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

The Correct Option is C

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 \([0, 1]\).