Relations & Functions
Composite Functions and Domain
Grade 12
Question:
<p>The domain of \(f(g(x))\) is</p>
<p>(a) \(x \in \mathbb{R}\)</p>
<p>(b) \(x \in \mathbb{R} \setminus \{1\}\)</p>
<p>(c) \(x \in \mathbb{R} \setminus \{0, 1\}\)</p>
<p>(d) \(x \in \mathbb{R} \setminus \{0, 1, -1\}\)</p>
Step-by-Step Solution
Key Concept: For composite functions, domain restrictions come from both the inner function g(x) and the outer function f applied to g(x). Check where each is undefined.
<p><strong>Solution:</strong> Given $f(x) = \frac{1}{1+x}$ for $x \notin \{0, 1\}$.</p><p>We computed: $g(x) = f(f(f(x))) = x$ for $x \in \mathbb{R} \setminus \{0, 1\}$.</p><p>For $f(g(x)) = f(x) = \frac{1}{1+x}$ to be defined:</p><p>1. $g(x)$ must be defined: $x \notin \{0, 1\}$</p><p>2. $f(g(x))$ must be defined: $g(x) \notin \{0, 1\}$</p><p>Since $g(x) = x$, we need $x \notin \{0, 1\}$.</p><p>∴ Domain of $f(g(x))$ is $\mathbb{R} \setminus \{0, 1\}$</p>
Correct Answer: c