Local Extrema of Trigonometric Rational Function

To analyze the function

$$f(x) = \begin{cases} \frac{6x + \sin x}{2x + \sin x} & \text{if } x \neq 0, \\[6pt] \frac{7}{3} & \text{if } x = 0 \end{cases}$$

we evaluate its behavior at $x = 0$ and its critical points.

First, using Taylor expansion near $x = 0$, we have $\sin x \approx x - \frac{x^3}{6}$, which gives: $f(x) = \frac{7x - \frac{x^3}{6}}{3x - \frac{x^3}{6}} = \frac{7 - \frac{x^2}{6}}{3 - \frac{x^2}{6}} \approx \frac{7}{3}\left(1 + \frac{2x^2}{63}\right) > \frac{7}{3} = f(0)$ Since $f(x) > f(0)$ in a neighborhood around $x = 0$, $x = 0$ is a point of local minima, making statement (B) TRUE and statement (A) FALSE.

For $x \neq 0$, the derivative of $f(x)$ is given by: $f'(x) = \frac{4(\sin x - x \cos x)}{(2x + \sin x)^2}$ The sign of $f'(x)$ is determined by $g(x) = \sin x - x \cos x$, with $g'(x) = x \sin x$.

• In intervals $((2k-1)\pi, 2k\pi)$ where $\sin x < 0$, $g'(x) < 0$, so $g(x)$ decreases from $(2k-1)\pi > 0$ to $-2k\pi < 0$, changing sign from positive to negative. This yields points of local maxima in $(\pi, 2\pi)$, $(3\pi, 4\pi)$, and $(5\pi, 6\pi)$. Thus, the number of local maxima in $[\pi, 6\pi]$ is $3$, so statement (C) is TRUE.

• In intervals $(2k\pi, (2k+1)\pi)$ where $\sin x > 0$, $g'(x) > 0$, so $g(x)$ increases from $-2k\pi < 0$ to $(2k+1)\pi > 0$, changing sign from negative to positive. This yields a point of local minima in $(2\pi, 3\pi)$. Thus, the number of local minima in $[2\pi, 4\pi]$ is $1$, so statement (D) is TRUE.

Hence, the correct options are (B), (C), and (D).