Consider the function `f(x)=max. ` `[(x^(2) , (1-x)^(2) , 2x(1-x))] , x in [0,1]` The interval in which f(x) is increasing, is

Options

A. `((1)/(3),(2)/(3))`
B. `((1)/(3),(1)/(2))`
C. `((1)/(3),(1)/(2))cup((1)/(2),(2)/(3))`
D. `((1)/(3),(1)/(2))cup((2)/(3),1)`

Answer

To solve the problem, we need to analyze the function \( f(x) = \max \{ x^2, (1-x)^2, 2x(1-x) \} \) for \( x \) in the interval \([0, 1]\) and determine the interval where \( f(x) \) is increasing.

### Step 1: Identify the functions
We have three functions to consider:
1. \( f_1(x) = x^2 \)
2. \( f_2(x) = (1-x)^2 \)
3. \( f_3(x) = 2x(1-x) \)

### Step 2: Find the critical points
To find where \( f(x) \) changes, we need to find the points where these functions intersect.

1. Set \( f_1(x) = f_2(x) \):
\[ x^2 = (1-x)^2 \]
Expanding and simplifying:
\[ x^2 = 1 - 2x + x^2 \implies 2x = 1 \implies x = \frac{1}{2} \]

2. Set \( f_1(x) = f_3(x) \):
\[ x^2 = 2x(1-x) \]
Simplifying gives:
\[ x^2 = 2x - 2x^2 \implies 3x^2 - 2x = 0 \implies x(3x - 2) = 0 \]
This gives \( x = 0 \) or \( x = \frac{2}{3} \).

3. Set \( f_2(x) = f_3(x) \):
\[ (1-x)^2 = 2x(1-x) \]
Simplifying gives:
\[ 1 - 2x + x^2 = 2x - 2x^2 \implies 3x^2 - 4x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} \]
This gives \( x = 1 \) and \( x = \frac{1}{3} \).

### Step 3: Determine intervals
Now we have the critical points \( x = 0, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 1 \). We will evaluate \( f(x) \) in the intervals \( [0, \frac{1}{3}) \), \( [\frac{1}{3}, \frac{1}{2}) \), \( [\frac{1}{2}, \frac{2}{3}) \), and \( [\frac{2}{3}, 1] \).

- **Interval \( [0, \frac{1}{3}) \)**:
- \( f_1(x) = x^2 \) is increasing.
- \( f_2(x) = (1-x)^2 \) is decreasing.
- \( f_3(x) = 2x(1-x) \) is increasing.
- Thus, \( f(x) = f_1(x) = x^2 \).

- **Interval \( [\frac{1}{3}, \frac{1}{2}) \)**:
- \( f_1(x) = x^2 \) is increasing.
- \( f_2(x) = (1-x)^2 \) is decreasing.
- \( f_3(x) = 2x(1-x) \) is increasing.
- Thus, \( f(x) = f_3(x) = 2x(1-x) \).

- **Interval \( [\frac{1}{2}, \frac{2}{3}) \)**:
- \( f_1(x) = x^2 \) is increasing.
- \( f_2(x) = (1-x)^2 \) is decreasing.
- \( f_3(x) = 2x(1-x) \) is decreasing.
- Thus, \( f(x) = f_1(x) = x^2 \).

- **Interval \( [\frac{2}{3}, 1] \)**:
- \( f_1(x) = x^2 \) is increasing.
- \( f_2(x) = (1-x)^2 \) is decreasing.
- \( f_3(x) = 2x(1-x) \) is decreasing.
- Thus, \( f(x) = f_2(x) = (1-x)^2 \).

### Step 4: Determine where \( f(x) \) is increasing
From our analysis:
- \( f(x) \) is increasing in the intervals \( [0, \frac{1}{3}) \) and \( [\frac{1}{2}, 1] \).

### Conclusion
The interval in which \( f(x) \) is increasing is:
\[ \boxed{[0, \frac{1}{3}) \cup [\frac{1}{2}, 1]} \]