Question

\( f(x) = \begin{cases} 3-x, & -1<x<0 \\ 1+\frac{5x}{3}, & -3 \le x \le 2 \end{cases} \) and \( g(x) = \begin{cases} -x, & -2 \le x \le 3 \\ x, & 0 \le x \le 1 \end{cases} \)
then range of (fog)(x) is }

• A. \( [1, \frac{8}{3}] \)
• B. \( [-4, \frac{8}{3}] \)
• C. \( [-4, \frac{13}{3}] \)
• D. \( [\frac{8}{3}, \frac{10}{3}] \)

Hint:

Range of $f(g(x))$ is found by taking the range of $g$ as the domain of $f$.

Answer 1

The Correct Option is C

Step 1: Find range of g(x)
For $-2 \le x \le 3$, $g(x) = -x \implies$ range is $[-3, 2]$.
Step 2: Substitute into f(x)
Input for $f$ is now $[-3, 2]$.
Case 1: $-1<x<0 \implies f(x) = 3-x$. Range is $(3, 4)$.
Case 2: $-3 \le x \le 2 \implies f(x) = 1 + \frac{5x}{3}$.
Step 3: Calculate extrema
$f(-3) = 1 + \frac{5(-3)}{3} = -4$.
$f(2) = 1 + \frac{5(2)}{3} = \frac{13}{3}$.
Step 4: Conclusion
The combined range is $[-4, \frac{13}{3}]$.
Final Answer:(C)