To find \((f \cdot g)(x)\), we first need to find the expressions for \(f(x)\) and \(g(x)\):
- \(f(x) = \sqrt{x - 2}\)
- \(g(x) = \sqrt{x + 7}\)
Now, we can compute the product:
\[ (f \cdot g)(x) = f(x) \cdot g(x) = \sqrt{x - 2} \cdot \sqrt{x + 7} \]
Using the property of square roots that \(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}\), we have:
\[ (f \cdot g)(x) = \sqrt{(x - 2)(x + 7)} \]
Next, we need to simplify \((x - 2)(x + 7)\):
\[ (x - 2)(x + 7) = x^2 + 7x - 2x - 14 = x^2 + 5x - 14 \]
So we can express \((f \cdot g)(x)\) as:
\[ (f \cdot g)(x) = \sqrt{x^2 + 5x - 14} \]
Considering the domain restrictions for \(f(x)\) and \(g(x)\):
- For \(f(x) = \sqrt{x - 2}\), the expression inside the square root must be non-negative, so \(x - 2 \geq 0\) implies \(x \geq 2\).
- For \(g(x) = \sqrt{x + 7}\), the expression inside the square root must be non-negative, so \(x + 7 \geq 0\) implies \(x \geq -7\).
However, since the more restrictive condition is \(x \geq 2\), that's the domain we will consider.
Thus, the final result is:
\[ (f \cdot g)(x) = \sqrt{x^2 + 5x - 14} \]
The matching response is:
\[ \text{left-parenthesis lower f dot lower g right-parenthesis left-parenthesis x right-parenthesis equals Start Root x squared plus 5 x minus 14 End Root} \]
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