lf `| f(x) + 6-x^2| =| f(x)|+|4-x^2| + 2`, then f(x) is necessarily non-negaive for

Options

A. `x in [-2,2]`
B. `x n (-oo, -2) uu (2,oo)`
C. `x in [-sqrt6, sqrt6]`
D. `x in [-5, -2] uu [2,5]`

Answer

To solve the equation \( |f(x) + 6 - x^2| = |f(x)| + |4 - x^2| + 2 \) and show that \( f(x) \) is necessarily non-negative for certain values of \( x \), we can follow these steps:

### Step 1: Analyze the given equation
We start with the equation:
\[ |f(x) + 6 - x^2| = |f(x)| + |4 - x^2| + 2 \]

### Step 2: Consider the cases for the absolute values
The equation involves absolute values, so we need to consider the cases based on the signs of the expressions inside the absolute values.

### Step 3: Identify the critical points
The critical points that affect the absolute values are when \( f(x) + 6 - x^2 = 0 \) and \( 4 - x^2 = 0 \).
- From \( 4 - x^2 = 0 \), we find \( x^2 = 4 \) or \( x = \pm 2 \).

### Step 4: Determine the intervals
The intervals to consider based on the critical points are:
- \( (-\infty, -2) \)
- \( [-2, 2] \)
- \( (2, \infty) \)

### Step 5: Analyze the interval \( [-2, 2] \)
In the interval \( [-2, 2] \):
- \( 4 - x^2 \geq 0 \) (since \( x^2 \leq 4 \))
- Therefore, \( |4 - x^2| = 4 - x^2 \).

### Step 6: Substitute into the equation
Substituting into the equation, we have:
\[ |f(x) + 6 - x^2| = |f(x)| + (4 - x^2) + 2 \]
This simplifies to:
\[ |f(x) + 6 - x^2| = |f(x)| + 6 - x^2 \]

### Step 7: Analyze the cases for \( f(x) \)
1. **Case 1**: If \( f(x) + 6 - x^2 \geq 0 \):
\[ f(x) + 6 - x^2 = f(x) + 6 - x^2 \]
This holds true for all \( f(x) \).

2. **Case 2**: If \( f(x) + 6 - x^2 < 0 \):
\[ -(f(x) + 6 - x^2) = |f(x)| + 6 - x^2 \]
This leads to:
\[ -f(x) - 6 + x^2 = |f(x)| + 6 - x^2 \]
Rearranging gives:
\[ -f(x) = |f(x)| + 12 - 2x^2 \]
Since \( |f(x)| \geq 0 \), this implies \( -f(x) \geq 12 - 2x^2 \).

### Step 8: Determine the non-negativity of \( f(x) \)
For \( f(x) \) to be non-negative, we need:
\[ f(x) \geq 0 \quad \text{and} \quad 12 - 2x^2 \leq 0 \]
This gives:
\[ x^2 \geq 6 \quad \Rightarrow \quad |x| \geq \sqrt{6} \]
However, within the interval \( [-2, 2] \), \( x^2 \) cannot exceed 4. Thus, \( f(x) \) must be non-negative in the interval \( [-2, 2] \).

### Conclusion
Thus, we conclude that \( f(x) \) is necessarily non-negative for \( x \) in the interval \( [-2, 2] \).

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