Problem 48 - College Algebra Essentials

Question

Find \(f+g, f-g,\) fg, and \(\frac{f}{g} .\) Determine the domain for each function. $$f(x)=\sqrt{x+6}, g(x)=\sqrt{x-3}$$

Step-by-Step Solution

Answer

The new functions \(f(x) + g(x), f(x) - g(x), f(x) \cdot g(x), \) and \( \frac{f(x)}{g(x)} \) are \(\sqrt{x+6} + \sqrt{x-3}, \sqrt{x+6} - \sqrt{x-3}, \sqrt{(x+6)(x-3)},\) and \(\frac{\sqrt{x+6}}{\sqrt{x-3}}\) respectively. They all have the domain [3, \(\infty)\), except for \( \frac{f(x)}{g(x)} \) which has the domain (3, \(\infty)\).

Step 1: Addition of the functions

To add the functions f and g we just add them directly. The result would be the function \(h(x) = f(x) + g(x) = \sqrt{x+6} + \sqrt{x-3}\). The domain of this function is the set of x values for which both \(x + 6\) and \(x - 3\) are non-negative, or \([3, \infty)\).

Step 2: Subtraction of the functions

Subtracting the functions is also straightforward. The result, \(h(x) = f(x) - g(x) = \sqrt{x+6} - \sqrt{x-3}\). The domain of this function is the set of x values for which \(x + 6\) is non-negative and \(x - 3\) is non-negative, or \([3, \infty)\).

Step 3: Multiplication of the functions

To multiply the functions, we calculate \(h(x) = f(x) \cdot g(x) = \sqrt{x+6} \cdot \sqrt{x-3}\). This simplifies to \(h(x) = \sqrt{(x+6)(x-3)}\). The domain of h(x) is the set of x values for which \((x + 6)(x - 3)\) is non-negative, or \([3, \infty)\).

Step 4: Division of the functions

To divide the functions we calculate \(h(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x+6}}{\sqrt{x-3}}\). This division is defined when \(x + 6 \geq 0\) and \(x-3 > 0\), which gives us the domain \((3, \infty)\).

Key Concepts

Domain of Functions

The domain of a function consists of all the possible input values (x-values) that allow the function to produce real values. When dealing with square roots like in our functions \( f(x) = \sqrt{x+6} \) and \( g(x) = \sqrt{x-3} \), the expressions inside the square roots must be non-negative. This is to ensure the result is a real number.

• For \( f(x) \), the domain is determined by \( x + 6 \geq 0 \), simplifying to \( x \geq -6 \).
• For \( g(x) \), the condition \( x - 3 \geq 0 \) means \( x \geq 3 \).

When combining functions through operations, such as addition, subtraction, multiplication, or division, the domain is the intersection of the individual domains. The intersection of the domains of \( f(x) \) and \( g(x) \) is \( [3, \infty) \). For division, however, we must exclude any x-values that cause division by zero, leading to \( (3, \infty) \).

Addition of Functions

Adding functions involves the sum of their outputs for each x-value in the domain. When adding \( f(x) = \sqrt{x+6} \) and \( g(x) = \sqrt{x-3} \), the new function is \( h(x) = f(x) + g(x) = \sqrt{x+6} + \sqrt{x-3} \).
The domain of this sum is determined by the most restrictive domain constraints from both functions, which means \( x \) must be greater than or equal to 3. Therefore, the domain is \( [3, \infty) \). This ensures both square root expressions are defined, producing real values for their sum.

Subtraction of Functions

Subtraction of functions follows the same principle as addition, but instead, we subtract the values obtained from each function. For \( f(x) \) and \( g(x) \), the result is \( h(x) = f(x) - g(x) = \sqrt{x+6} - \sqrt{x-3} \).
Like addition, the domain for this subtraction is the overlapping domain of both functions, \( x \geq 3 \), resulting in the domain \( [3, \infty) \). As in the case of addition, each square root must be valid to ensure subtracting them results in real numbers.

Multiplication of Functions

The multiplication of functions \( f(x) \) and \( g(x) \) creates a product function. In our case, this looks like \( h(x) = f(x) \cdot g(x) = \sqrt{x+6} \cdot \sqrt{x-3} \).
This expression simplifies to \( h(x) = \sqrt{(x+6)(x-3)} \). To identify the domain, we check when the product \( (x+6)(x-3) \geq 0 \). This can be simplified, and by checking values, we establish the domain is \( [3, \infty) \), ensuring that the product within the square root is non-negative, thus producing a real output.

Division of Functions

Dividing one function by another involves ensuring the denominator is never zero. Here we're dividing \( f(x) \) by \( g(x) \), giving the function \( h(x) = \frac{\sqrt{x+6}}{\sqrt{x-3}} \).
The condition \( x-3 > 0 \) must be met to have a valid denominator. Consequently, \( x \) must be strictly greater than 3, resulting in a domain of \( (3, \infty) \). This domain maintains that the square root function \( g(x) \) isn't zero or undefined when performing the division. Therefore, ensuring all provided values for \( x \) lead to a defined and real result.