Rental Car Cost: A Piecewise Function Guide

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Rental Car Cost: A Piecewise Function Guide

Hey guys, let's dive into a super common problem that pops up in math class: piecewise functions, especially when we're talking about something relatable like rental car costs. You know how sometimes you rent a car and the price changes depending on how much you drive? That's exactly what a piecewise function is all about! It's like a set of rules that apply to different 'pieces' or ranges of your driving. So, if you're looking to understand how these functions work, especially with real-world examples, you've come to the right spot. We're going to break down how a rental car company might structure its pricing and represent that with a mathematical function that has different rules for different mileages. This isn't just about crunching numbers; it's about understanding how math helps us model and predict costs in everyday situations. We'll explore the base fee, the per-mile charges, and how those charges can shift, all while keeping it easy to understand. Get ready to see how a simple rental car scenario can turn into a cool math problem that's actually pretty useful!

Understanding the Basics of Rental Car Pricing

Alright, let's get down to business with this rental car scenario, guys. The core idea here is that the cost isn't a simple, flat rate. It's dynamic, meaning it changes based on how much you use the car. Our specific rental car company has a pretty standard pricing model that many of you might encounter. They start with a base fee, which is like your entry ticket to renting the car. In this case, that base fee is a solid $40. Think of this as the price you pay just for having the car available to you, regardless of whether you drive it a mile or a hundred miles. But that's not all, right? You also pay for the miles you actually drive. For the first 100 miles, the company charges an additional $0.25 per mile. This is the initial rate, and it's pretty straightforward. If you drive 10 miles, you pay $40 + (10 * $0.25). If you drive 50 miles, you pay $40 + (50 * $0.25). This rate is designed to cover their costs for wear and tear, fuel adjustments, and general operational expenses for the initial usage. It's a common practice to have a higher rate for the initial portion of usage because it accounts for depreciation and the initial setup costs associated with renting out a vehicle.

Now, here's where it gets interesting and where our piecewise function really starts to take shape. The company recognizes that if you're driving a significant distance, they might want to offer a slightly more attractive rate to encourage longer rentals or to remain competitive. So, for any miles driven over the initial 100 miles, the rate changes. They keep the same base fee of $40 – you don't pay a new base fee every time you hit a mileage threshold, which is important to note. However, the per-mile charge drops to a reduced price of $0.18 per mile for those extra miles. This means if you drive 150 miles, the first 100 miles are charged at $0.25/mile, and the remaining 50 miles are charged at $0.18/mile. The total cost will be the base fee plus the cost of the first 100 miles at the higher rate, plus the cost of the miles exceeding 100 at the lower rate. This tiered pricing strategy is super common in many service industries, not just car rentals. Think about utility bills or phone plans; they often have different rates for different usage tiers. It’s a smart way for companies to structure pricing to capture different customer segments and usage patterns. Understanding this distinction is key to setting up the correct piecewise function.

So, to recap the pricing structure: You pay $40 upfront. Then, for every mile up to 100, you add $0.25. If you go beyond 100 miles, you still pay the $40 base fee, and you still pay for the first 100 miles at $0.25 each, but then you pay only $0.18 for each additional mile after that. This transition point at 100 miles is the crucial factor that defines the different 'pieces' of our function. It’s this shift in the per-mile rate that necessitates a piecewise approach to accurately model the total cost. Without this change in rate, we could simply use a linear equation. But because the rate itself changes, we need a function that can handle these different rates across different mileage intervals. This is precisely where the power and utility of piecewise functions become apparent. They allow us to represent real-world scenarios with changing conditions in a precise and mathematical way, making complex pricing structures understandable and calculable.

Defining the Piecewise Function

Now, guys, let's translate this rental car pricing into the language of mathematics: a piecewise function. A piecewise function is basically a function defined by multiple sub-functions, each applying to a certain interval of the independent variable. In our case, the independent variable is the number of miles driven, let's call it 'mm'. The dependent variable is the total cost, let's call it 'C(m)C(m)'. We need to define our function C(m)C(m) based on the rules we just discussed. Remember, there's a critical point at 100 miles where the pricing structure changes.

