Chapter 6.5, Problem 4QR

Solution

To find:The absolute minimum and maximum value of $y=3-3\sin x$ in the interval $[0,2\pi ]$ and the location where the maxima and minima occur.

The absolute maximum is $6$ at $x=\frac{3\pi }{2}$ .

The absolute minimum is $0$ at $x=\frac{\pi }{2}$ .

Given information:

The given function is $y=3-3\sin x$ .

Calculation:

Range for the function $\sin x$ is $-1\le \sin x\le 1$ .

Thus, the minimum value for $\sin x$ is $-1$ and the maximum value is $1$ since negative values can be included in the square-root.

Determine the maximumvalue is for the function $y=3-3\sin x$ .

  $\begin{array}{c}y=3-3\sin x\\ =3-3\left(-1\right)\\ =6\end{array}$

This occurs for $x=\frac{3\pi }{2}+2\pi n$ .

For the range $\left[0,2\pi \right]$, the minimum value occurs at: $x=\frac{3\pi }{2}$ .

Determine the minimumvalue is for the function $y=3-3\sin x$ .

  $\begin{array}{c}y=3-3\sin x\\ =3-3\left(1\right)\\ =0\end{array}$

This occurs for $x=\frac{\pi }{2}+2\pi n$ .

For the range $\left[0,2\pi \right]$, the minimum value occurs at: $x=\frac{\pi }{2}$ .

Therefore, the absolute maximum is $6$ at $x=\frac{3\pi }{2}$ and absolute minimum is $0$ at $x=\frac{\pi }{2}$ .