Consider $f(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$ Number of critical points of $f(x)$ in the interval $(-2\pi, 2\pi)$ is:
Step 1: Understanding the Concept:
Critical points occur where the derivative is zero or does not exist. The function is continuous and differentiable at \( x = 0 \) (\( f'(0) = 0 \)). Step 2: Key Formula or Approach:
For \( x \neq 0 \), differentiate using quotient rule:
\( f'(x) = \frac{x \cos x - \sin x}{x^2} \).
Set \( f'(x) = 0 \implies x \cos x - \sin x = 0 \implies \tan x = x \). Step 3: Detailed Explanation:
We need to find the number of solutions to \( \tan x = x \) in \( (-2\pi, 2\pi) \).
By checking the graphs of \( y = x \) and \( y = \tan x \):
1. At \( x = 0 \), both are equal. So \( x = 0 \) is one solution (and a critical point).
2. In \( (0, 2\pi) \): The graph of \( \tan x \) has a branch in \( (\pi, 3\pi/2) \) where it will cross the line \( y = x \) exactly once.
3. In \( (-2\pi, 0) \): Due to odd symmetry, there is exactly one solution on the negative branch as well.
Total critical points = 1 (at zero) + 1 (positive) + 1 (negative) = 3. Step 4: Final Answer:
The total number of critical points in the given interval is 3.