So, we'll have two main 'pieces' or rules for our function. The first rule applies when the number of miles driven, 'mm', is less than or equal to 100. For this first 'piece', the cost is straightforward: the base fee plus the per-mile charge at the higher rate. The base fee is $40, and the charge per mile is $0.25. So, for mgtr100m gtr 100, the cost is represented by the equation:

C(m)=40+0.25mC(m) = 40 + 0.25m

This equation holds true for all mileages from 0 miles up to and including 100 miles. If you drive 50 miles, you plug in m=50m=50: C(50)=40+0.25(50)=40+12.50=52.50C(50) = 40 + 0.25(50) = 40 + 12.50 = 52.50. If you drive exactly 100 miles, you plug in m=100m=100: C(100)=40+0.25(100)=40+25=65C(100) = 40 + 0.25(100) = 40 + 25 = 65. This part is pretty simple and represents the initial cost structure.

Now, for the second 'piece' of our function, this applies when the number of miles driven, 'mm', is greater than 100. This is where the reduced rate kicks in for the additional miles. For these situations, the total cost includes the base fee ($40), the cost of the first 100 miles at the higher rate ($0.25 per mile), and then the cost of the miles exceeding 100 at the lower rate (0.18permile).Tocalculatethemilesexceeding100,wetakethetotalmiles′0.18 per mile). To calculate the miles exceeding 100, we take the total miles 'm

and subtract 100. So, the number of miles charged at the lower rate is (m−100)(m - 100).

Therefore, for m>100m > 100, the cost is calculated as follows:

C(m)=extBaseFee+(extCostoffirst100miles)+(extCostofmilesover100)C(m) = ext{Base Fee} + ( ext{Cost of first 100 miles}) + ( ext{Cost of miles over 100})

C(m)=40+(100imes0.25)+((m−100)imes0.18)C(m) = 40 + (100 imes 0.25) + ((m - 100) imes 0.18)

Let's simplify this second equation:

C(m)=40+25+0.18m−18C(m) = 40 + 25 + 0.18m - 18

C(m)=47+0.18mC(m) = 47 + 0.18m

Wait a minute, guys, let me double-check that simplification. Ah, I made a small mistake in the algebraic simplification. Let's re-calculate carefully.

The cost for miles over 100 is (m−100)imes0.18(m-100) imes 0.18. This means we add the $40 base fee, plus the cost of the first 100 miles (which is 100imes0.25=25100 imes 0.25 = 25), plus the cost of the miles over 100.

So the full expression for m>100m > 100 is:

C(m)=40+(100imes0.25)+(m−100)imes0.18C(m) = 40 + (100 imes 0.25) + (m - 100) imes 0.18

C(m)=40+25+0.18m−18C(m) = 40 + 25 + 0.18m - 18

C(m)=65+0.18m−18C(m) = 65 + 0.18m - 18

C(m)=47+0.18mC(m) = 47 + 0.18m

My apologies, guys, I made the same mistake twice! Let's think this through logically without just simplifying blindly. The $40 base fee is always there. For miles over 100, the additional miles are charged at $0.18. So, if you drive 150 miles:

Cost = Base Fee + Cost of first 100 miles + Cost of next 50 miles Cost = 40+(100imes0.25)+(50imes0.18)40 + (100 imes 0.25) + (50 imes 0.18) Cost = 40+25+940 + 25 + 9 Cost = 7474

Now let's use the formula for m>100m > 100: C(m)=40+(100imes0.25)+(m−100)imes0.18C(m) = 40 + (100 imes 0.25) + (m - 100) imes 0.18 Let's try plugging in m=150m=150: C(150)=40+(100imes0.25)+(150−100)imes0.18C(150) = 40 + (100 imes 0.25) + (150 - 100) imes 0.18 C(150)=40+25+(50)imes0.18C(150) = 40 + 25 + (50) imes 0.18 C(150)=40+25+9C(150) = 40 + 25 + 9 C(150)=74C(150) = 74

This matches our manual calculation! So the correct expression for the second piece is indeed:

C(m)=40+(100imes0.25)+(m−100)imes0.18C(m) = 40 + (100 imes 0.25) + (m - 100) imes 0.18

This formula correctly accounts for the base fee, the cost of the first 100 miles, and the reduced rate for miles exceeding 100. It's crucial to understand that the (m−100)(m-100) part correctly isolates only those miles that are charged at the lower rate. The term (100imes0.25)(100 imes 0.25) represents the fixed cost for the first 100 miles, which is always $25, regardless of how many miles over 100 you drive. So, the total cost for m>100m > 100 is the base fee plus the cost of the first 100 miles plus the cost of the additional miles.

