Questions: Given f(x)=3/(x-3), g(x)=sqrt(x), and (f ∘ g)(x)=3/(sqrt(x)-3), find the domain of f ∘ g. a. (-∞, 0] ∪ (0,9) ∪ (9, ∞) b. [0,9) ∪ (9, ∞) c. [9, ∞]

Transcript text: Given $f(x)=\frac{3}{x-3}, g(x)=\sqrt{x}$, and $(f \circ g)(x)=\frac{3}{\sqrt{x}-3}$, find the domain of $f \circ g$. a. $(-\infty, 0] \cup(0,9) \cup(9, \infty)$ b. $[0,9) \cup(9, \infty)$ c. $[9, \infty]$

Solution

Solution Steps

To find the domain of the composite function $(f \circ g)(x) = \frac{3}{\sqrt{x} - 3}$, we need to consider the domains of both $f(x)$ and $g(x)$ and ensure that the composition is defined. Specifically, we need to ensure that:

$g(x) = \sqrt{x}$ is defined, which requires $x \geq 0$.
$f(g(x)) = \frac{3}{\sqrt{x} - 3}$ is defined, which requires $\sqrt{x} \neq 3$.

Solution Approach

Determine the domain of $g(x) = \sqrt{x}$, which is $x \geq 0$.
Determine where $\sqrt{x} - 3 \neq 0$, which means $\sqrt{x} \neq 3$ or $x \neq 9$.
Combine these conditions to find the domain of $f \circ g$.

Step 1: Determine the Domain of $ g(x) $

The function $ g(x) = \sqrt{x} $ is defined for $ x \geq 0 $. Therefore, the domain of $ g $ is given by the interval: \[ [0, \infty) \]

Step 2: Identify Points of Discontinuity in $ f(g(x)) $

Next, we analyze the function $ f(g(x)) = \frac{3}{\sqrt{x} - 3} $. This function is undefined when the denominator is zero, which occurs when: \[ \sqrt{x} - 3 = 0 \implies \sqrt{x} = 3 \implies x = 9 \]

Step 3: Combine the Domains

The domain of $ f \circ g $ must satisfy both the domain of $ g $ and the condition that $ f(g(x)) $ is defined. Thus, we exclude $ x = 9 $ from the domain of $ g $: \[ \text{Domain of } f \circ g = [0, 9) \cup (9, \infty) \]