The area of the bounded region enclosed by the curve $$y = 3 - \left|x - \frac{1}{2}\right| - |x + 1|$$ and the $$x$$-axis is

We need to find the area enclosed by $$y = 3 - \left|x - \frac{1}{2}\right| - |x+1|$$ and the x-axis.

Identify critical points:

The absolute values change at $$x = \frac{1}{2}$$ and $$x = -1$$. We analyze three regions.

Simplify $$y$$ in each region:

Region 1: $$x \geq \frac{1}{2}$$

$$y = 3 - (x - \frac{1}{2}) - (x+1) = 3 - x + \frac{1}{2} - x - 1 = \frac{5}{2} - 2x$$

Setting $$y = 0$$: $$x = \frac{5}{4}$$

Region 2: $$-1 \leq x < \frac{1}{2}$$

$$y = 3 - (\frac{1}{2} - x) - (x+1) = 3 - \frac{1}{2} + x - x - 1 = \frac{3}{2}$$

This is a constant, so $$y = \frac{3}{2} > 0$$ throughout this interval.

Region 3: $$x < -1$$

$$y = 3 - (\frac{1}{2} - x) - (-x - 1) = 3 - \frac{1}{2} + x + x + 1 = \frac{7}{2} + 2x$$

Setting $$y = 0$$: $$x = -\frac{7}{4}$$

Compute the area:

The curve is above the x-axis for $$x \in [-\frac{7}{4}, \frac{5}{4}]$$.

Area in Region 3 ($$x \in [-\frac{7}{4}, -1]$$):

$$A_1 = \int_{-7/4}^{-1} \left(\frac{7}{2} + 2x\right) dx = \left[\frac{7x}{2} + x^2\right]_{-7/4}^{-1}$$

$$= \left(-\frac{7}{2} + 1\right) - \left(-\frac{49}{8} + \frac{49}{16}\right) = -\frac{5}{2} - \left(-\frac{49}{16}\right) = -\frac{5}{2} + \frac{49}{16} = \frac{-40 + 49}{16} = \frac{9}{16}$$

Area in Region 2 ($$x \in [-1, \frac{1}{2}]$$):

$$A_2 = \frac{3}{2} \cdot \left(\frac{1}{2} - (-1)\right) = \frac{3}{2} \cdot \frac{3}{2} = \frac{9}{4}$$

Area in Region 1 ($$x \in [\frac{1}{2}, \frac{5}{4}]$$):

$$A_3 = \int_{1/2}^{5/4} \left(\frac{5}{2} - 2x\right) dx = \left[\frac{5x}{2} - x^2\right]_{1/2}^{5/4}$$

$$= \left(\frac{25}{8} - \frac{25}{16}\right) - \left(\frac{5}{4} - \frac{1}{4}\right) = \frac{25}{16} - 1 = \frac{9}{16}$$

Total area:

$$A = A_1 + A_2 + A_3 = \frac{9}{16} + \frac{9}{4} + \frac{9}{16} = \frac{9}{16} + \frac{36}{16} + \frac{9}{16} = \frac{54}{16} = \frac{27}{8}$$

The correct answer is Option C: $$\boxed{\dfrac{27}{8}}$$.