Redefine the function `f(x)=|x-2|+|2+x|,-3lt=xlt=3`

Answer

To redefine the function \( f(x) = |x-2| + |2+x| \) for the interval \(-3 < x < 3\), we need to analyze the points where the expressions inside the absolute values change their signs. The critical points are \( x = 2 \) and \( x = -2 \).

### Step 1: Identify intervals based on critical points
The critical points divide the interval \(-3 < x < 3\) into three segments:
1. \( -3 < x < -2 \)
2. \( -2 < x < 2 \)
3. \( 2 < x < 3 \)

### Step 2: Evaluate the function in each interval

#### Interval 1: \( -3 < x < -2 \)
In this interval:
- \( x - 2 < 0 \) (thus, \( |x - 2| = -(x - 2) = -x + 2 \))
- \( 2 + x < 0 \) (thus, \( |2 + x| = -(2 + x) = -2 - x \))

So,
\[ f(x) = |x - 2| + |2 + x| = (-x + 2) + (-2 - x) = -2x \]

#### Interval 2: \( -2 < x < 2 \)
In this interval:
- \( x - 2 < 0 \) (thus, \( |x - 2| = -x + 2 \))
- \( 2 + x > 0 \) (thus, \( |2 + x| = 2 + x \))

So,
\[ f(x) = |x - 2| + |2 + x| = (-x + 2) + (2 + x) = 4 \]

#### Interval 3: \( 2 < x < 3 \)
In this interval:
- \( x - 2 > 0 \) (thus, \( |x - 2| = x - 2 \))
- \( 2 + x > 0 \) (thus, \( |2 + x| = 2 + x \))

So,
\[ f(x) = |x - 2| + |2 + x| = (x - 2) + (2 + x) = 2x \]

### Step 3: Combine results
Now we can redefine the function \( f(x) \) based on the intervals we analyzed:

\[ f(x) = \begin{cases} -2x & \text{for } -3 < x < -2 \\ 4 & \text{for } -2 < x < 2 \\ 2x & \text{for } 2 < x < 3 \end{cases} \]

### Final Result
Thus, the redefined function is:
\[ f(x) = \begin{cases} -2x & \text{for } -3 < x < -2 \\ 4 & \text{for } -2 < x < 2 \\ 2x & \text{for } 2 < x < 3 \end{cases} \]