Chapter 6.5, Problem 1QR

Solution

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

The absolute minimum is $-3$ at $x=\left\{\frac{\pi }{2},\frac{3\pi }{2}\right\}$ .

The absolute maximum is $3$ at $x=\{0,\pi ,2\pi \}$ .

Given information:

The given function is $y=3\cos 2x$ .

Calculation:

Range for the function $\cos 2x$ is $-1\le \cos 2x\le 1$ .

Thus, the minimum value for $\cos 2x$ is $-1$ and the maximum value is $1$ .

Determine the minimum value is for the function $y=3\cos 2x$ .

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

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

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

Determine the maximumvalue is for the function $y=3\cos 2x$ .

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

This occurs for $x=\pi n$ .

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

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