Problem 13 - Precalculus Essentials

Question

Find the average rate of change of the function from \(x_{1}\) to \(x_{2}\). $$f(x)=3 x \text { from } x_{1}=0 \text { to } x_{2}=5$$

Step-by-Step Solution

Answer

The average rate of change of the function \(f(x)=3x\) from \(x_{1}=0\) to \(x_{2}=5\) is 3.

Step 1: Identify the Function and Interval

The function given is \(f(x)=3x\). The interval for which the average rate of change is to be found ranges from \(x_{1}=0\) to \(x_{2}=5\).

Step 2: Calculate the Change in 'y'

The 'y' values for the function at \(x_{1}\) and \(x_{2}\) can be found by plugging these 'x' values into the function \(f(x)\). These are: \(f(x_{1}) = f(0) = 3*0 =0\) and \(f(x_{2})=f(5)=3*5=15\). So, the change in 'y' or \(\Delta y\) is \(y_{2}-y_{1} = f(x_{2})-f(x_{1}) = 15-0 = 15\).

Step 3: Calculate the Change in 'x'

The change in 'x' or \(\Delta x\) is \(x_{2}-x_{1} = 5-0 = 5\).

Step 4: Calculate the Average Rate of Change

The average rate of change is defined as \(\Delta y / \Delta x\). Hence, the average rate of change for this interval is \(15/5 = 3\).

Key Concepts

Rate of Change

The concept of the rate of change is foundational in the study of functions in precalculus. It measures how much a quantity, such as 'y' in a function 'y=f(x)', changes with respect to the change in another quantity, such as 'x'. The average rate of change is particularly useful in understanding the behavior of a function over a specific interval.

For example, if we consider the function in the textbook exercise, which is a linear function, the average rate of change is consistent throughout its domain. This means that for every unit increase in 'x', 'y' increases by a constant amount. In contrast, for non-linear functions, the rate of change can vary at different points on the curve.

Function Evaluation

Evaluating a function is a primary skill in precalculus that involves finding the value of the function 'f(x)' at a particular 'x'. This is essential for calculating the average rate of change. When we say 'f(x_{1})' or 'f(x_{2})', we are specifically finding the 'y' values (output of the function) when 'x' is 'x_{1}' and 'x_{2}', respectively.

• Finding the output for a point on the function helps us determine the change in 'y' (\( \Delta y \)).
• In our textbook solution, function evaluation at 'x_{1}=0' and 'x_{2}=5' gives us the two 'y' values needed to find \( \Delta y \).
• Evaluating functions also helps with understanding the graph of the function, predicting trends, and solving real-world problems.

Precalculus

Precalculus is an advanced form of secondary education mathematics, focusing on the concepts and skills that prepare students for calculus. It includes studies on various types of functions, their properties, and graphs, as well as tools for analyzing these functions such as the rate of change.

Precalculus also introduces the idea of limits, which is crucial for understanding instantaneous rates of change in calculus. In layman's terms, you can consider precalculus as the building block that paves the way for the advanced studies of change and motion, commonly called calculus.

Change in Variable

In the context of functions, 'change in variable' typically refers to the difference in the value of the independent variable 'x' across an interval (\( \Delta x \)) and the corresponding difference in the value of the dependent variable 'y' (\( \Delta y \)).

• The notation \( \Delta \) (Greek letter Delta) signifies 'change in.' So when we write \( \Delta x \), we're talking about the change in 'x', and similarly, \( \Delta y \) for the change in 'y'.
• In the textbook exercise, we calculated the change in 'x' from 'x_{1}=0' to 'x_{2}=5', which was 5 units. Identifying changes in variables is critical for various applications in science, economics, and mathematics itself, especially for analyzing trends and rates.