Question

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

• A. (0, 1]
• B. [0, 3]
• C. [0, 1)
• D. [0, 1]

Answer 1

Answer: (D)

[0, 1]

Answer 2

The range of $g(x)$ is $[0, 3]$. For $y \in [0, 3]$, $f(y) = 1 - \frac{y}{3}$. The range of $f(y)$ for $y \in [0, 3]$ is $[f(3), f(0)] = [1 - \frac{3}{3}, 1 - \frac{0}{3}] = [0, 1]$. Thus, the range of $(fog)(x)$ is $[0, 1]$.