Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 38

Use the guidelines of curve sketching to sketch the curve. $\displaystyle y = \frac{\sin x}{2 + \cos x}$

The guidelines of Curve Sketching
A. Domain.
The domain is $(-\infty, \infty)$

B. Intercepts.
Solving for $y$-intercept, when $x = 0$
$\displaystyle y = \frac{\sin 0}{2 + \cos 0} = 0$
Solving for $x$-intercept, when $y = 0$
$\displaystyle 0 = \frac{\sin x}{2 + \cos x}$

C. Symmetry.
Since $f(-x) = -f(x)$, the function is symmetric to origin.

D. Asymptotes.
Since the domain is $(-\infty,\infty)$, we can say that the function has no vertical asymptote.

E. Intervals of Increase or Decrease.

$
\begin{equation}
\begin{aligned}
\text{if } f(x) &= \frac{\sin x}{2 + \cos x} \quad \text{, then by using Quotient Rule}\\
\\
f'(x) &= \frac{(2 + \cos x)(\cos x)-(\sin x)(-\sin x)}{(2+\cos x)^2} = \frac{2\cos x + \cos^2 x + \sin^2 x}{(2 + \cos x)^2}\\
\\
f'(x) &= \frac{2 \cos x + 1}{(2 + \cos x)^2} \quad \text{; recall that } \sin^2x + \cos^2 x = 1
\end{aligned}
\end{equation}
$

when $f'(x) = 0$,
The critical numbers are, $\displaystyle x = \pm \frac{2\pi}{3} + 2 \pi n \quad$ where $n$ is any integer


F. Local Maximum and Minimum Values.
Since $f'(x)$ changes from positive to negative at $\displaystyle x = \frac{2\pi(3n+1)}{3}, \quad f \left( \frac{2\pi(3n +1)}{3} \right) = \frac{\sqrt{3}}{3}$ is a local maxima. On the other hand, since $f'(x)$ changes from negative to positive at $\displaystyle x = \frac{2\pi}{3} (3n + 2), \quad f \left( \frac{2\pi(3n+2)}{3} \right) = -\frac{\sqrt{3}}{3}$ is a local minima.

G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= \frac{2 \cos x + 1}{(2 + \cos x)^2} \quad \text{, then by using Quotient Rule and Chain Rule}\\
\\
f''(x) &= \frac{(2 + \cos x)^2 (-2\sin x)- (2 \cos x +1) \left(2 (2 + \cos x)(-\sin x) \right)}{\left[ (2 + \cos x)^2 \right]^2}
\end{aligned}
\end{equation}
$

Which can be simplified as...
$\displaystyle f''(x) = \frac{2 \sin x(\cos x -1)}{(2+ \cos x)^3}$
So, $f''(x) < 0 \text{ on interval } \left( \pi (2 n - 1) , 2\pi n\right)$
H. Sketch the Graph.