Problem 20 - College Algebra

Question

Find the domain of each function. $$ g(x)-\frac{1}{\sqrt{x+2}} $$

Step-by-Step Solution

Answer

The domain of the function \( g(x)=\frac{1}{\sqrt{x+2}} \) is all real numbers greater than or equal to -2.

Step 1: Define the function

Here is the function given : \( g(x)=\frac{1}{\sqrt{x+2}} \)

Step 2: Identify the restrictions

By knowing that there cannot be a negative number inside the square root and we cannot divide by zero, we identify 2 possible restrictions to the function.

Step 3: Eliminate negative radicands

For the radicand to be non-negative, we must have \(x+2 >= 0\). This gives us \(x >= -2\). This means that the function is undefined for values of \(x < -2\).

Step 4: Elimate zero denominators

From the given function, we see that the denominator can't be zero for the entire function not to be undefined. However, we've already established that x must be greater than or equal to -2 for the square root to work, so this is not really a restriction. It would have been, if we hadn't already established the previous condition.

Step 5: Combine the restrictions

By combining the restrictions we are able to find the domain of the function. In this case, with only one condition, \(x >= -2\), the domain of \( g(x) \) is \( x | x >= -2 \), which is the set of all real numbers greater than or equal to -2.

Key Concepts

College Algebra

College algebra is an essential foundation in higher mathematics and covers a variety of topics including functions, polynomials, and rational expressions. One of the key focuses in college algebra is understanding the concept of a function and its domain. A function can be thought of as a machine that takes an input, performs some operations, and then gives an output. However, not all inputs are allowed for every function. There are certain rules that must be adhered to, and this is where domain determination comes in.

In the case of the function given, \( g(x)=\frac{1}{\sqrt{x+2}} \), college algebra teaches us to look closely at the restrictions that come with square roots and division. It's a perfect example of how algebraic knowledge is applied to ensure we are working within the realms of mathematical correctness. When identifying the domain of a function, we’re essentially asking, 'What are the possible values that we can plug into this function without breaking any mathematical rules?'

Square Root Restrictions

Square root restrictions play a major role when determining the domain of a function that involves a radical. According to the mathematical rules, you cannot take the square root of a negative number when dealing with real numbers. Therefore, when you encounter a function with a square root, like \(g(x)\), you must ensure that the radicand (the number inside the square root) is always non-negative. That's why in our exercise, the expression underneath the square root, \(x+2\), must be greater than or equal to zero.

Understanding Radicands

In our exercise, we set \(x+2 \geq 0\) to ensure that we won't end up with an imaginary number, leading to the conclusion that \(x \geq -2\). This part of algebra often confuses students because it involves both algebraic manipulation and understanding the fundamental properties of square roots. Remember that this restriction is paramount, as it's integral to determining which values of \(x\) make the function \(g(x)\) work in the real number system.

Function Domain Determination

Function domain determination is like solving a puzzle – certain pieces cannot fit due to specific constraints. In this context, the 'pieces' are the input values for the function, and the 'puzzle' is the mathematical rules that dictate which of these values are permissible. When you're working with a function's domain, you're essentially identifying all the legal inputs that can be used within the function without causing any undefined behaviors like dividing by zero or taking the square root of a negative number.

Combining Conditions to Find the Domain

As we saw in the provided solution, after determining that the radicand must be non-negative, we ultimately found that the domain for \(g(x)\) includes all real numbers where \(x\) is greater than or equal to -2. It means we're excluding the part of the number line that would cause the function to be undefined. Remember, it is the combination of these constraints that truly defines the function's domain and ensures it's properly understood within the scope of college algebra.