The area of the bounded region enclosed by the curve `y= 3 - |x- (1)/(2)|- |x+1|` and the x-axis is:

Options

A. `(9)/(4)`
B. `(45)/(16)`
C. `(27)/(8)`
D. `(63)/(16)`

Answer

To find the area of the bounded region enclosed by the curve \( y = 3 - |x - \frac{1}{2}| - |x + 1| \) and the x-axis, we will follow these steps:

### Step 1: Analyze the function
The function \( y = 3 - |x - \frac{1}{2}| - |x + 1| \) involves absolute values, so we need to determine the points where the expressions inside the absolute values change sign.

The critical points are:
- \( x = -1 \) (where \( |x + 1| \) changes)
- \( x = \frac{1}{2} \) (where \( |x - \frac{1}{2}| \) changes)

### Step 2: Break the function into intervals
We will analyze the function on the intervals defined by the critical points:
1. \( x < -1 \)
2. \( -1 \leq x < \frac{1}{2} \)
3. \( x \geq \frac{1}{2} \)

### Step 3: Evaluate the function in each interval

**Interval 1: \( x < -1 \)**
- Here, both \( |x - \frac{1}{2}| = -x + \frac{1}{2} \) and \( |x + 1| = -x - 1 \).
- Thus, the function becomes:
\[ y = 3 - (-x + \frac{1}{2}) - (-x - 1) = 3 + x - \frac{1}{2} + x + 1 = 2x + \frac{5}{2} \]

**Interval 2: \( -1 \leq x < \frac{1}{2} \)**
- Here, \( |x - \frac{1}{2}| = -x + \frac{1}{2} \) and \( |x + 1| = x + 1 \).
- Thus, the function becomes:
\[ y = 3 - (-x + \frac{1}{2}) - (x + 1) = 3 + x - \frac{1}{2} - x - 1 = \frac{5}{2} \]

**Interval 3: \( x \geq \frac{1}{2} \)**
- Here, both \( |x - \frac{1}{2}| = x - \frac{1}{2} \) and \( |x + 1| = x + 1 \).
- Thus, the function becomes:
\[ y = 3 - (x - \frac{1}{2}) - (x + 1) = 3 - x + \frac{1}{2} - x - 1 = -2x + \frac{5}{2} \]

### Step 4: Find the points where the function intersects the x-axis
To find the area, we need to determine where \( y = 0 \) in each interval.

**For Interval 1:**
\[ 2x + \frac{5}{2} = 0 \implies 2x = -\frac{5}{2} \implies x = -\frac{5}{4} \]

**For Interval 2:**
\[ \frac{5}{2} = 0 \implies \text{No intersection} \]

**For Interval 3:**
\[ -2x + \frac{5}{2} = 0 \implies 2x = \frac{5}{2} \implies x = \frac{5}{4} \]

### Step 5: Determine the area
The area can be calculated as the sum of the areas of the regions formed in each interval.

1. **Area in Interval 1** (from \( -\frac{5}{4} \) to \( -1 \)):
- The height is \( y = 0 \) to \( y = 2x + \frac{5}{2} \).
- Area of the triangle:
\[ \text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \left(-1 + \frac{5}{4}\right) \times \left(2(-1) + \frac{5}{2}\right) = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{2} = \frac{3}{16} \]

2. **Area in Interval 3** (from \( \frac{1}{2} \) to \( \frac{5}{4} \)):
- The height is \( y = 0 \) to \( y = -2x + \frac{5}{2} \).
- Area of the triangle:
\[ \text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \left(\frac{5}{4} - \frac{1}{2}\right) \times \left(-2\left(\frac{5}{4}\right) + \frac{5}{2}\right) = \frac{1}{2} \times \frac{3}{4} \times \frac{5}{4} = \frac{15}{32} \]

### Step 6: Total Area
The total area is:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{3}{16} + \frac{15}{32} = \frac{6}{32} + \frac{15}{32} = \frac{21}{32} \]

### Final Answer
The area of the bounded region enclosed by the curve and the x-axis is \( \frac{21}{32} \).