Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 62

a.) Given the function $f(x) = x - 2 \cos x, \quad -2 \leq x \leq 0$, use a graph to estimate the absolute maximum and minimum values.
b.) Use calculus to find the exact maximum and minimum values.

a.)


Based from the graph, the absolute minimum value is approximately $f(-0.5) \approx -2.20$ and the absolute maximum value is approximately $f(-2) = -1.20$

b.) To find the exact value, we take the derivative of the function.


$\begin{equation} \begin{aligned} f'(x) &= \frac{d}{dx} (x) - 2 \frac{d}{dx} ( \cos x )\\ \\ f'(x) &= 1 + 2 \sin x \end{aligned} \end{equation}$


when $f'(x) = 0$


$\begin{equation} \begin{aligned} 0 &= 1 + 2 \sin x\\ \\ 2 \sin x &= -1 \\ \\ x &= \sin^{-1} \left[ - \frac{1}{2} \right]\\ \\ x &= \frac{-\pi}{6} \end{aligned} \end{equation}$


We have either maximum or minimum at $\displaystyle x = \frac{-\pi}{6}$

when $\displaystyle x = \frac{-\pi}{6}$


$\begin{equation} \begin{aligned} f \left(\frac{-\pi}{6} \right) &= \frac{-\pi}{6} -2 \left(\cos \left( \frac{-\pi}{6} \right) \right)\\ \\ f \left(\frac{-\pi}{6} \right) &= -2.2556 \end{aligned} \end{equation}$


Evaluating the function of its interval $[-2,0]$.

When $x = 0$,


$\begin{equation} \begin{aligned} f(0) =& 0 - 2 \cos (0) \\ \\ f(0) =& -2 \end{aligned} \end{equation}$



When $x = -2$,


$\begin{equation} \begin{aligned} f(-2) =& -2 - 2 \cos (-2) \\ \\ f(-2) =& -1.1677 \end{aligned} \end{equation}$


Therefore, the absolute minimum is exactly at $\displaystyle f \left(\frac{-\pi}{6} \right) = -2.2556$ and the absolute maximum is exactly at $f(-2) = -1.1677$.