Question

Let \[ f(x)=\sqrt{7-x} \] and \[ g(x)=\sqrt{x-5}. \] Then the domain of the function \[ h(x)=f(x)g(x) \] is:

• A. [5, 7]
• B. \((-\infty, 7]\)
• C. [5, \(\infty\))
• D. [-7, -5]
• E. \((-\infty, \infty)\)

Hint:

Whenever you see a product of multiple square roots, list all conditions \(radicand \geq 0\) and find the overlapping region on a number line. If the region is empty, the function is never defined.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The function \(h(x)\) is the product of two square root functions.
For a square root function \(\sqrt{u(x)}\) to be defined in the set of real numbers, the radicand \(u(x)\) must be non-negative (\(u(x) \geq 0\)).
The domain of \(h(x) = f(x)g(x)\) is the intersection of the domains of \(f(x)\) and \(g(x)\).

Step 2: Key Formula or Approach:
1. For \(f(x) = \sqrt{7-x}\), we solve \(7 - x \geq 0\).
2. For \(g(x) = \sqrt{x-5}\), we solve \(x - 5 \geq 0\).

Step 3: Detailed Explanation:
Finding domain of \(f(x)\):
\[ 7 - x \geq 0 \implies x \leq 7 \]
So, Domain of \(f = (-\infty, 7]\).
Finding domain of \(g(x)\):
\[ x - 5 \geq 0 \implies x \geq 5 \]
So, Domain of \(g = [5, \infty)\).
Finding intersection of domains:
The domain of \(h(x)\) is where both conditions are satisfied simultaneously:
\[ x \in (-\infty, 7] \cap [5, \infty) \]
This gives the closed interval \([5, 7]\).

Step 4: Final Answer:
The domain of the function \(h(x)\) is \([5, 7]\).