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.