Problem 14 - Algebra and Trigonometry

Question

Find the domain of the function. $$ k(x)=\frac{\sqrt{x+3}}{x-1} $$

Step-by-Step Solution

Answer

The domain of the function is \([-3, 1) \cup (1, \infty)\).

Step 1: Identify Domain Restrictions for the Numerator

The numerator of the function is \(\sqrt{x+3}\). A square root function is defined when its argument is non-negative. Therefore, we need to find the values of \(x\) such that:\[ x + 3 \geq 0 \]Solving for \(x\), we find:\[ x \geq -3 \]This means \(x\) must be at least \(-3\) for the square root to be defined.

Step 2: Identify Domain Restrictions for the Denominator

The denominator of the function is \(x-1\). A fraction is undefined when its denominator is zero. Therefore, the denominator \(x-1\) cannot equal zero. Thus, we find:\[ x - 1 eq 0 \]Solving for \(x\), we find:\[ x eq 1 \]This means \(x\) cannot be equal to 1.

Step 3: Combine Restrictions

Now, we combine the restrictions from Step 1 and Step 2. From the numerator, we have \(x \geq -3\), and from the denominator, \(x eq 1\). Therefore, the domain is all values of \(x\) such that:\[ x \geq -3, x eq 1 \]In interval notation, this is expressed as:\[ [-3, 1) \cup (1, \infty) \]

Key Concepts

Interval Notation

Interval notation is a convenient way of expressing a range of values, particularly for defining the domain of a function. It uses brackets and parentheses to describe the set of numbers that a variable, like \(x\), can take.

• Brackets \([a, b]\): Indicate that the endpoint numbers \(a\) and \(b\) are included in the set. This is known as a closed interval.
• Parentheses \((a, b)\): Indicate the endpoint numbers \(a\) and \(b\) are not included. This represents an open interval.

When specifying domain, you often encounter intervals that are joined by a union symbol \(\cup\). This symbol indicates that the domain includes values from both intervals. For example, the interval notation \([-3, 1) \cup (1, \infty)\) indicates the set of all numbers from \(-3\) to \(1\) (excluding \(1\)), and from any number greater than \(1\), towards infinity. This notation succinctly expresses multiple, potentially non-continuous segments of a domain.

Square Root Function

Square root functions involve the operation of taking the square root of an expression, which introduces certain mathematical considerations.To preserve real-number results, the expression inside the square root, often referred to as the radicand, must be non-negative. This is because the square root of a negative number results in an imaginary number, which is outside the scope of real-number domains often considered in basic algebra.Consider the radicand \(x + 3\) in the expression \(\sqrt{x+3}\). To keep the radicand non-negative:

• Set up an inequality: \(x + 3 \geq 0\).
• Solve it: \(x \geq -3\).

This inequality tells us that \(x\) must be greater than or equal to \(-3\) for the square root to be defined. This condition must be considered carefully when establishing the domain of a function that incorporates a square root.

Denominator Restrictions

When dealing with fractions in functions, the most critical factor for the function's domain is ensuring the denominator is never zero. A zero denominator would make the function undefined, as division by zero is not possible in mathematics.For example, in the function \(k(x)=\frac{\sqrt{x+3}}{x-1}\), the denominator is \(x - 1\). To avoid dividing by zero, set the denominator not equal to zero:

• Equation: \(x - 1 eq 0\)
• Solution: \(x eq 1\)

This implies that \(x\) must not equal \(1\) for the function to remain valid. This constraint, combined with other conditions like non-negative radicands, helps comprehensively determine the domain of a function.