Question 3. If f(x) and g(x) are defined as
f(x)= \begin{cases} 2+2x, & -1 \le x < 0 \ 1-\frac{x}{3}, & 0 \le x \le 3 \end{cases}and
g(x)= \begin{cases} -x, & -3 \le x \le 0 \ x, & 0 < x \le 1 \end{cases}then the range of (f \circ g)(x) is
(1) (0,1]
(2) [0,3)
(3) [0,1]
(4) [0,1)
Answer (3)
Explanation:
By definition, (f \circ g)(x)=f(g(x))
So,
f(g(x))= \begin{cases} 2+2g(x), & -1 \le g(x) < 0 \ 1-\frac{g(x)}{3}, & 0 \le g(x) \le 3 \end{cases}From the definition of g(x), g(x) \in [0,3]
For 0 \le g(x) \le 3,
f(g(x)) = 1-\frac{g(x)}{3}At g(x)=0, f(g(x))=1
At g(x)=3, f(g(x))=0
Hence, the range of f(g(x)) is [0, 1]
