find the point of local maxima and minima of f(x)= sin2x - x , -pi/2 < x < pi/2 by first derivative test.

To find the local maxima and minima of the function \( f(x) = \sin(2x) - x \) in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2} \right) \), we'll use the **first derivative test**. Let's go through the process step by step:

### 1. Find the first derivative of \( f(x) \)
First, we calculate the derivative of the given function with respect to \( x \):

\[
f'(x) = \frac{d}{dx} [\sin(2x) - x]
\]

Using the chain rule:

\[
f'(x) = 2\cos(2x) - 1
\]

### 2. Set the derivative equal to zero
To find the critical points, we set \( f'(x) = 0 \):

\[
2\cos(2x) - 1 = 0
\]

Solving for \( \cos(2x) \):

\[
2\cos(2x) = 1
\]
\[
\cos(2x) = \frac{1}{2}
\]

Now, solve for \( x \). We know that \( \cos(2x) = \frac{1}{2} \) occurs when:

\[
2x = \pm \frac{\pi}{3}
\]
\[
x = \pm \frac{\pi}{6}
\]

Thus, the critical points are \( x = \frac{\pi}{6} \) and \( x = -\frac{\pi}{6} \), both of which lie within the given interval \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).

### 3. Use the first derivative test
Now, we will apply the first derivative test by checking the sign of \( f'(x) \) around these critical points:

- **For \( x < -\frac{\pi}{6} \)** (e.g., \( x = -\frac{\pi}{3} \)):
  \[
  f'(-\frac{\pi}{3}) = 2\cos\left( -\frac{2\pi}{3} \right) - 1 = 2 \times \left( -\frac{1}{2} \right) - 1 = -1 - 1 = -2
  \]
  So, \( f'(x) < 0 \), meaning \( f(x) \) is decreasing in this region.

- **For \( x = 0 \)** (between \( -\frac{\pi}{6} \) and \( \frac{\pi}{6} \)):
  \[
  f'(0) = 2\cos(0) - 1 = 2 \times 1 - 1 = 1
  \]
  So, \( f'(x) > 0 \), meaning \( f(x) \) is increasing in this region.

- **For \( x > \frac{\pi}{6} \)** (e.g., \( x = \frac{\pi}{3} \)):
  \[
  f'(\frac{\pi}{3}) = 2\cos\left( \frac{2\pi}{3} \right) - 1 = 2 \times \left( -\frac{1}{2} \right) - 1 = -2
  \]
  So, \( f'(x) < 0 \), meaning \( f(x) \) is decreasing in this region.

### 4. Conclusion: Points of local maxima and minima
- At \( x = -\frac{\pi}{6} \), \( f'(x) \) changes from negative to positive, indicating a **local minimum**.
- At \( x = \frac{\pi}{6} \), \( f'(x) \) changes from positive to negative, indicating a **local maximum**.

Thus, the function has:
- A **local minimum** at \( x = -\frac{\pi}{6} \)
- A **local maximum** at \( x = \frac{\pi}{6} \)

### Do you want further details or have any questions?

Here are 5 related questions:
1. What is the second derivative of \( f(x) \), and how does it confirm concavity at critical points?
2. How do boundary conditions affect local extrema in an open interval?
3. What would happen if the function \( f(x) \) were defined on a larger interval?
4. How would you use the first derivative test for functions involving products or quotients?
5. Can a function have both local and absolute maxima/minima in the same region?

**Tip**: Always verify critical points with a sign change test for the first derivative to confirm whether they are local maxima or minima.

Discover how to find the local maxima and minima of f(x) = sin(2x) - x in the interval (-π/2, π/2) using the first derivative test. This solution includes critical points, first derivative analysis, and sign change tests. Ideal for Grades 11-12 or early college students.

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