Chapter 4.2, Problem 60E

a)

To determine

To estimate: The absolute maximum and minimum values of the function $f\left(x\right)=x-2\cos x,-2\le x\le 0$ to two decimal places.

Answer to Problem 60E

The absolute maximum value of $f\left(x\right)$ is $f\left(-2\right)\approx -1.171$.

The absolute minimum value of $f\left(x\right)$ is $f\left(\frac{-\pi }{6}\right)\approx -2.256$.

Explanation of Solution

Graph:

Use the graphing calculator to draw the graph of the function $f\left(x\right)=x-2\cos x,-2\le x\le 0$ as shown below in Figure 1.

From Figure 1, it is observed that $f\left(-2\right)\ge f\left(x\right)$ for all values of x in the domain. Therefore, the absolute maximum value of $f\left(x\right)$ occurs at $x=-2$ and the value is $-1.171$

Also from Figure 1, it is observed that $f\left(-\frac{\pi }{6}\right)\le f\left(x\right)$ for all values of x in the domain. Therefore, the absolute minimum value of $f\left(x\right)$ occurs at $x=-\frac{\pi }{6}$ and the value is $-2.256$.

Thus, The absolute maximum value of $f\left(x\right)$ is $f\left(-2\right)\approx -1.171$ and the absolute minimum value of $f\left(x\right)$ is $f\left(\frac{-\pi }{6}\right)\approx -2.256$.

(b)

To determine

To find: The exact maximum and minimum values of the given function $f\left(x\right)=x-2\cos x,-2\le x\le 0$.

Answer to Problem 60E

The absolute maximum value of $f\left(x\right)$ is. $f\left(-2\right)\approx -1.1677$.

The absolute minimum value of $f\left(x\right)$ is. $f\left(\frac{-\pi }{6}\right)\approx -2.2557$.

Explanation of Solution

Obtain the first derivative of the given function.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(x-2\cos x\right)\\ =1-2\left(-\sin x\right)\left[\because {\left(\cos x\right)}^{\prime }=-\sin x\right]\\ =1+2\sin x\end{array}$

Set $f^{\prime }\left(x\right)=0$ and obtain the critical number.

$\begin{array}{l}1+2\sin x=0\\ 2\sin x=-1\\ \sin x=\frac{-1}{2}\\ x=\frac{-\pi }{6}\left[\because \sin \left(\frac{\pi }{6}\right)=\frac{1}{2}\right]\end{array}$

Hence, the critical numbers is $x=\frac{-\pi }{6}\approx -0.52$ which lies in the given interval $-2\le x\le 0$

Apply the critical points and the interval values in the function $f\left(x\right)$.

When $x=-2$,

$\begin{array}{c}f\left(-2\right)=\left(-2\right)-2\cos \left(-2\right)\\ =-2-2\cos 2\left[\because \cos \left(-\theta \right)=\cos \theta \right]\\ \approx -1.1677\end{array}$

When $x=0$,

$\begin{array}{c}f\left(0\right)=\left(0\right)-2\cos \left(0\right)\\ =0-2\left(1\right)\left[\because \cos \left(0\right)=1\right]\\ =-2\end{array}$

When $x=\frac{-\pi }{6}$,

$\begin{array}{c}f\left(\frac{-\pi }{6}\right)=\left(\frac{-\pi }{6}\right)-2\cos \left(\frac{-\pi }{6}\right)\\ =\frac{-\pi }{6}-2\cos \left(\frac{\pi }{6}\right)\left[\because \cos \left(-\theta \right)=\cos \theta \right]\\ =\frac{-\pi }{6}-2\left(\frac{\sqrt{3}}{2}\right)\left[\because \cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\right]\\ =\frac{-\pi }{6}-\sqrt{3}\end{array}$

               $\approx -2.2557$

Recall the fact that the largest functional value is absolute maximum and smallest functional value is absolute minimum.

Thus, the absolute maximum value of $f\left(x\right)$ is $f\left(-2\right)\approx -1.1677$ and the absolute minimum value of $f\left(x\right)$ is $f\left(\frac{-\pi }{6}\right)\approx -2.2557$.