Complete solution set of the inequality `|x^(2)-x-2|+|x+1|le0` is

Options

A. `(-oo, -1)`
B. `(-1,0)`
C. `{-1, 0}`
D. `{-1}`

Explanation

To solve the inequality \( |x^2 - x - 2| + |x + 1| \leq 0 \), we need to analyze the expressions inside the absolute values.

### Step 1: Identify the expressions inside the absolute values
The expressions we have are:
1. \( x^2 - x - 2 \)
2. \( x + 1 \)

### Step 2: Find the points where each expression equals zero
We set each expression to zero to find the critical points.

1. For \( x^2 - x - 2 = 0 \):
- This can be factored as \( (x - 2)(x + 1) = 0 \).
- Thus, the solutions are \( x = 2 \) and \( x = -1 \).

2. For \( x + 1 = 0 \):
- This gives us \( x = -1 \).

### Step 3: List the critical points
The critical points we found are:
- \( x = -1 \)
- \( x = 2 \)

### Step 4: Analyze the intervals defined by the critical points
The critical points divide the real number line into the following intervals:
1. \( (-\infty, -1) \)
2. \( [-1, 2] \)
3. \( (2, \infty) \)

### Step 5: Test each interval
We need to check the sign of \( |x^2 - x - 2| + |x + 1| \) in each interval.

1. **Interval \( (-\infty, -1) \)**:
- Choose \( x = -2 \):
\[ |(-2)^2 - (-2) - 2| + |(-2) + 1| = |4 + 2 - 2| + |-1| = |4| + |1| = 4 + 1 = 5 > 0 \]

2. **Interval \( [-1, 2] \)**:
- Choose \( x = 0 \):
\[ |0^2 - 0 - 2| + |0 + 1| = |-2| + |1| = 2 + 1 = 3 > 0 \]
- Check at \( x = -1 \):
\[ |(-1)^2 - (-1) - 2| + |(-1) + 1| = |1 + 1 - 2| + |0| = |0| + |0| = 0 \]

3. **Interval \( (2, \infty) \)**:
- Choose \( x = 3 \):
\[ |3^2 - 3 - 2| + |3 + 1| = |9 - 3 - 2| + |4| = |4| + |4| = 4 + 4 = 8 > 0 \]

### Step 6: Compile the results
From our analysis:
- In the interval \( (-\infty, -1) \), the expression is greater than 0.
- In the interval \( [-1, 2] \), the expression is greater than 0 except at \( x = -1 \) where it equals 0.
- In the interval \( (2, \infty) \), the expression is greater than 0.

### Conclusion
The only solution to the inequality \( |x^2 - x - 2| + |x + 1| \leq 0 \) is:
\[ \boxed{-1} \]