Problem 45 - Precalculus with Limits

Question

In Exercises 37-52, evaluate the function at each specified value of the independent variable and simplify. \(q(x) = 1/(x^2-9)\) (a) \(q(0)\) (b) \(q(3)\) (c) \(q(y+3)\)

Step-by-Step Solution

Answer

For the given values, the function evaluates to: \(q(0) = -1/9\), \(q(3)\) is undefined, and \(q(y+3) = 1/(y^2+6y)\).

Step 1: Substituting \(x = 0\)

First, substitute \(x = 0\) into the given function, which results into function \(q(x) = 1/(0^2-9) = 1/(-9)\). So, \(q(0) = -1/9\).

Step 2: Substituting \(x = 3\)

Next, substitute \(x = 3\) into the function, which results into function \(q(x) = 1/(3^2-9) = 1/(9-9) = 1/0\). The function is undefined at \(x = 3\) because division by zero is undefined.

Step 3: Substituting \(x = y+3\)

Lastly, substitute \(x = y+3\) into the function, which results into function \(q(x) = 1/((y+3)^2-9) = 1/(y^2+6y+9-9) = 1/(y^2+6y)\). So, \(q(y+3) = 1/(y^2+6y)\).

Key Concepts

Independent Variable

In a mathematical function, the independent variable is the input value that determines the outcome or output of the function. For instance, in the function \( q(x) = \frac{1}{x^2-9} \), \( x \) is the independent variable. It is the value we can change to see how it affects the function's result.
When evaluating a function, we substitute different values into the place of the independent variable. This allows us to understand how the function behaves and what outputs it produces. For example, in the given exercise, substituting \( x = 0 \), \( x = 3 \), and \( x = y+3 \) helps us discover the behavior of the function \( q \) for these particular inputs.

Undefined Function

A function becomes undefined when it encounters a situation that does not have a meaningful output. In our context, this often occurs during division when the denominator is zero.
For the function \( q(x) = \frac{1}{x^2-9} \), it becomes undefined at \( x = 3 \). This results from substituting \( x = 3 \) into the formula, leading to a denominator of zero, as shown by \( (3^2-9) = 0 \).
When a function is undefined, it highlights a limitation or a point where the mathematical expression does not provide a valid result. Understanding where functions become undefined is important because it affects graph behavior, especially when considering real-world applications or modeling scenarios.

Division by Zero

Division by zero is a concept most students encounter early in mathematics. It is crucial to understand because it renders a function undefined. Simply put, dividing a number by zero does not produce a meaningful result.
When evaluating functions like \( q(x) = \frac{1}{x^2-9} \), checking for zero in the denominator ensures we avoid undefined values. For instance, the function becomes \( \frac{1}{0} \) at \( x = 3 \), which is mathematically undefined.
Avoiding division by zero is essential in algebra and calculus because calculations and theorems often assume valid, defined operations. This knowledge helps prevent errors in mathematical reasoning and solution finding.

Algebraic Simplification

Algebraic simplification involves reducing expressions to their simplest form, making them easier to understand and work with. In the exercise, simplifying \( q(x) = \frac{1}{(y+3)^2-9} \) allows us to see its simplified form \( q(y+3) = \frac{1}{y^2+6y} \).
During simplification, like terms are combined, and expressions are reduced. This process involves applying algebraic rules and operations, such as expanding squares or factoring out common factors.
Simplification helps not only in making expressions less complex but also in identifying key features, such as points where the function is defined or undefined, as well as simplifying calculations required to find solutions or make predictions.