Chapter 6.5, Problem 3QR

Solution

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

The absolute minimum is $0$ at $x=\left\{\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\right\}$ .

The absolute maximum is $2$ at $x=\{0,\pi ,2\pi \}$ .

Given information:

The given function is $y=2\sqrt{\cos 2x}$ .

Calculation:

Range for the function $\sqrt{\cos 2x}$ is $0\le \sqrt{\cos 2x}\le 1$ .

Thus, the minimum value for $\cos 2x$ is $0$ and the maximum value is $1$ since negative values can be included in the square-root.

Determine the minimum value is for the function $y=2\sqrt{\cos 2x}$ .

  $\begin{array}{c}y=2\sqrt{0}\\ =0\end{array}$

This occurs for $x=\frac{\pi }{4}+\frac{\pi }{2}n$ .

For the range $\left[0,2\pi \right]$, the minimum value occurs at: $x=\left\{\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\right\}$ .

Determine the maximum value is for the function $y=2\sqrt{\cos 2x}$ .

  $\begin{array}{c}y=2\sqrt{1}\\ =2\end{array}$

This occurs for $x=\pi n$ .

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

Therefore, The absolute minimum is $0$ at $x=\left\{\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\right\}$ and the absolute maximum is $2$ at $x=\{0,\pi ,2\pi \}$ .