Problem 15 - College Algebra

Question

Find \(f+g, f-g, f g,\) and \(\frac{f}{g}\). Determine the domain for each function. $$f(x)=\sqrt{x+4}, g(x)=\sqrt{x-1}$$

Step-by-Step Solution

Answer

The sum of the functions \(f+g = \sqrt{x+4} + \sqrt{x-1}\) with the domain [1, +inf). The difference \(f-g = \sqrt{x+4} - \sqrt{x-1}\) with the domain [1, +inf). The product \(fg = \sqrt{(x+4)(x-1)}\) with the domain [1, +inf). The division \(f/g = (\sqrt{x+4}) / (\sqrt{x-1})\) with the domain (1, +inf).

Step 1: Calculate f+g

Adding f and g gives: \(f(x) + g(x) = \sqrt{x+4} + \sqrt{x-1}\). The domain for this function is the intersection of the domains of f and g, which is \([1, +\infty)\)

Step 2: Calculate f-g

Subtracting g from f gives: \(f(x) - g(x) = \sqrt{x+4} - \sqrt{x-1}\). The domain for this function is also the intersection of the domains of f and g that is \([1, +\infty)\)

Step 3: Calculate fg

Multiplying f and g gives: \(f(x) \cdot g(x) = \sqrt{x+4} \cdot \sqrt{x-1} = \sqrt{(x+4)(x-1)}\). The domain of this function is \([1, +\infty)\), since both expressions under square root must be non-negative.

Step 4: Calculate f/g

Dividing f by g results in: \(f(x) / g(x) = (\sqrt{x+4}) / (\sqrt{x-1})\). Considering that we cannot divide by zero, we must exclude the values for which g(x) = 0 from the domain of f(x). Therefore, the domain of \(f(x) / g(x)\) is \((1, +\infty)\)

Key Concepts

Domain of Functions

When working with functions, the domain is the set of all possible inputs (values of \(x\)) for which the function is defined. For example, in the functions \(f(x) = \sqrt{x+4}\) and \(g(x) = \sqrt{x-1}\), the expressions \(x+4\) and \(x-1\) must be greater than or equal to zero because square roots of negative numbers are not real numbers.

• For \(f(x)\), \(x+4 \geq 0\) implies \(x \geq -4\).
• For \(g(x)\), \(x-1 \geq 0\) implies \(x \geq 1\).

Therefore, when considering the domain for operations involving both \(f\) and \(g\), like addition or multiplication, it's crucial to use the intersection of the individual domains. Here, the intersection is \([1, +\infty)\), as both functions must be simultaneously defined.

Function Addition and Subtraction

Function operations enable us to combine or compare functions in meaningful ways. When you add or subtract functions, you simply perform the addition or subtraction on the output values of the functions for each input \(x\).
In our example:

• Addition: \(f(x) + g(x) = \sqrt{x+4} + \sqrt{x-1}\)
• Subtraction: \(f(x) - g(x) = \sqrt{x+4} - \sqrt{x-1}\)

The domains for both of these operations derive from the need for both functions to be defined. Thus, both have the domain \([1, +\infty)\). This ensures that only values of \(x\) that work in both expressions are used.

Function Multiplication and Division

Multiplying and dividing functions works with the respective outputs for each \(x\) as well. However, division requires a bit more caution since division by zero makes functions undefined.
Here's how it works in our example:

• Multiplication: \(f(x) \cdot g(x) = \sqrt{x+4} \cdot \sqrt{x-1} = \sqrt{(x+4)(x-1)}\)
• Division: \(f(x) / g(x) = \frac{\sqrt{x+4}}{\sqrt{x-1}}\)

The domain for multiplication remains \([1, +\infty)\) since each function must be defined. However, for division, \(g(x)\) must not be zero. So, \(xeq 1\) needs to be excluded, giving us the domain \((1, +\infty)\) for division.

Radical Functions

Radical functions often involve roots, such as square roots in this example. These functions have specific domain considerations because the expression under a root sign, known as the radicand, cannot be negative if we want real-number outputs.
For \(f(x) = \sqrt{x+4}\) and \(g(x) = \sqrt{x-1}\), the square root signs indicate that:

• The radicand \(x+4\) must be \(\geq 0\), therefore \(x \geq -4\) for \(f\).
• The radicand \(x-1\) must be \(\geq 0\), thus \(x \geq 1\) for \(g\).

This constraint is why their combined operations' domain is \([1, +\infty)\). Radical functions require careful attention to their domain to ensure real and defined outcomes.