Question

If $f(x) = \begin{cases} 2+2x & ;x∈(-1,0)\\\ 1-\frac x3 & ;x∈[0,3) \end{cases}$ and $g(x) = \begin{cases} x & ;x∈[0,1)\\\ -x & ;x∈(-3,0) \end{cases}$. Then range of $fog(x)$ is

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

Answer 1

Answer: (C)

(0,1]

Answer 2

To find the range of $f(g(x))$, we start by evaluating $g(x)$ and then substitute it into $f(x)$ according to the intervals provided.

Evaluate $g(x)$:
$g(x) = \begin{cases} x, & x \in [0, 1] \\\ -x, & x \in (-3, 0) \end{cases}$
For $x \in [0, 1]$, $g(x) = x$ which gives $g(x) \in [0, 1]$.
For $x \in (-3, 0)$, $g(x) = -x$ which gives $g(x) \in (0, 3]$.
Therefore, the range of $g(x)$ is $(0, 3]$.

Since $g(x) \in (0, 3]$, we use the definition of $f(x)$ for $x \in [0, 3]$:
$f(g(x)) = 1 - \frac{g(x)}{3}$

Determine the range of $f(g(x))$ by substituting values from the range of $g(x)$. For $g(x) = 0$, $f(g(x)) = 1 - \frac{0}{3} = 1$. For $g(x) = 3$, $f(g(x)) = 1 - \frac{3}{3} = 0$. Thus, as $g(x)$ varies over the interval $(0, 3]$, $f(g(x))$ varies over the interval $[0, 1]$.

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