To find the point of local minimum of the function $f(x) = \sin(2x)$ on the interval $0 < x < \pi$, follow these steps:

Step 1: Find the derivative of $f(x)$

First, compute the derivative of the function: $f'(x) = \frac{d}{dx}[\sin(2x)] = 2\cos(2x)$

Step 2: Set the derivative equal to zero

To find critical points, set the derivative equal to zero: $2\cos(2x) = 0$ $\cos(2x) = 0$

Step 3: Solve for $x$

We solve $\cos(2x) = 0$. The cosine function equals zero at: $2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$ Dividing by 2: $x = \frac{\pi}{4}, \frac{3\pi}{4}, \dots$

Since we are restricted to the interval $0 < x < \pi$, the possible critical points are: $x = \frac{\pi}{4}, \frac{3\pi}{4}$

Step 4: Analyze the second derivative to classify the critical points

The second derivative of $f(x) = \sin(2x)$ is: $f''(x) = -4\sin(2x)$ Now evaluate $f''(x)$ at the critical points:

• At $x = \frac{\pi}{4}$: $f''\left(\frac{\pi}{4}\right) = -4\sin\left(\frac{\pi}{2}\right) = -4 \cdot 1 = -4$ Since $f''\left(\frac{\pi}{4}\right) < 0$, this is a point of local maximum.

• At $x = \frac{3\pi}{4}$: $f''\left(\frac{3\pi}{4}\right) = -4\sin\left(\frac{3\pi}{2}\right) = -4 \cdot (-1) = 4$ Since $f''\left(\frac{3\pi}{4}\right) > 0$, this is a point of local minimum.

Conclusion:

The point of local minimum is at $x = \frac{3\pi}{4}$.

Would you like further details or have any questions?

Here are 5 related questions for practice:

1. Find the local minimum of $f(x) = \cos(2x)$ on $0 < x < 2\pi$.
2. Determine the critical points of $f(x) = \sin(3x)$ in the interval $0 < x < \pi$.
3. Find the second derivative of $f(x) = \sin(2x)$ and interpret its behavior.
4. What is the maximum value of $f(x) = \sin(2x)$ on $0 < x < \pi$?
5. Solve $\sin(2x) = 0$ for $x$ in the interval $0 < x < \pi$.

Tip: Always use the second derivative test to confirm whether a critical point is a minimum or maximum.