Problem 3 - College Algebra

Question

Let \(f(x)=\sqrt{x-6}\) and \(g(x)=\sqrt{x+2},\) find: a. \((f+g)(x)\) b. the domain of \(f+g .\)

Step-by-Step Solution

Answer

The result of \((f+g)(x)\) is \(\sqrt{x-6} + \sqrt{x+2}\) and the domain of \(f+g\) is all real numbers x such that \(x \geq 6\).

Step 1: Compute (f+g)(x)

Add together the functions f and g. This process involves substituting into the equation, which gives \((f+g)(x) = \sqrt{x-6} + \sqrt{x+2} .\)

Step 2: Find the Domain of f+g

To find the domain of \(f+g\), it's needed to solve for x in the following inequalities to ensure that the values of x won't make the quantity in the square root negative: \n 1. \(x-6 \geq 0\). Solving for x gives \(x \geq 6\). This means that all values for x should be 6 or greater. \n2. \(x+2 \geq 0\). Solving for x gives \(x \geq -2\). This means that all values for x should be -2 or greater. \nSince we need to satisfy both conditions, we take the intersection, which gives \(x \geq 6\).

Step 3: State the Result

After calculations, the function (f+g)(x) and its domain have been found to be \((f+g)(x) = \sqrt{x-6} + \sqrt{x+2}\) and \(x \geq 6\), respectively.

Key Concepts

Domain of a Function

The domain of a function is a set of all possible input values that a function can accept. For any function, these are the values you can plug into the function without encountering mathematical inconsistencies like division by zero or taking the square root of a negative number. Understanding the domain is crucial, as it helps us know where the function is defined and can operate properly.
For instance, consider our function \(f(x) = \sqrt{x-6}\). To determine this function's domain, we need to ensure that whatever is inside the square root, known as the radicand, is not negative. That's because the square root of a negative number is not defined in the real number system. Therefore, for \(f(x)\), we set up the inequality \(x - 6 \geq 0\), solving gives us that \(x \geq 6\).
This tells us that the domain is from 6 to infinity in interval notation or \([6, \infty)\). It means any x-value less than 6 does not work in this function.

Square Root Function

A square root function is fundamentally a radical function. The general form of a square root function is \(f(x) = \sqrt{x}\). These functions exhibit unique properties. They are always non-negative for real values of x and start from a particular point called the starting point or endpoint. This starting point is determined by the values that make the expression under the square root non-negative.
Let's apply this to our function \(f(x)=\sqrt{x-6}\) and \(g(x)=\sqrt{x+2}\):

• For \(f(x)\), x must be at least 6, i.e., \(x \geq 6\).
• For \(g(x)\), x must be at least -2, i.e., \(x \geq -2\).

The graph of these functions depicts how they start from a certain x-value (known as the domain starting point) and rise, illustrating an increase in output as input values increase.
These properties of square root functions make them useful in modeling situations where the growth rate slows down as x increasesdue to their gentle upward slope.

Intersection of Domains

When dealing with multiple functions, like \(f(x)=\sqrt{x-6}\) and \(g(x)=\sqrt{x+2}\), the domain of their combined function is influenced by the domains of both individual functions. In scenarios where you must perform operations like addition on multiple functions, finding an intersection of their domains becomes necessary.
The intersection of domains means finding the set of x-values that satisfy the domain conditions of all involved functions simultaneously. This is crucial because the combined function, such as \(f+g=(f+g)(x) = \sqrt{x-6} + \sqrt{x+2}\), must be defined wherever both \(f\) and \(g\) are defined.
For the discussed functions, the domain of \(f(x)\) is \(x \geq 6\) and for \(g(x)\) is \(x \geq -2\). To ensure both components are valid for \(f+g(x)\), the domain must satisfy x-values for both which leads to the intersection domain: \(x \geq 6\).
This means that the function \( (f+g)(x) \) will only be valid for x-values greater than or equal to 6. By determining the intersection of the domains, we ensure that every square root operation is within the realm of real numbers.