Real-World Applications of Linear Functions

Linear functions are widely applicable in everyday life, from managing finances to solving engineering problems. The following examples demonstrate how linear functions and their inverses can be used to model and solve practical problems.

1. Managing Expenses: Budgeting

Sarah's monthly expenses include a fixed rent of $600 and weekly expenses of $20 for groceries and $15 for transportation.

Problem:

1. Write a function \( f(x) \) to represent Sarah's total monthly expenses, where \( x \) is the number of weeks.
2. Find the inverse function \( f^{-1}(x) \) to determine how many weeks it will take for her total expenses to reach $1,000.
3. If her total expenses increase to $1,200, calculate the number of weeks required.
4. Discuss the domain and range of the function.

Solution:

1. Function:

2. Inverse Function:

3. Substitute \( f(x) = 1,200 \):

4. Domain and Range:

- Domain: \( x \geq 0 \) (weeks).
- Range: \( f(x) \geq 600 \) (dollars).

2. Loan Repayment: A Financial Commitment

Emma borrowed $10,000 and repays $500 monthly, with an additional $50 monthly interest.

Problem:

1. Write a function \( f(x) \) for the remaining loan balance after \( x \) months.
2. Find the inverse \( f^{-1}(x) \) to calculate the time required to reduce her balance to $3,000.
3. Calculate the time required to repay the loan in full.
4. Discuss the domain and range of the function.

Solution:

1. Function:

2. Inverse Function:

3. Substitute \( f(x) = 3,000 \):

4. Domain and Range:

- Domain: \( x \in [0, 18.18] \) (months).
- Range: \( f(x) \in [0, 10,000] \) (dollars).

3. Selling Products: Revenue vs. Costs

A company produces widgets. The fixed production cost is $5,000, and each widget costs $12 to produce. Widgets are sold for $20 each.

Problem:

1. Write functions for the revenue \( R(x) \) and cost \( C(x) \).
2. Determine the break-even point where \( R(x) = C(x) \).
3. If the revenue is $40,000, calculate the number of widgets sold using the inverse of \( R(x) \).
4. Discuss the domain and range of each function.

Solution:

1. Functions:

2. Break-Even Point:

3. Inverse Function for Revenue:

Substitute \( R(x) = 40,000 \):

4. Domain and Range:

- Domain of \( R(x) \): \( x \geq 0 \).
- Range of \( R(x) \): \( R(x) \geq 0 \).
- Domain of \( C(x) \): \( x \geq 0 \).
- Range of \( C(x) \): \( C(x) \geq 5,000 \).

4. Depreciation of an Asset

A car costing $25,000 depreciates by $2,000 annually.

Problem:

1. Write a function \( f(t) \) for the car's value after \( t \) years.
2. Find the inverse function \( f^{-1}(x) \) to calculate how many years it will take for the value to depreciate to $15,000.
3. Discuss the domain and range of both the function and its inverse.

Solution:

1. Function:

2. Inverse Function:

3. Substitute \( f(x) = 15,000 \):

4. Domain and Range:

- Domain of \( f(t) \): \( t \in [0, 12.5] \).
- Range of \( f(t) \): \( f(t) \in [0, 25,000] \).
- Domain of \( f^{-1}(x) \): \( x \in [0, 25,000] \).
- Range of \( f^{-1}(x) \): \( t \in [0, 12.5] \).

    C(t) = 
    { 
      20, if 0 < t ≤ 1
      20 + 15(t - 1), if t > 1
    }
  

    C(5.5) = 20 + 15(5.5 - 1)
    C(5.5) = 20 + 67.5 = 87.5
  

5. Speed and Time: A Travel Scenario

A car travels at a constant speed of 60 km/h. The driver takes a 15-minute break every 2 hours.

Problem:

1. Write a piecewise function \( f(t) \) for the distance traveled, accounting for breaks.
2. Determine the inverse \( f^{-1}(x) \) to calculate the time taken to travel 300 km.

Solution:

Part 1: Writing the Piecewise Function \( f(t) \)

Given:

• The car travels at a constant speed of \( 60 \, \text{km/h} \).
• The driver takes a \( 15 \, \text{minute break} = 0.25 \, \text{hours} \) every 2 hours.

Approach:

• The total time \( t \) includes both driving time and break time.
• For every 2 hours of driving, an additional \( 0.25 \, \text{hours} \) is added due to breaks.
• The distance traveled \( f(t) \) depends on whether the break period is factored in or not.

Function \( f(t) \):

Here, \( n = \left\lfloor \frac{t}{2} \right\rfloor \): The number of breaks is the integer part of \( \frac{t}{2} \), i.e., one break occurs for every 2 hours of driving.

Part 2: Finding the Inverse \( f^{-1}(x) \)

Given:

• Distance traveled \( x = 300 \, \text{km} \).
• We need to find \( t \) (time taken), including breaks.

Steps to Derive the Inverse \( f^{-1}(x) \):

If no breaks occur:

If breaks occur:

Substitute \( n = \left\lfloor \frac{t}{2} \right\rfloor \), iteratively solving for \( t \).

Calculation for \( x = 300 \, \text{km} \)

Step 1: Compute driving time ignoring breaks:

Step 2: Determine breaks required:

Step 3: Account for break time:

Step 4: Total time:

Final Answers:

Piecewise Function:

Inverse Function:

Time to travel 300 km:

6. Foreign Exchange: USD to British Pounds

A forex bureau converts USD to British Pounds at a rate of $1 = £0.75.

Problem:

1. Write a function \( f(x) \) to represent the amount in British Pounds for \( x \), where \( x \) is the amount in USD.
2. Find the inverse function \( f^{-1}(x) \) to calculate the amount in USD required for a specific amount in Pounds.
3. If a customer wants £300, how much USD will they need?

Solution:

1. Function:

2. Inverse Function:

3. Calculate USD for £300:

4. Domain and Range:

- Domain of \( f(x) \): \( x \geq 0 \) (USD).
- Range of \( f(x) \): \( f(x) \geq 0 \) (Pounds).
- Domain of \( f^{-1}(x) \): \( x \geq 0 \) (Pounds).
- Range of \( f^{-1}(x) \): \( f^{-1}(x) \geq 0 \) (USD).

7. Simple Interest: Saving for a Goal

James deposits $5,000 in a savings account that offers simple interest at an annual rate of 5%.

Problem:

1. Write a function \( f(t) \) to represent the total amount in the account after \( t \) years.
2. Find the inverse function \( f^{-1}(x) \) to calculate the time required to achieve a specific savings goal.
3. If James wants $7,000, how many years must he leave the money in the account?

Solution:

1. Function:

2. Inverse Function:

3. Time for $7,000:

4. Domain and Range:

- Domain of \( f(t) \): \( t \geq 0 \).
- Range of \( f(t) \): \( f(t) \geq 5,000 \).
- Domain of \( f^{-1}(x) \): \( x \geq 5,000 \).
- Range of \( f^{-1}(x) \): \( f^{-1}(x) \geq 0 \).

Practice Questions

1. A savings account has an initial deposit of $3,000 with a simple interest rate of 6% per year. If the account grows to $4,200, how many years will it take?
2. Emily deposits $8,000 in a bank offering a simple interest rate of 4% annually. She wants to save $10,000. How long will it take?
3. A principal amount of $12,000 earns a simple interest rate of 7% annually. If the total amount is $16,200, how many years has the money been in the account?
4. Michael deposits $15,000 at an annual simple interest rate of 5%. He plans to save $18,000. Calculate the time required to reach his goal.

Answers:

1. 6.67 years
2. 6.25 years
3. 5 years
4. 4 years