[tex]$
\begin{array}{l}
f(x)=\sqrt{32x} \\
g(x)=\sqrt{2x}
\end{array}
$[/tex]

Find [tex]$(f \cdot g)(x)$[/tex]. Assume [tex]$x \geq 0$[/tex].

A. [tex]$(f \cdot g)(x)=8 \sqrt{x}$[/tex]

B. [tex]$(f \cdot g)(x)=\sqrt{34x}$[/tex]

C. [tex]$(f \cdot g)(x)=8x$[/tex]

D. [tex]$(f \cdot g)(x)=32x$[/tex]



Answer :

To solve for [tex]\((f \cdot g)(x)\)[/tex] where [tex]\( f(x) = \sqrt{32x} \)[/tex] and [tex]\( g(x) = \sqrt{2x} \)[/tex], follow these steps:

### Step-by-Step Solution:

1. Understand the Functions:
- We have [tex]\( f(x) = \sqrt{32x} \)[/tex].
- We also have [tex]\( g(x) = \sqrt{2x} \)[/tex].

2. Define the Product of the Functions:
- The product of two functions, [tex]\( (f \cdot g)(x) \)[/tex], is calculated as [tex]\( f(x) \cdot g(x) \)[/tex].

3. Set Up the Expression:
- [tex]\( (f \cdot g)(x) = \sqrt{32x} \cdot \sqrt{2x} \)[/tex].

4. Combine the Square Roots:
- Recall the property of square roots: [tex]\( \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \)[/tex]:
[tex]\[ (f \cdot g)(x) = \sqrt{32x} \cdot \sqrt{2x} = \sqrt{(32x) \cdot (2x)}. \][/tex]

5. Simplify the Expression:
- Simplify inside the square root:
[tex]\[ (32x) \cdot (2x) = 64x^2. \][/tex]
- Therefore:
[tex]\[ (f \cdot g)(x) = \sqrt{64x^2}. \][/tex]

6. Evaluate the Square Root:
- Simplify the square root of [tex]\( 64x^2 \)[/tex]:
[tex]\[ \sqrt{64x^2} = \sqrt{64} \cdot \sqrt{x^2} = 8 \cdot x = 8x. \][/tex]

### Conclusion:
The function that describes [tex]\((f \cdot g)(x)\)[/tex] is:
[tex]\[ (f \cdot g)(x) = 8x. \][/tex]

Thus, the correct answer is:

[tex]\[ \boxed{8x} \][/tex]

Therefore, the correct option is C. [tex]\((f \cdot g)(x)=8x\)[/tex].