suppose f is a differentiable function with f'(x) = 2sin(pix). At what value of x in the interval 1 < x < 2 does f take its absolute maximum? please explain.

To find the value of \( x \) in the interval \( 1 < x < 2 \) where \( f(x) \) attains its absolute maximum, we proceed by analyzing the behavior of the derivative \( f'(x) = 2\sin(\pi x) \).

### Step 1: Identify critical points
A critical point occurs where \( f'(x) = 0 \). To find these points, set the derivative equal to zero:

\[
f'(x) = 2\sin(\pi x) = 0
\]

This simplifies to:

\[
\sin(\pi x) = 0
\]

The sine function equals zero at integer multiples of \( \pi \), i.e., when \( \pi x = n\pi \), where \( n \) is an integer. Thus:

\[
x = n
\]

In the interval \( 1 < x < 2 \), the only integer is \( x = 1 \), but since this is not strictly inside the interval, we exclude it. However, we do need to check for points where the derivative changes sign within this interval.

### Step 2: Analyze the behavior of \( f'(x) \) in \( 1 < x < 2 \)
We will check the sign of \( f'(x) \) in the interval \( 1 < x < 2 \) by evaluating \( \sin(\pi x) \) at some representative points.

- At \( x = 1.5 \):
  \[
  \sin(\pi \times 1.5) = \sin\left(\frac{3\pi}{2}\right) = -1
  \]
  So \( f'(1.5) = 2 \times (-1) = -2 \), which means the function is decreasing at \( x = 1.5 \).

Since \( \sin(\pi x) \) is negative in the interval \( 1 < x < 2 \) (excluding \( x = 1 \)), the derivative is negative throughout this interval, indicating that \( f(x) \) is decreasing.

### Step 3: Evaluate the endpoints
The function could attain its absolute maximum at the boundary points. To determine this, we check the values of \( f(x) \) at the endpoints of the interval \( 1 < x < 2 \). However, since the function is decreasing in the entire interval, the absolute maximum will occur at the left endpoint of the interval, which is just after \( x = 1 \).

Therefore, the absolute maximum of \( f(x) \) in the interval \( 1 < x < 2 \) occurs **as \( x \) approaches 1 from the right**.

### Final Answer:
The absolute maximum occurs as \( x \to 1^+ \) in the interval \( 1 < x < 2 \).

---

Would you like further clarification on any of the steps?

Here are 5 related questions to expand your understanding:
1. How would the result change if the interval were \( 0 < x < 2 \)?
2. What is the general behavior of the function \( f(x) \) given its derivative?
3. How would you determine if the maximum is absolute or local based on \( f'(x) \)?
4. What role do critical points outside the interval play in determining maxima or minima?
5. How does the periodic nature of sine affect the behavior of the derivative?

**Tip:** To analyze critical points, always check the derivative's sign changes, as they indicate where a function increases or decreases.

Suppose f is a differentiable function with f'(x) = 2sin(pix). At what value of x in the interval 1 < x < 2 does f take its absolute maximum? Please explain.

Discover how to find the absolute maximum of the function f'(x) = 2sin(πx) within the interval 1 < x < 2 using critical points and the first derivative test. This calculus problem involves analyzing trigonometric functions and is ideal for students in Grades 11-12.

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