Find the equivalent definition of `f(x)=m a xdot{x^2,(-x)^2,2x(1-x)}where 0lt=xlt=1`

Explanation

To find the equivalent definition of the function \( f(x) = \max\{x^2, (1-x)^2, 2x(1-x)\} \) for \( 0 < x < 1 \), we will analyze each of the three functions involved and determine their behavior within the interval.

### Step 1: Define 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 Points of Intersection
To find the maximum value among these functions, we need to find the points where they intersect.

**Intersection of \( f_1(x) \) and \( f_2(x) \):**
Set \( x^2 = (1-x)^2 \):
\[ x^2 = 1 - 2x + x^2 \implies 2x = 1 \implies x = \frac{1}{2} \]

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

**Intersection of \( f_2(x) \) and \( f_3(x) \):**
Set \( (1-x)^2 = 2x(1-x) \):
\[ 1 - 2x + x^2 = 2x - 2x^2 \implies 3x^2 - 4x + 1 = 0 \]
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: Evaluate the Functions at Critical Points
Now we will evaluate \( f_1, f_2, \) and \( f_3 \) at the critical points \( x = 0, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 1 \).

1. At \( x = \frac{1}{3} \):
- \( f_1\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \)
- \( f_2\left(\frac{1}{3}\right) = \left(1 - \frac{1}{3}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \)
- \( f_3\left(\frac{1}{3}\right) = 2 \cdot \frac{1}{3} \cdot \left(1 - \frac{1}{3}\right) = 2 \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{4}{9} \)

2. At \( x = \frac{1}{2} \):
- \( f_1\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
- \( f_2\left(\frac{1}{2}\right) = \left(1 - \frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
- \( f_3\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{2} \cdot \left(1 - \frac{1}{2}\right) = 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2} \)

3. At \( x = \frac{2}{3} \):
- \( f_1\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \)
- \( f_2\left(\frac{2}{3}\right) = \left(1 - \frac{2}{3}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \)
- \( f_3\left(\frac{2}{3}\right) = 2 \cdot \frac{2}{3} \cdot \left(1 - \frac{2}{3}\right) = 2 \cdot \frac{2}{3} \cdot \frac{1}{3} = \frac{4}{9} \)

### Step 4: Determine the Maximum Function
From the evaluations:
- For \( x \in (0, \frac{1}{3}) \): \( f(x) = f_2(x) = (1-x)^2 \)
- For \( x \in [\frac{1}{3}, \frac{2}{3}] \): \( f(x) = f_3(x) = 2x(1-x) \)
- For \( x \in (\frac{2}{3}, 1) \): \( f(x) = f_1(x) = x^2 \)

### Final Equivalent Definition
Thus, the equivalent definition of the function \( f(x) \) can be expressed as:
\[ f(x) = \begin{cases} (1-x)^2 & \text{for } 0 < x < \frac{1}{3} \\ 2x(1-x) & \text{for } \frac{1}{3} \leq x \leq \frac{2}{3} \\ x^2 & \text{for } \frac{2}{3} < x < 1 \end{cases} \]