Problem 11 - College Algebra with Corequisite Support

Question

For the following exercises, find the average rate of change of each function on the interval specified for real numbers \(b\) or \(h\) in simplest form. \(a(t)=\frac{1}{t+4}\) on \([9,9+h]\)

Step-by-Step Solution

Answer

The average rate of change is \(-\frac{1}{13(13+h)}\).

Step 1: Understand the Average Rate of Change Formula

The average rate of change of a function between two points \( a \) and \( b \) is calculated as \( \frac{f(b) - f(a)}{b - a} \). Here, the function is \( a(t) = \frac{1}{t+4} \), and we calculate the change from \( t = 9 \) to \( t = 9 + h \).

Step 2: Calculate Function Values at the Endpoints

Evaluate \( a(t) \) at \( t = 9 \) to get \( a(9) = \frac{1}{9 + 4} = \frac{1}{13} \). Now, evaluate \( a(t) \) at \( t = 9 + h \), so \( a(9 + h) = \frac{1}{9 + h + 4} = \frac{1}{13 + h} \).

Step 3: Apply Values to the Formula

Substitute the evaluated values into the average rate of change formula: \[ \frac{a(9 + h) - a(9)}{(9 + h) - 9} = \frac{\frac{1}{13 + h} - \frac{1}{13}}{h} \].

Step 4: Simplify the Expression

Find a common denominator for the expression in the numerator: \( \frac{1}{13 + h} - \frac{1}{13} = \frac{13 - (13 + h)}{(13 + h) \cdot 13} = \frac{-h}{(13 + h) \cdot 13} \). Thus, the expression becomes \( \frac{-h}{h \cdot (13 + h) \cdot 13} \).

Step 5: Simplify the Result

Cancel \( h \) in the numerator and the denominator: \( \frac{-h}{h \cdot (13 + h) \cdot 13} = \frac{-1}{(13 + h) \cdot 13} \). Therefore, the average rate of change is \(-\frac{1}{13(13+h)}\).

Key Concepts

Understanding Functions

In mathematics, a function is a relation that uniquely associates members of one set with members of another set. More specifically, it maps each element from a set of inputs (called the domain) to a set of outputs (called the codomain or range). Functions are a foundational concept in algebra and calculus because they describe how one quantity changes with another. In our exercise, we are dealing with a specific function defined by the equation \(a(t) = \frac{1}{t+4}\).
This function takes an input \(t\) and returns an output which is the reciprocal of \(t + 4\). This is an example of a rational function, as it involves a fraction with polynomials in the numerator and the denominator. Understanding how functions behave is crucial for calculating changes, such as finding the average rate of change.

The Role of Algebra in Calculating Changes

Algebra provides the tools needed to manipulate and solve equations and expressions. It enables us to rearrange, simplify and solve complex math problems effectively. In problems like finding the average rate of change, algebra becomes essential.
We start by plugging our function values into the average rate of change formula, \(\frac{f(b) - f(a)}{b - a}\). Here, algebra helps us simplify the expression by finding a common denominator for the fraction \(\frac{1}{13 + h} - \frac{1}{13}\).
Algebraic simplification involves canceling terms and reducing complex expressions into simpler forms. In this case, by finding the common denominator and then dividing through by \(h\), algebra lets us arrive at \(-\frac{1}{13(13+h)}\) as the simplified form for the average rate of change.

Real Numbers and Their Application

Real numbers form the backbone of our mathematical universe. They include all the numbers on the number line, encompassing both rational and irrational numbers. When dealing with functions, it is crucial to understand that each variable like \(t\), \(h\), and their outcomes are real numbers.
In the exercise, when calculating the average rate of change for \(a(t) = \frac{1}{t+4}\), both \(9\) and \(9 + h\) lie within the realm of real numbers. It ensures that the inputs and outputs of our function remain consistent and graphable.
Understanding real numbers help us appreciate the continuity and behavior of functions over different intervals. This reinforces the importance of domains and valid input values for functions when evaluating mathematical expressions.