91Ó°ÊÓ

The first derivative of a function helps us find its critical points, which are crucial for identifying relative maxima and minima. In the given problem, we start with the function: \( f(x) = \cos^{2}(x) \). To find the first derivative, we differentiate using the chain rule: \( f'(x) = -2 \cos(x) \sin(x) \). This derivative shows the rate of change of the original function. By setting the first derivative to zero: \( -2 \cos(x) \sin(x) = 0 \), we solve for \( x \) and find the critical points within the interval \([0, 2 \pi] \). Because we are analyzing trigonometric functions, the critical points occur when either \( \cos(x) = 0 \) or \( \sin(x) = 0 \). Solving these within the interval, we find: \( x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \).

The second derivative provides insight into the concavity of the function and aids in classifying the nature of the critical points. Following from the first derivative \( f'(x) = -2 \cos(x) \sin(x) \), we use the product rule to find the second derivative: \( f''(x) = -2 (\cos^{2}(x) - \sin^{2}(x)) = -2 \cos(2x) \). This second derivative is instrumental for the Second-Derivative Test. It helps determine whether the critical points are relative maxima, minima, or neither by indicating changes in concavity. For instance, positive values of \( f''(x) \) at critical points suggest a relative minimum, while negative values indicate a relative maximum.

To identify relative maxima, we use the second derivative at the critical points. A point where the second derivative is negative indicates a local peak. In our example, by evaluating \( f''(x) \) at the critical points:

• For \( x = 0 \), \( f''(0) = -2 \cos(0) = -2 \), indicating a relative maximum.
• For \( x = \pi \), \( f''(\pi) = -2 \cos(2\pi) = -2 \), another relative maximum.
• For \( x = 2\pi \), \( f''(2\pi) = -2 \cos(4\pi) = -2 \), yet another relative maximum.

This means the function \(\cos^{2}(x) \) has relative maxima at these points within the given interval.

A relative minimum occurs where the second derivative is positive at a critical point, suggesting a local trough in the function. Using our second derivative formula, we evaluate:

• For \( x = \frac{\pi}{2} \), \( f''(\frac{\pi}{2}) = -2 \cos(\pi) = 2 \), indicating a relative minimum.
• For \( x = \frac{3\pi}{2} \), \( f''(\frac{3\pi}{2}) = -2 \cos(3\pi) = 2 \), another relative minimum.

This means the function \(\cos^{2}(x) \) has relative minima at these points within the given interval.

Inflection points occur where the second derivative changes sign, indicating a change in concavity. To find these points, we solve \( f''(x) = -2 \cos(2x) = 0 \Rightarrow \cos(2x) = 0 \). Solving this within the interval \([0, 2 \pi] \), we get:

• \( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).

We verify these points by checking the sign of \( f''(x) \) around these values to ensure the concavity changes. These inflection points are crucial as they indicate where the function changes from concave up to concave down or vice versa, adding detail to the understanding of the function's graph.