Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 14

Suppose that $f(x) = \cos^2 x - 2 \sin x, \quad 0 \leq x \leq 2\pi$
a.) Determine the intervals on which $f$ is increasing or decreasing.
b.) Find the local maximum and minimum values of $f$.
c.) Find the intervals of concavity and the inflection points

a.) If $f(x) = \cos^2x - 2 \sin x$; then
Recall that $\cos^2 x = 1 - \sin^2x$
So, $f(x) = 1 - \sin ^2 x - 2 \sin x$


$
\begin{equation}
\begin{aligned}
f'(x) &= -2 \sin x \cos x - 2 \cos x && \Longleftarrow \text{(By using Chain Rule)}\\
\\
f''(x) &= 2 \left[ \sin x - (-\sin x) + \cos x (\cos x)\right] && \Longleftarrow \text{(By using product Rule)}\\
\\
f''(x) &= 2 \sin^2 x - 2 \cos^2 x + 2 \sin x
\end{aligned}
\end{equation}
$

Again, recall that


$
\begin{equation}
\begin{aligned}
f''(x) &= 2 \sin^2 x - 2 ( 1 - \sin^2 x ) + 2 \sin x\\
\\
f''(x) &= 2 \sin^2 x - 2 + 2 \sin^2 x + 2\sin x\\
\\
f''(x) &= 4 \sin^2 x + 2 \sin x - 2
\end{aligned}
\end{equation}
$


To find the critical numbers, we set $f'(x)= 0$, so...

$
\begin{equation}
\begin{aligned}
0 &= - 2 \sin x \cos x - 2 \cos x\\
\\
\cancel{2} \sin x \cancel{\cos x} &= \cancel{2}\cancel{\cos x}\\
\\
0 &= -2 \cos x ( \sin x + 1 )
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\text{we have } \cos x &= 0 && \text{and} & \sin x + 1 &= 0\\
\\
x &= \frac{\pi}{2} + 2 \pi n && \text{and} & x &= \frac{-\pi}{2} + 2 \pi n
\end{aligned}
\end{equation}
$

Where $n$ is any integer.

For the interval $0 \leq x \leq 2 \pi$, the critical number are $\displaystyle x = \frac{\pi}{2} \text{ and } x = \frac{3\pi}{2}$
Hence, we can divide the interval by...

$
\begin{array}{|c|c|c|}
\hline\\

\text{Interval} & f'(x) & f\\
\hline\\
\\
\displaystyle 0 < x < \frac{\pi}{2} & - & \displaystyle \text{decreasing on } \left( 0, \frac{\pi }{2}\right)\\
\hline\\
\\
\displaystyle \frac{\pi}{2} < x < \frac{3\pi}{2} & + & \displaystyle \text{increasing on } \left(\frac{\pi}{2}, \frac{3\pi}{2} \right)\\
\hline\\
\\
\\
\\
\\
\displaystyle \frac{3\pi}{2} < x < 2\pi & - & \displaystyle \text{decreasing on } \left( \frac{3\pi}{2}, 2\pi \right)\\
\\
\hline
\end{array}
$


These data are obtained by substituting any values of $x$ to $f'(x)$ within the specified interval. Check its sign, if its positive, it means that the curve is increasing on that interval. On the other hand, if the sign is negative, it means that the curve is decreasing on that interval.

b.) We will use the Second Derivative test to determine the points of inflections...
So when $\displaystyle x = \frac{\pi}{2}$

$
\begin{equation}
\begin{aligned}
f'' \left( \frac{\pi}{2} \right) &= 4 \sin^2 \left( \frac{\pi}{2} \right) + 2 \sin \left( \frac{\pi}{2} \right) - 2\\
\\
f''\left( \frac{\pi}{2} \right) &= 4
\end{aligned}
\end{equation}
$


when $\displaystyle x =\frac{3\pi}{2} $

$
\begin{equation}
\begin{aligned}
f'' \left( \frac{3\pi}{2} \right) &= 4 \sin^2 \left( \frac{3\pi}{2} \right) + 2 \sin \left( \frac{3\pi}{2} \right) - 2\\
\\
f'' \left( \frac{3\pi}{2} \right) &= 0
\end{aligned}
\end{equation}
$


Since $\displaystyle f'\left( \frac{\pi}{2} \right) = 0$ and $\displaystyle f'' \left( \frac{\pi}{2} \right) > 0, f \left( \frac{\pi}{2} \right) = -2$ is a local minimum. On the other hand, since $\displaystyle f'\left( \frac{3\pi}{2} \right) = 0, \left( \frac{3\pi}{2} \right) = 2$ is a local maximum

c.) We set $f''(x) = 0$, to determine the point of inflections...

$
\begin{equation}
\begin{aligned}
f''(x) = 0 &= 4 \sin^2 x + 2 \sin x - 2\\
\\
0 &= 4 \sin^2 x + 2 \sin x - 2
\end{aligned}
\end{equation}
$

By factoring,
$0 = (4 \sin x - 2) (\sin x + 1)$
We have,

$
\begin{equation}
\begin{aligned}
4 \sin x - 2 &= 0 &&\text{and}& \sin x + 1 &=0\\
\\
\sin x &= \frac{1}{2} &&& \sin x &= -1\\
\\
x &= \sin^{-1} \left[ \frac{1}{2} \right] &&& x &= \sin^{-1} [-1]\\
\\
x &= \frac{\pi}{6} + 2\pi n &&& x &= \frac{-\pi}{2} + 2 \pi n \quad; \text{where } n \text{ is any integer}
\end{aligned}
\end{equation}
$


For the interval $0 \leq x \leq 2 \pi$, the point of inflections are...
$\displaystyle x = \frac{\pi}{6}, x = \frac{5\pi}{6} \text{ and } x = \frac{3\pi}{6}$

Let's divide the interval to determine the concavity...


$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity}\\
\hline\\
0 < x < \frac{\pi}{6} & - & \text{Downward}\\
\hline\\
\displaystyle \frac{\pi}{6} < x < \frac{5\pi}{6} & + & \text{Upward}\\
\hline\\
\displaystyle \frac{5\pi}{6} < x < \frac{3\pi}{6} & - & \text{Downard}\\
\hline\\
\displaystyle \frac{\pi}{6} < x < 2 \pi & + & \text{Upward}\\
\hline
\end{array}
$


These values are obtained by evaluating $f''(x)$ within the specified interval. The concavity is upward when the sign of $f''(x)$ is positive. On the other hand, the concavity is downward when the sign of $f''(x)$ is negative.