Problem 77 - Beginning and Intermediate Algebra

Question

Determine the domain of each function. $$k(x)=\frac{1}{x^{2}+11 x+24}$$

Step-by-Step Solution

Answer

The domain of the function \(k(x) = \frac{1}{x^2 + 11x + 24}\) is all real numbers except for x = -3 and x = -8, which can be written in interval notation as: \((-\infty, -8) \cup (-8, -3) \cup (-3, \infty)\).

Step 1: Identify the denominator of the function

We are given the function \(k(x) = \frac{1}{x^2 + 11x + 24}\). The denominator of this function is \(x^2 + 11x + 24\).

Step 2: Find the values of x that make the denominator equal to zero

To find the values of \(x\) that make the denominator equal to zero, we need to solve the following equation: \(x^2 + 11x + 24 = 0\). To solve this quadratic equation, we can either use factoring or the quadratic formula. Let's try factoring first: \((x^2 + 11x + 24) = (x+3)(x+8)\) Now, set each factor equal to zero: \(x + 3 = 0 => x = -3\) \(x + 8 = 0 => x = -8\) So, there are two values of \(x\) that make the denominator equal to zero: \(x = -3\) and \(x = -8\).

Step 3: Determine the domain of the function

Since the denominator cannot be equal to zero, we have to exclude the values of \(x\) that we found in step 2 from the domain of the function. Therefore, the domain of the function \(k(x)\) is all real numbers except for \(x = -3\) and \(x = -8\). We can write this in interval notation as: Domain: \((-\infty, -8) \cup (-8, -3) \cup (-3, \infty)\)

Key Concepts

Quadratic Equation

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). This type of equation is fundamental in algebra and describes a parabolic relationship within a function.

• It features the square of the unknown variable \(x\), hence "quadratic" from the Latin word "quadratus," meaning square.
• The general solutions for a quadratic equation can be found using the quadratic formula, factoring, or completing the square.

In our exercise, the quadratic equation given in the denominator of the function is \(x^2 + 11x + 24\). Solving this equation is crucial as it determines the x-values where the function is undefined.

Factoring

Factoring involves breaking down an expression into simpler "factors" or components that, when multiplied together, give back the original expression. It is a useful technique for solving quadratic equations.

• This involves rewriting the quadratic equation in the form \((x+p)(x+q)\), where \(p\) and \(q\) are numbers such that \(p+q = b\) and \(pq = c\) in the standard form \(ax^2 + bx + c\).
• When we factor \(x^2 + 11x + 24\), it becomes \((x+3)(x+8)\).

By setting each factor equal to zero, we solve for the roots of the equation, which are \(x = -3\) and \(x = -8\). These roots tell us where the denominator equals zero and are crucial in determining the function's domain.

Interval Notation

Interval notation is a way of writing subsets of the real number line. It is very handy when expressing the domain or range of a function.

• Closed intervals [a, b] include the endpoints \(a\) and \(b\), meaning every number between \(a\) and \(b\) is included.
• Open intervals \((a, b)\) do not include the endpoints a and b, thus including only the numbers strictly between \(a\) and \(b\).
• Union of intervals, denoted by \(\cup\), is used to combine disjoint intervals necessary to express a complete domain.

For example, for the function \(k(x)\), the domain is all real numbers except where the denominator is zero. The domain can therefore be written in interval notation as \((-\infty, -8)\cup(-8,-3)\cup(-3,\infty)\). This clearly shows that the function includes all real numbers except \(x = -8\) and \(x = -3\).

Real Numbers

Real numbers include all the numbers on the number line. This set encompasses both rational numbers (such as 2, 1/2, 0.75) and irrational numbers (such as \(\pi\), \(\sqrt{2}\)). Real numbers are essential when discussing the domain of functions because they describe all possible values that \(x\) can take under typical circumstances.

• The domain of a typical function, unless specified, is usually the real numbers, excluding any restrictions like division by zero or the square roots of negative numbers.

In the context of our function \(k(x) = \frac{1}{x^2 + 11x + 24}\), we state that the domain is "all real numbers" except for the values \(x = -3\) and \(x = -8\) because these values would make the function's denominator zero, making the function undefined.