Question

If \(f(x)=\sqrt{3 x-1}\) and \(g(x)=\sqrt{3 x+1}\) find \((f \cdot g)(x)\) and state its domain.

Short answer

(f ⋅ g)(x) = \(\sqrt{9x^2 - 1}\). The domain is \((-\infty, -\frac{1}{3}] \cup [\frac{1}{3}, \infty)\).

Step-by-step solution

Identify the given functions

The functions provided are:\[f(x) = \sqrt{3x - 1}\]\[g(x) = \sqrt{3x + 1}\]

Define the product of the functions

The product of the functions \(f(x)\) and \(g(x)\) is given by:\((f \cdot g)(x) = f(x) \cdot g(x)\)

Substitute the functions into the product

Substitute \(f(x)\) and \(g(x)\) into the product formula:\[(f \cdot g)(x) = \sqrt{3x - 1} \cdot \sqrt{3x + 1}\]

Simplify the product

Using the property of square roots \(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}\), we can simplify:\[(f \cdot g)(x) = \sqrt{(3x - 1)(3x + 1)}\]Next, expand the product inside the square root:\[(3x - 1)(3x + 1) = 9x^2 - 1\]Thus,\[(f \cdot g)(x) = \sqrt{9x^2 - 1}\]

Determine the domain

The expression under the square root, \(9x^2 - 1\), must be non-negative since the square root of a negative number is not defined in real numbers.Therefore, we need to solve the inequality:\[9x^2 - 1 \geq 0\]Rewriting it as:\[9x^2 \geq 1\]Divide both sides by 9:\[x^2 \geq \frac{1}{9}\]So,\[x \leq -\frac{1}{3} \text{ or } x \geq \frac{1}{3}\]Thus, the domain of the function \((f \cdot g)(x)\) is:\[(-\infty, -\frac{1}{3}] \cup [\frac{1}{3}, \infty)\]

Key concepts

domain of a function

The domain of a function refers to all the possible input values (x-values) that the function can accept. For functions involving radical expressions, like square roots, we can't take the square root of a negative number within the real number system. This limitation greatly influences the domain. For instance, consider the functions:

• f(x) = \(\sqrt{3x - 1}\)
• g(x) = \(\sqrt{3x + 1}\)

To determine the domain of the product \((f \cdot g)(x)\), we first look at the expressions under the square roots individually:

• The expression under f(x) is \(3x - 1\). For the square root to be defined, \(3x - 1 \geq 0\). Solving this inequality, we get \(x \geq \frac{1}{3}\).
• The expression under g(x) is \(3x + 1\). For the square root to be defined, \(3x + 1 \geq 0\). Solving this, we get \(x \geq -\frac{1}{3}\).

Next, we consider the product \((f \cdot g)(x) = \sqrt{(3x - 1)(3x + 1)}\). Here, \(3x - 1\) multiplied by \(3x + 1\) needs to be non-negative. This gives us the inequality \(9x^2 - 1 \geq 0\). Solving gives \(x \leq - \frac{1}{3} \text{ or } x \geq \frac{1}{3}\). Thus, the domain is:
\(( -\infty , - \frac{1}{3}] \cup [ \frac{1}{3}, \infty )\).

properties of square roots

Understanding the properties of square roots is crucial in algebra. Square roots possess several important properties that simplify computations and expressions. Specifically:

• The square root of a product: \(\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}\). This allows us to separate products under a single square root.
• The square root of a square: \(\sqrt{a^2} = |a|\). This tells us that taking the square root of a squared number results in the absolute value of the original number.
• Non-negativity: Since squaring any real number results in a non-negative number, the square root function is only defined for non-negative values.
• Inequality handling: When dealing with inequalities involving square roots, ensure that expressions inside the square root are non-negative.

These properties enable the simplification of \(\sqrt{(3x - 1)(3x + 1)}\). By expanding the product inside, we use:
\[(3x - 1)(3x + 1) = 3x \cdot 3x + 3x \cdot 1 - 1 \cdot 3x - 1 \cdot 1 = 9x^2 - 1\]
So, \(\sqrt{(3x - 1)(3x + 1)} = \sqrt{9x^2 - 1}\). This follows directly from applying the properties of square roots.

inequality solutions

Solving inequalities is a key concept in algebra. To determine where an expression is non-negative or non-positive, we must solve the inequality step-by-step. For the problem \(9x^2 - 1 \geq 0\), we proceed as follows:

• Start by isolating x: \(9x^2 \geq 1\).
• Divide both sides by 9: \(x^2 \geq \frac{1}{9}\).
• Take the square root of both sides, remembering to consider both positive and negative roots: \(|x| \geq \frac{1}{3}\).
• This results in: \(x \leq -\frac{1}{3}\) or \(x \geq \frac{1}{3}\).

Thus, the solution to the inequality \(9x^2 - 1 \geq 0\) is:
\(x \leq -\frac{1}{3}\) or \(x \geq \frac{1}{3}\). This aligns with our earlier domain finding for the function \((f \cdot g)(x)\), which is: \(( -\infty , - \frac{1}{3}] \cup [ \frac{1}{3}, \infty )\).