Question:medium

\( f(x) = \begin{cases} 3-x, & -1<x<0 \\ 1+\frac{5x}{3}, & -3 \le x \le 2 \end{cases} \) 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 }

Show Hint

Range of $f(g(x))$ is found by taking the range of $g$ as the domain of $f$.
Updated On: May 16, 2026
  • \( [1, \frac{8}{3}] \)
  • \( [-4, \frac{8}{3}] \)
  • \( [-4, \frac{13}{3}] \)
  • \( [\frac{8}{3}, \frac{10}{3}] \)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
We are given two piecewise functions, \(f(x)\) and \(g(x)\), and we need to find the range of their composition, \(f(g(x))\). The range of \(f(g(x))\) is the set of all possible output values of \(f\) when the input is taken from the range of \(g\).
Step 2: Key Formula or Approach:
1. Determine the domain of the composite function \(f(g(x))\). This is the set of \(x\) in the domain of \(g\) for which \(g(x)\) is in the domain of \(f\).
2. Determine the range of the inner function, \(g(x)\), over its domain. Let's call this Range(\(g\)).
3. Determine the range of the outer function, \(f(x)\), for the inputs that come from Range(\(g\)).
Step 3: Detailed Explanation:
1. Analyze the functions and their domains:
Domain of \(f\), \(D_f = [-3, 2] \cup (-1, 0) = [-3, 2]\).
Domain of \(g\), \(D_g = [-2, 3] \cup [0, 1] = [-2, 3]\).
The domain of \(f(g(x))\) is the domain of \(g\), which is \(x \in [-2, 3]\).
2. Find the range of g(x):
We need to find the set of all values \(g(x)\) can take for \(x \in [-2, 3]\).
- For \(x \in [-2, 3]\), \(g(x) = -x\). The range for this part is \(g(-2) = 2\) and \(g(3) = -3\). Since it's a line, the range is \([-3, 2]\).
- For \(x \in [0, 1]\), \(g(x) = x\). The range for this part is \([0, 1]\).
The overall range of \(g(x)\) is the union of these ranges: \([-3, 2] \cup [0, 1] = [-3, 2]\).
So, Range(\(g\)) = \([-3, 2]\).
3. Find the range of f(y) for y \(\in\) Range(g):
The range of \(g(x)\) becomes the set of inputs for \(f(x)\). So we need to find the range of \(f(y)\) for \(y \in [-3, 2]\).
Let's look at the definition of \(f(x)\):
\( f(x) = \begin{cases} 3-x, & -1<x<0
1+\frac{5x}{3}, & -3 \le x \le 2 \end{cases} \)
The input domain we are interested in is \(y \in [-3, 2]\). This entire interval falls into the second piece of the definition of \(f(x)\).
So, we need to find the range of the function \(h(y) = 1 + \frac{5y}{3}\) on the interval \(y \in [-3, 2]\).
Since \(h(y)\) is a linear function with a positive slope (5/3), it is monotonically increasing. Its minimum and maximum values will occur at the endpoints of the interval.
- At the minimum input \(y = -3\):
\(f(-3) = 1 + \frac{5(-3)}{3} = 1 - 5 = -4\).
- At the maximum input \(y = 2\):
\(f(2) = 1 + \frac{5(2)}{3} = 1 + \frac{10}{3} = \frac{3+10}{3} = \frac{13}{3}\).
Since the function is continuous on this interval, the range of \(f(g(x))\) is the closed interval from the minimum to the maximum value.
Range of \(f(g(x))\) = \([-4, \frac{13}{3}]\).
Step 4: Final Answer:
The range of the composite function (fog)(x) is \( [-4, \frac{13}{3}] \), which matches option (C).
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