Problem 55 - Contemporary Precalculus

Question

Determine the domain of the function according to the usual convention. $$g(u)=\frac{u^{2}+1}{u^{2}-u-6}$$

Step-by-Step Solution

Answer

Answer: The domain of the function $$g(u)=\frac{u^{2}+1}{u^{2}-u-6}$$ is $$(-\infty, -2) \cup (-2, 3) \cup (3, \infty)$$.

Step 1: Rewrite the function

We're given the function $$g(u)=\frac{u^{2}+1}{u^{2}-u-6}$$. Our goal is to find the domain of this function.

Step 2: Find values of u that make the denominator zero

We need to determine all values of $$u$$ which make the denominator equal to zero. Therefore, we need to solve the equation $$u^{2}-u-6=0$$. This equation can be factored into $$(u-3)(u+2) = 0$$. So, the values of $$u$$ that make the denominator equal to zero are $$u=3$$ and $$u=-2$$.

Step 3: Exclude the values from the domain

A function is undefined when its denominator is equal to zero. Thus, we exclude the values $$u=3$$ and $$u=-2$$ from the domain. The domain of the function $$g(u)$$ is all real numbers except $$u=3$$ and $$u=-2$$.

Step 4: Write the domain using interval notation

In interval notation, we can write the domain of the function as $$(-\infty, -2) \cup (-2, 3) \cup (3, \infty)$$. This notation represents all real numbers except $$u=3$$ and $$u=-2$$. So the domain of the function $$g(u)=\frac{u^{2}+1}{u^{2}-u-6}$$ is $$(-\infty, -2) \cup (-2, 3) \cup (3, \infty)$$.

Key Concepts

Real Numbers

Real numbers are a crucial building block of mathematics. They include all numbers that you can think of. Whether they are whole numbers like 3, decimals like 2.5, or even irrational numbers like \( \pi \), they all fall under the category of real numbers.

Real numbers are continuous. This means there are no gaps or missing parts on the number line. Between any two real numbers, there is an infinite set of numbers. This characteristic makes them so essential for various fields, including calculus, physics, and engineering.

When determining the domain of functions, we usually look for the set of real numbers that do not break the rules of mathematics, like dividing by zero. Every value that doesn't cause such a problem is part of the domain of that function. In the case of our given function \( g(u)=\frac{u^{2}+1}{u^{2}-u-6} \), the real numbers must exclude values that make the denominator zero to maintain the function's validity.

Rational Functions

Rational functions are a type of mathematical expression that involves ratios of polynomials. They take the form of \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) \) is not zero.

The domain of a rational function consists of all real numbers except for those that make the denominator zero. This is because dividing by zero is undefined in mathematics. If you have a rational function like \( g(u)=\frac{u^{2}+1}{u^{2}-u-6} \), you need to calculate which values for \( u \) make the denominator \( u^2-u-6 \) equal to zero.

Solving \( u^2-u-6=0 \) by factoring results in \( (u-3)(u+2)=0 \), giving \( u=3 \) and \( u=-2 \). Excluding these values ensures the function remains well-defined and free from undefined operations.

Interval Notation

Interval notation is a mathematical shorthand used to express ranges of values. It is especially useful for describing the domain of a function in a clear and concise way.

In interval notation, the domain of all real numbers except specific values is written using parentheses and brackets. Parentheses \((\) represent values that are not included in the interval, while brackets \([\) mean the values are included. For example, the function \( g(u)=\frac{u^{2}+1}{u^{2}-u-6} \) has excluded values \( u=3 \) and \( u=-2 \).

To express this in interval notation, you write \((-\infty, -2) \cup (-2, 3) \cup (3, \infty)\). This notation shows clearly that the domain consists of all real numbers except where \( u \) equals \( -2 \) and \( 3 \), using the union symbol \( \cup \) to combine non-overlapping intervals.