Putting it all together, our piecewise function for the rental car cost C(m)C(m) is:

C(m)={40+0.25mif 0≤m≤10040+(100×0.25)+(m−100)×0.18if m>100 C(m) = \begin{cases} 40 + 0.25m & \text{if } 0 \le m \le 100 \\ 40 + (100 \times 0.25) + (m - 100) \times 0.18 & \text{if } m > 100 \end{cases}

Let's simplify the second line slightly for clarity, keeping in mind the breakdown: the base fee is $40. The first 100 miles cost $100 imes 0.25 = $25. The miles over 100 cost (m−100)imes0.18(m-100) imes 0.18. So, the second part could also be written as:

C(m)=40+25+0.18(m−100)C(m) = 40 + 25 + 0.18(m - 100)

C(m)=65+0.18m−18C(m) = 65 + 0.18m - 18

C(m)=47+0.18mC(m) = 47 + 0.18m

Ah, it seems I was getting confused with the algebraic simplification again! The key is that the entire expression 40+(100imes0.25)+(m−100)imes0.1840 + (100 imes 0.25) + (m - 100) imes 0.18 correctly represents the cost for m>100m > 100. The $40 is the base, the $25 is for the first 100 miles, and 0.18(m−100)0.18(m-100) is for the miles beyond 100. There's no need to simplify it further into 47+0.18m47 + 0.18m if it leads to confusion or errors. Sticking to the form that clearly shows the components of the cost is often best.

So, the piecewise function definition, using the original breakdown for clarity, is:

C(m)={40+0.25mif 0≤m≤10040+(100×0.25)+(m−100)×0.18if m>100 C(m) = \begin{cases} 40 + 0.25m & \text{if } 0 \le m \le 100 \\ 40 + (100 \times 0.25) + (m - 100) \times 0.18 & \text{if } m > 100 \end{cases}

This function accurately captures the two different pricing tiers based on the number of miles driven. It’s a perfect example of how piecewise functions are used to model situations with changing rates or conditions.

Calculating Costs with the Piecewise Function

Now that we've got our piecewise function defined, guys, let's put it to the test! This is where the math becomes super practical. We can use our function C(m)C(m) to calculate the exact cost for any number of miles driven. Remember our function:

C(m)={40+0.25mif 0≤m≤10040+(100×0.25)+(m−100)×0.18if m>100 C(m) = \begin{cases} 40 + 0.25m & \text{if } 0 \le m \le 100 \\ 40 + (100 \times 0.25) + (m - 100) \times 0.18 & \text{if } m > 100 \end{cases}

Let's try a few scenarios to see how it works. Suppose you rent a car and drive exactly 50 miles. Since 50 is between 0 and 100 (inclusive), we use the first part of the function:

C(50)=40+0.25imes50C(50) = 40 + 0.25 imes 50 C(50)=40+12.50C(50) = 40 + 12.50 C(50)=$52.50C(50) = \$52.50

So, for 50 miles, the total cost is $52.50. Pretty straightforward, right?

What if you drive exactly 100 miles? Again, 100 falls into the first condition (0gtrmgtr1000 gtr m gtr 100):

C(100)=40+0.25imes100C(100) = 40 + 0.25 imes 100 C(100)=40+25C(100) = 40 + 25 C(100)=$65.00C(100) = \$65.00

This means driving up to 100 miles costs $65.00.

