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Interval notation is a way of writing subsets of the real number line. It’s a concise way to express an interval utilizing parentheses and brackets. Parentheses \( ( \) or \( ) \) indicate that an endpoint is not included, while brackets \( [ \) or \( ] \) indicate that an endpoint is included.
In the context of the function \(k(x) = \frac{1}{49 - x^2}\), we exclude points where the denominator is zero. These are the points where the function is undefined.
Here’s a step-by-step guide to using interval notation:

• Identify the points to exclude or include.
• Use ∪ (union) to combine multiple intervals.
• Use (a, b) for intervals where a and b are excluded.
• Use [a, b] for intervals where a and b are included.

Applying this to our function, and given that \(x = 7\) and \( x = -7\) make the function undefined, we write the domain in interval notation as: \( (-∞, -7) \cup (-7, 7) \cup (7, -∞) \). This means all real numbers excluding -7 and 7.

Undefined points in a function are values that cause the function to have no output or result in an undefined mathematical operation, like division by zero.
For the function \(k(x) = \frac{1}{49 - x^2}\), the denominator is zero when \(x = 7\) and \( x = -7\). These are our undefined points because:

• When the denominator is zero, the function value would be undefined as division by zero is not possible in real number arithmetic.
• To find undefined points, set the denominator equal to zero and solve for x.

By solving \(49 - x^2 = 0\), we get \(x^2 = 49\). Taking the square root results in \( x = 7 \) and \( x = -7 \). Thus, the domain excludes these points.

Real numbers include all numbers that can be found on the number line. This includes both rational numbers (fractions) and irrational numbers (non-repeating, non-terminating decimals).
In terms of the function domain, real numbers are crucial because we need to identify which real numbers are allowed. For \(k(x) = \frac{1}{49 - x^2}\), the function is defined for all real numbers except where the denominator is zero.

• Identify real number exclusions by setting the denominator equal to zero and solving for x.
• Our original problem excludes 7 and -7 from the domain, as these points make the function undefined.

Therefore, the domain of our function is all real numbers except -7 and 7: \( (-∞, -7) \cup (-7, 7) \cup (7, -∞) \).