Problem 12 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The domain is \([-1, 0) \cup (0, \infty)\).

Step 1: Identify Restrictions in the Square Root

The expression inside the square root, \(x+1\), must be non-negative for \(g(x)\) to be defined. This means \(x+1 \geq 0\). Solving this inequality gives us \(x \geq -1\).

Step 2: Identify Restrictions in the Denominator

Since there is a division by \(x\), we need \(xeq 0\) to ensure the function does not have a division by zero.

Step 3: Combine Restrictions

From Step 1, we know \(x \geq -1\). From Step 2, we know \(x eq 0\). Combining these, we find the domain of \(g(x)\) is \([-1, 0) \cup (0, \infty)\). This means \(x\) can be any real number greater than or equal to -1, except zero.

Key Concepts

Square Root Restrictions

When dealing with a function that includes a square root, it's crucial to remember that the expression inside the root must be non-negative. This is because you cannot take the square root of a negative number without involving complex numbers, which changes the function's scope.
For the function given here, we have the square root term as \( \sqrt{x+1} \). To ensure that this term is defined for real numbers:

• The expression \( x+1 \) must be greater than or equal to zero. This translates to the inequality \( x+1 \geq 0 \).
• Solving \( x+1 \geq 0 \) gives \( x \geq -1 \).

Thus, the domain of the function considering only the square root is all real numbers \( x \) such that \( x \geq -1 \). This does not account for any potential restrictions from the other part of the function, such as division by zero.

Division by Zero

The rule of thumb is that you can never divide by zero in mathematics because it yields undefined or infinite values. In the function \( g(x) = \sqrt{x+1} - \frac{1}{x} \), there is a division by \( x \) in the second term.
To avoid making a division by zero:

• We label \( x eq 0 \) as a restriction. This means \( x \) cannot be zero because substituting zero would cause the function to attempt division by zero.

This restriction divides our potential domain into two segments: any number greater than or equal to -1, and avoiding zero entirely. We must view each part of the function for both root and denominator restrictions together.

Inequalities

Inequalities help us express conditions or restrictions on variables within verbal or algebraic statements. In domain problems, they show us the permissible "range" where a function remains valid and does not reach undefined values.
In the example function:

• The inequality \( x+1 \geq 0 \) arises from the square root restriction, giving \( x \geq -1 \).
• The division restriction \( x eq 0 \), as discussed, prevents division by zero.

When we combine these restrictions using inequalities, we look for \( x \) values that satisfy both conditions:

• \( x \geq -1 \) from the square root.
• \( x eq 0 \) from the denominator.

Merging these solutions effectively divides space into two intervals for the domain: \( [-1, 0) \) and \( (0, \infty) \), ensuring the function remains defined across these regions without causing undefined situations.