Now, let's say you go a bit further and drive 120 miles. Since 120 is greater than 100, we need to use the second part of our piecewise function:

C(120)=40+(100imes0.25)+(120−100)imes0.18C(120) = 40 + (100 imes 0.25) + (120 - 100) imes 0.18

First, let's break down the components:

Now, add them all up: C(120)=40+25+3.60C(120) = 40 + 25 + 3.60 C(120)=$68.60C(120) = \$68.60

So, for 120 miles, the total cost is $68.60. Notice how this is only slightly more expensive than driving 100 miles ($65.00) because the rate for the extra miles is lower. This illustrates the benefit of the tiered pricing.

Let's try one more example, maybe a longer trip of 250 miles. Again, since 250 > 100, we use the second part of the function:

C(250)=40+(100imes0.25)+(250−100)imes0.18C(250) = 40 + (100 imes 0.25) + (250 - 100) imes 0.18

Let's calculate:

Now, add them all up: C(250)=40+25+27.00C(250) = 40 + 25 + 27.00 C(250)=$92.00C(250) = \$92.00

For a 250-mile trip, the total cost is $92.00. This shows how the function allows us to calculate costs for any mileage, demonstrating the practical application of piecewise functions in everyday financial scenarios. By understanding these calculations, you can better predict your expenses and make informed decisions when renting vehicles.

Visualizing the Piecewise Function: The Graph

Guys, to really get a solid grasp on piecewise functions, visualizing them with a graph is super helpful. It makes it clear how the cost changes at different mileage points. Remember our function for the rental car cost, C(m)C(m):

C(m)={40+0.25mif 0≤m≤10040+(100×0.25)+(m−100)×0.18if m>100 C(m) = \begin{cases} 40 + 0.25m & \text{if } 0 \le m \le 100 \\ 40 + (100 \times 0.25) + (m - 100) \times 0.18 & \text{if } m > 100 \end{cases}

When we graph this, the horizontal axis (the x-axis) will represent the number of miles driven, 'mm', and the vertical axis (the y-axis) will represent the total cost, 'C(m)C(m)'.

Let's look at the first piece: C(m)=40+0.25mC(m) = 40 + 0.25m for 0gtrmgtr1000 gtr m gtr 100. This is a linear equation, meaning its graph will be a straight line. The y-intercept (where m=0m=0) is $40. The slope is $0.25, meaning for every mile driven, the cost increases by $0.25. At the point m=100m=100, we calculated the cost to be C(100)=40+0.25(100)=40+25=65C(100) = 40 + 0.25(100) = 40 + 25 = 65. So, the first part of our graph will be a line segment starting at the point (0, 40) and ending at the point (100, 65). Both of these points are included in this piece of the function.

Now for the second piece: C(m)=40+(100imes0.25)+(m−100)imes0.18C(m) = 40 + (100 imes 0.25) + (m - 100) imes 0.18 for m>100m > 100. This is also a linear equation, but it starts after 100 miles. Let's evaluate the cost at the boundary point where this new rule starts to apply, which is just after 100 miles. Technically, this rule applies for m>100m > 100. However, to see where the line begins, we can think about what the cost would be if this rate applied from mile 100 onwards. Using the simplified form C(m)=47+0.18mC(m) = 47 + 0.18m (even though we noted potential confusion, it helps see the slope), or better yet, using the original definition with m=100m=100:

C(100)=40+(100imes0.25)+(100−100)imes0.18=40+25+0=65C(100) = 40 + (100 imes 0.25) + (100 - 100) imes 0.18 = 40 + 25 + 0 = 65.

This means that at exactly 100 miles, the cost is $65.00. The second part of the function starts from this point but with a different slope. The slope for m>100m > 100 is $0.18. Since $0.18 is less than $0.25, the line segment for m>100m > 100 will be less steep than the first segment. It will start right where the first segment ended, at (100, 65), but it will continue upwards with a gentler slope. So, the graph consists of two connected line segments. The first segment goes from (0, 40) to (100, 65), and the second segment starts at (100, 65) and continues upwards with a slope of 0.18 for all miles greater than 100.

When you visualize this, you'll see a graph that looks like a bent line. It starts at $40, goes up linearly until it reaches 100 miles at a cost of $65, and then it continues to go up, but at a slower rate. This visual representation clearly shows the piecewise nature of the cost function. The