Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 44

a.) Estimate the maximum and minimum values by using the graph of $f(x) = x+ 2 \cos x, \quad 0 \leq x \leq 2\pi$. Then, find the exact values.
b.) Estimate the value of $x$ at which $f$ increases most rapidly. Then determine the exact value.

a.)


Based from the graph, the local maximum is $f(0.50) \approx 2.20$ and local minimum $f(2.60) \approx 0.90$

To find for the exact values, we set $f'(x) = 0$ and some for the critical numbers

$
\begin{equation}
\begin{aligned}
\text{if } f(x) &= x + 2 \cos x, \quad \text{then}\\
\\
f'(x) &= 1 - 2 \sin x \\
\\
\\
\text{when } f'(x) &= 0 \\
\\
0 &= 1 - 2 \sin x \\
\\
\sin x &= \frac{1}{2}\\
\\
x &= \sin^{-1} \left[ \frac{1}{2} \right]\\
\\
x &= \frac{\pi}{6} + 2 \pi n \qquad \text{ or } \qquad x = \frac{5\pi}{6}+2\pi n; \quad \text{where } n \text{ is any integer }

\end{aligned}
\end{equation}
$

For the interval of $0 \leq x \leq 2 \pi$, the critical number are $\displaystyle x = \frac{\pi}{6} \text{ and } x = \frac{5\pi}{6}$

$
\begin{equation}
\begin{aligned}
\text{so when } x &= \frac{\pi}{6}, &&& \text{when } x &= \frac{5\pi}{6},\\
\\
f \left( \frac{\pi}{6} \right) & = \frac{\pi}{6} + 2 \cos f \left( \frac{\pi}{6} \right) &&& f \left( \frac{5\pi}{6} \right) &= \frac{5\pi}{6} + 2 \cos \left( \frac{5\pi}{6} \right)\\
\\
f \left( \frac{\pi}{6} \right) & = 2.2556 &&& f \left( \frac{5\pi}{6} \right) &= 0.8859
\end{aligned}
\end{equation}
$

Therefore, the exact value of local maximum is $\displaystyle f \left( \frac{\pi}{6} \right) = 2.2556$. While the local minimum is $\displaystyle f \left( \frac{5\pi}{6} \right) = 0.8859$


Based from the graph, the value of $x$ which $f$ increases rapidly is somewhere in $\displaystyle \left( \frac{5\pi}{4}, \frac{7\pi}{4} \right)$
To solve for the exact value, we set $f''(x) = 0$ and determine the inflection points.

$
\begin{equation}
\begin{aligned}
\text{so if } f'(x) &= 1 - 2 \sin x, \text{ then}\\
\\
f''(x) &= - 2 \cos x\\
\\
\\
\text{when } f''(x) &= 2 \cos x,\\
\\
0 &= -2 \cos x\\
\\
\cos x &= 0 \\
\\
x &= \cos^{-1} [0]\\
\\
x &= \frac{\pi}{2} + 2 \pi n \qquad \text{or} \qquad x = \frac{\pi}{2} + 2 \pi n \text{ ;where } n \text{ is any integer}
\end{aligned}
\end{equation}
$


For interval $0 \leq x \leq 2 \pi$, the inflection points are...
$\displaystyle x = \frac{\pi}{2} \text{ and } x = \frac{3\pi}{2}$

$
\begin{equation}
\begin{aligned}
\text{so when } x &= \frac{\pi}{2} &&& \text{when } x &= \frac{3\pi}{2} ,\\
\\
f' \left( \frac{\pi}{2} \right) &= 1 - 2 \sin \left( \frac{\pi}{2} \right) &&& f'\left( \frac{3\pi}{2} \right) &= 1 - 2 \sin \left( \frac{3\pi}{2} \right) \\
\\
f' \left( \frac{\pi}{2} \right) &= -1 &&& f'\left( \frac{3\pi}{2} \right) &= 3
\end{aligned}
\end{equation}
$

Therefore, the function increases most rapidly at $\displaystyle x = \frac{3\pi}{2}$