

• A = 3
• B = 2

Period

`\textsf{Period} = (2 \pi)/(|B|)=(2 \pi)/(|2|)=\pi`

Amplitude

The amplitude of the given function is 3, so:

• minimum points are when y = 3.

• maximum points are when y = -3.

x-values of minimum points

`\begin{aligned}3\csc(2x)=3 \implies \csc(2x)&=1\\(1)/(\sin(2x))&=1\\\sin(2x)&=1\\2x&=(\pi)/(2)+2\pi n\\x&=(\pi)/(4)+\pi n \end{aligned}`

`\implies \textsf{Minimum points}: \left((\pi)/(4)\pm \pi n,3 \right)`

x-values of maximum points

`\begin{aligned}3\csc(2x)=-3 \implies \csc(2x)&=-1\\(1)/(\sin(2x))&=-1\\\sin(2x)&=-1\\2x&=-(\pi)/(2)+2\pi n\\x&=-(\pi)/(4)+\pi n \end{aligned}`

`\implies \textsf{Maximum points}: \left(-(\pi)/(4)\pm \pi n,-3 \right)`

Asymptotes:

`\implies x=(\pi)/(|B|)=(\pi)/(|2|)=(1)/(2) \pi`

Therefore, there are vertical asymptotes every ¹/₂π.

To graph the given function:

• 
  `\textsf{Draw vertical asymptotes at} \;\;x = 0 + (\pi)/(2) n`
• 
  `\textsf{Plot minimum points at}\; \left((\pi)/(4)\pm \pi n,3 \right)`
• 
  `\textsf{Plot maximum points at}\; \left(-(\pi)/(4)\pm \pi n,-3 \right)`
• Draw curves between the asymptotes with the given min/max points.