Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

Given the functions:
[tex]\[
\begin{array}{l}
f(x) = \sqrt{3x} \\
g(x) = \sqrt{48x}
\end{array}
\][/tex]

Find \((f \cdot g)(x)\). Assume \(x \geq 0\).

A. \((f \cdot g)(x) = \sqrt{51x}\)

B. \((f \cdot g)(x) = 72x\)

C. \((f \cdot g)(x) = 12\sqrt{x}\)

D. [tex]\((f \cdot g)(x) = 12x\)[/tex]


Sagot :

To find the product of the functions \( f(x) \) and \( g(x) \), we need to compute \( (f \cdot g)(x) = f(x) \cdot g(x) \).

Given the functions:
[tex]\[ f(x) = \sqrt{3x} \][/tex]
[tex]\[ g(x) = \sqrt{48x} \][/tex]

Let's determine \( (f \cdot g)(x) \):

Step 1: Write down the product \( f(x) \cdot g(x) \):
[tex]\[ (f \cdot g)(x) = \left(\sqrt{3x}\right) \cdot \left(\sqrt{48x}\right) \][/tex]

Step 2: Use the property of square roots \(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}\):
[tex]\[ (f \cdot g)(x) = \sqrt{(3x) \cdot (48x)} \][/tex]

Step 3: Simplify the expression inside the square root:
[tex]\[ (f \cdot g)(x) = \sqrt{3x \cdot 48x} \][/tex]
[tex]\[ (f \cdot g)(x) = \sqrt{3 \cdot 48 \cdot x \cdot x} \][/tex]
[tex]\[ (f \cdot g)(x) = \sqrt{144x^2} \][/tex]

Step 4: Simplify the square root:
[tex]\[ \sqrt{144x^2} = \sqrt{144} \cdot \sqrt{x^2} \][/tex]
[tex]\[ \sqrt{144x^2} = 12 \cdot x \][/tex]

So, \( (f \cdot g)(x) = 12x \).

Thus, the correct answer is:
[tex]\[ \boxed{12x} \][/tex]

The answer is Option D:
[tex]\[ (f \cdot g)(x) = 12x \][/tex]
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.