The Monthly Cost Of Driving A Car Depends On The Number Of M
The Monthly Cost Of Driving A Car Depends On the Number of Miles Dr
The assignment involves analyzing a real-world scenario where the monthly cost of driving a car depends on the number of miles driven, using linear models to make predictions, interpret the model, and analyze the function's behavior through graphing and limits. The core tasks include forming a linear function based on given data points, predicting costs for future miles, graphing the function, interpreting its slope and intercept, and understanding the suitability of linear models in this context. Additionally, the assignment explores evaluating expressions from a table, analyzing given functions’ limits, and investigating a piecewise-defined function through domain, intercepts, graphing, range, and continuity. Lastly, it involves applying the concept of limits and understanding how to determine the proximity of values using epsilon-intervals.
Paper For Above instruction
Understanding the relationship between the monthly cost of driving a car and the distance driven entails applying fundamental concepts of linear functions, modeling, and calculus. Each component of this assignment offers an opportunity to explore different mathematical principles via practical and theoretical lenses.
Part A: Formulating the Linear Model
Given the data points: in May, Juan paid \$380 for 480 miles, and in June, \$460 for 800 miles, we can model this relationship with a linear function of the form:
f(m) = a * m + b
where 'f(m)' represents the monthly cost, and 'm' is the miles driven.
To determine the slope (a), we calculate the rate of change of cost with respect to miles:
a = (cost2 - cost1) / (miles2 - miles1) = (460 - 380) / (800 - 480) = 80 / 320 = 0.25
Thus, the cost increases by \$0.25 for each additional mile traveled. Next, to find the y-intercept (b), substitute m=480 and f(480)=380 into the linear equation:
380 = 0.25 * 480 + b => 380 = 120 + b => b = 260
Therefore, the linear function describing the monthly cost as a function of miles driven is:
f(m) = 0.25m + 260
Part B: Cost Prediction for 1500 Miles
Using the derived model to predict the cost for 1500 miles:
f(1500) = 0.25 * 1500 + 260 = 375 + 260 = \$635
This prediction suggests that driving 1500 miles in a month would cost approximately \$635, assuming the linear relationship holds.
Part C: Graphing the Linear Function and Interpreting the Slope
The graph of the function f(m) = 0.25m + 260 is a straight line with a slope of 0.25, which indicates that for each additional mile driven, the monthly cost increases by \$0.25. The slope represents the rate of change of cost with respect to miles — essentially, the variable cost per mile driven.
Plotting the graph involves marking the y-intercept at (0, 260), and drawing a straight line with slope 0.25 passing through this point. The line extends through the points like (480, 380) and (800, 460), confirming the linear relationship.
Part D: Interpretation of the Intercept
The intercept, \$260, represents the fixed monthly cost incurred regardless of miles driven. It could include expenses such as insurance, registration, or other fees that are not dependent on usage.
Part E: Suitability of a Linear Model
A linear function offers a suitable model here because the relationship between miles driven and cost appears to be proportional, with a fixed per-mile cost and an initial fixed fee. Given the data points, the cost increases consistently with miles, and the assumption of linearity simplifies the modeling process. In real-world scenarios, fuel costs, maintenance, and other factors often contribute to near-linear increases over moderate ranges, making linear modeling an appropriate approximation.
Additional Analyses: Evaluating Expressions, Limits, and Piecewise Functions
In the subsequent parts of the assignment, evaluating expressions from tables involves substituting known values into given functions and simplifying, which solidifies understanding of function evaluation. Analyzing the limits of functions—for example, understanding the behavior as input approaches a specific value—requires applying limit laws or constructing tables to observe function behavior near certain points. When dealing with piecewise functions, identifying the domain, intercepts, and points of discontinuity helps understand their structure and graph.
Particularly, the evaluation of limits, for example, using limit laws, grants insight into the behavior of functions at boundary points or infinity. Determining whether limits exist or do not exist informs about discontinuities or asymptotic behavior. For piecewise functions, the graphing process illustrates how functions might change across different intervals, with potential points of discontinuity or jumps, which are crucial in calculus for understanding continuity and differentiability.
Finally, epsilon-delta definitions of limits provide a rigorous foundation for understanding how close a function value can be to a particular point, given any degree of accuracy. This conceptual understanding underpins much of calculus and analysis.
Conclusion
This assignment combines foundational algebraic modeling with analytical tools from calculus, emphasizing the importance of understanding relationships between variables, limits, and functions’ behavior. The approach illustrates how simple linear models can effectively describe real-world phenomena, while also highlighting the importance of limits and function analysis for a deeper understanding of mathematical behavior across different contexts.
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