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Let's break down the exercise step by step.

### Given:
The function \( f(x) = x(1 - x) \).

#### Part 1: 
**Show that for all \( x \in \mathbb{R} \), \( f(x) \leq \frac{1}{4} \).**

- First, express the function clearly:
  \[
  f(x) = x(1 - x) = x - x^2
  \]
  
  To find the maximum value of \( f(x) \), we can take the derivative and find its critical points.

  \[
  f'(x) = 1 - 2x
  \]
  
  Set \( f'(x) = 0 \) to find the critical point:
  \[
  1 - 2x = 0 \implies x = \frac{1}{2}
  \]
  
  Now evaluate \( f(x) \) at \( x = \frac{1}{2} \):
  \[
  f\left( \frac{1}{2} \right) = \frac{1}{2} \left( 1 - \frac{1}{2} \right) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
  \]

  Since the parabola \( f(x) = x - x^2 \) opens downward (as the coefficient of \( x^2 \) is negative), this is the maximum value. Therefore, for all \( x \in \mathbb{R} \), \( f(x) \leq \frac{1}{4} \).

#### Part 2: 
**Deduce that the function \( f \) attains a maximum at \( x = \frac{1}{2} \).**

From Part 1, we have already found that the maximum value of \( f(x) \) is \( \frac{1}{4} \), and this occurs at \( x = \frac{1}{2} \). Therefore, \( f(x) \) has a maximum at \( x = \frac{1}{2} \).

#### Part 3: 
**Demonstrate that \( f(x) = \frac{1}{4}(x - \frac{1}{2})^2 \) and deduce that the function is increasing on \( \left( -\infty, \frac{1}{2} \right] \) and decreasing on \( \left[ \frac{1}{2}, \infty \right) \).**

- Let's rewrite \( f(x) = x - x^2 \) in the form \( f(x) = \frac{1}{4} - \frac{1}{4}(x - \frac{1}{2})^2 \).

  \[
  f(x) = x - x^2 = \frac{1}{4} - \frac{1}{4}(x - \frac{1}{2})^2
  \]

  This is a standard form of a downward-opening parabola. The vertex form of the parabola shows that \( f(x) \) reaches its maximum when \( x = \frac{1}{2} \), and the value of the function decreases as we move away from \( x = \frac{1}{2} \).

- Since the parabola opens downward, the function is:
  - **Increasing** on the interval \( \left( -\infty, \frac{1}{2} \right] \),
  - **Decreasing** on the interval \( \left[ \frac{1}{2}, \infty \right) \).

### Summary:
- The maximum value of \( f(x) \) is \( \frac{1}{4} \), and it occurs at \( x = \frac{1}{2} \).
- The function is increasing on \( \left( -\infty, \frac{1}{2} \right] \) and decreasing on \( \left[ \frac{1}{2}, \infty \right) \).

Do you want more details on any step, or explanations of certain concepts?

### Related Questions:
1. How do you find the maximum or minimum of a quadratic function?
2. What does the derivative tell us about the increasing or decreasing nature of a function?
3. Can the maximum value of a quadratic function occur at a point other than the vertex?
4. How can you express a quadratic function in vertex form?
5. How does completing the square help in finding the maximum or minimum of a function?

**Tip:** For quadratic functions, the sign of the \( x^2 \)-coefficient determines whether the parabola opens upwards or downwards (positive opens upwards, negative opens downwards).

Exercise N°4: We consider the function f defined by f(x) = x(1 - x). 1) Show that for all x in R: f(x) ≤ 1/4. 2) Deduce that the function f has a maximum at x = 1/2. 3) Demonstrate that f(x) = 1/4 - 1/4(x - 1/2)^2 and deduce that the function f is increasing on the interval (-∞, 1/2] and decreasing on the interval [1/2, ∞).

Learn how to analyze the quadratic function f(x) = x(1 - x). This problem covers finding the maximum value, demonstrating inequalities, and understanding increasing and decreasing intervals. Suitable for students in Grades 10-12.

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