If $$f(x) = \begin{cases} 2 + 2x, & -1 \leq x < 0 \\ 1 - \frac{x}{3}, & 0 \leq x \leq 3 \end{cases}$$; $$g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}$$, then range of $$(f \circ g(x))$$ is

We need to find the range of the composite function $$f \circ g(x) = f(g(x))$$.

Determine the domain and range of $$g(x)$$.

$$g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}$$

For $$x \in [-3, 0]$$: $$g(x) = -x \in [0, 3]$$

For $$x \in (0, 1]$$: $$g(x) = x \in (0, 1]$$

So the range of $$g$$ is $$[0, 3]$$.

Evaluate $$f(g(x))$$ where $$g(x) \in [0, 3]$$.

Since $$g(x) \in [0, 3]$$, we use the second piece of $$f$$:

$$f(t) = 1 - \frac{t}{3}$$ for $$0 \leq t \leq 3$$

When $$t = 0$$: $$f(0) = 1$$

When $$t = 3$$: $$f(3) = 1 - 1 = 0$$

Since $$f(t) = 1 - t/3$$ is continuous and decreasing on $$[0, 3]$$, its range on this interval is $$[0, 1]$$.

Verify that all values in $$[0, 1]$$ are achieved.

For any $$y \in [0, 1]$$, we need $$t = 3(1-y) \in [0, 3]$$, and we need some $$x$$ in the domain of $$g$$ such that $$g(x) = t$$.

Since $$g$$ maps $$[-3, 0]$$ onto $$[0, 3]$$ (via $$g(x) = -x$$), every value $$t \in [0, 3]$$ is achieved.

Therefore, $$f(g(x))$$ achieves all values in $$[0, 1]$$, including both endpoints.

The range of $$f \circ g(x)$$ is $$[0, 1]$$.

The correct answer is Option (3): $$[0, 1]$$.