find the point of local minimum of f(x)= sin2x , 0<x<pi.

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.

Learn how to find the point of local minimum of the function f(x) = sin(2x) in the interval 0 < x < pi using the derivative and second derivative test. This calculus problem involves critical points, trigonometric functions, and the second derivative test. Suitable for high school students in Grades 11-12.

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