Applications of Function Inverses

Function and Its Inverse

Let’s consider the function f(x) = 9/5x + 32, which converts Celsius to Fahrenheit.

To find the inverse of this function, we follow these steps:

Replace f(x) with y, so the equation becomes y = 9/5x + 32.
Swap x and y, resulting in x = 9/5y + 32.
Solve for y to get the inverse function. In this case, the inverse function is f^-1(x) = 5/9(x – 32).

In the function f(x) = 9/5x + 32:

x represents the temperature in Celsius.
f(x) represents the temperature in Fahrenheit.

In the inverse function f^-1(x) = 5/9(x – 32):

x represents the temperature in Fahrenheit.
f^-1(x) represents the temperature in Celsius.

Here is the plot of the function and its inverse:

The graph shows both the function f(x) = 9/5x + 32 and its inverse f^-1(x) = 5/9(x – 32). The function is represented by a line that slopes upwards from left to right, showing the relationship between the temperature in Celsius (x-axis) and the temperature in Fahrenheit (y-axis). The inverse function, represented by the same line, shows the relationship between the temperature in Fahrenheit (x-axis) and the temperature in Celsius (y-axis).

Function and Its Inverse

Let’s consider the function f(x) = 9/5x + 32, which converts Celsius to Fahrenheit.

To find the inverse of this function, we follow these steps:

Replace f(x) with y, so the equation becomes y = 9/5x + 32.
Swap x and y, resulting in x = 9/5y + 32.
Solve for y to get the inverse function. In this case, the inverse function is f^-1(x) = 5/9(x – 32).

In the function f(x) = 9/5x + 32:

x represents the temperature in Celsius.
f(x) represents the temperature in Fahrenheit.

In the inverse function f^-1(x) = 5/9(x – 32):

x represents the temperature in Fahrenheit.
f^-1(x) represents the temperature in Celsius.

The blue line represents the function f(x) = 9/5x + 32, which is the conversion from Celsius to Fahrenheit. The orange line represents the inverse function f^-1(x) = 5/9(x – 32), which is the conversion from Fahrenheit to Celsius. They intersect at the point (-40, -40), which is the temperature that is the same in both Celsius and Fahrenheit. The blue line has a steeper slope than the orange line because each degree change in Celsius corresponds to a 1.8 degree change in Fahrenheit.

Application of Linear Functions and Their Inverses

Let’s consider a small business scenario where the profit can be modeled by a linear function. Suppose the profit P in dollars is given by the function P(x) = 2x + 3, where x is the number of products sold. This means that for every product sold, the profit increases by $2. Additionally, there is a base profit of $3, which could be from other sources of income.

To find the inverse of this function, we follow these steps:

Replace P(x) with y, so the equation becomes y = 2x + 3.
Swap x and y, resulting in x = 2y + 3.
Solve for y to get the inverse function. In this case, the inverse function is x(P) = 0.5(x – 3).

Now, suppose the goal is to find out how many products need to be sold to reach a certain profit goal. This is where the inverse function comes in handy. The inverse function tells us the number of products that need to be sold to reach a given profit.

For example, if the profit goal is $100, the inverse function can be used to find out how many products need to be sold:

x(100) = 0.5(100 – 3) = 48.5

So, approximately 49 products (since we can’t sell half a product) need to be sold to reach a profit goal of $100.

This is a simple example of how linear functions and their inverses can be used in real-world applications.

Application of Linear Functions and Their Inverses

Consider a shipping company that charges a base fee of $5 for any shipment, and then adds an additional cost of $4 per mile traveled. This can be represented by the linear function $f(x) = 4x + 5$, where $x$ is the distance traveled in miles and $f(x)$ is the total cost of the shipment.

To find the inverse of this function, we follow these steps:

Replace $f(x)$ with $y$, so the equation becomes $y = 4x + 5$.
Swap $x$ and $y$, resulting in $x = 4y + 5$.
Solve for $y$ to get the inverse function. In this case, the inverse function is $f^{-1}(x) = \frac{1}{4}(x – 5)$.

Now, suppose the goal is to find out how far a package traveled given the total cost of the shipment. This is where the inverse function comes in handy. The inverse function tells us the distance traveled given the total cost.

For example, if the total cost was $21, the inverse function can be used to find out how far the package traveled:

$x(21) = \frac{1}{4}(21 – 5) = 4$ miles

So, the package traveled 4 miles to reach a total shipment cost of $21.

This is a simple example of how linear functions and their inverses can be used in real-world applications.

Application of Linear Functions and Their Inverses

Consider a store that offers a discount scheme where they first take off 15% from the original price of an item, and then take off an additional $10. This can be represented by the linear function $f(x) = 0.85x – 10$, where $x$ is the original price of the item and $f(x)$ is the final price after the discounts.

To find the inverse of this function, we follow these steps:

Replace $f(x)$ with $y$, so the equation becomes $y = 0.85x – 10$.
Swap $x$ and $y$, resulting in $x = 0.85y – 10$.
Solve for $y$ to get the inverse function. In this case, the inverse function is $f^{-1}(x) = \frac{1}{0.85}(x + 10)$.

Now, suppose the goal is to find out the original price of an item given the final price after discounts. This is where the inverse function comes in handy. The inverse function tells us the original price given the final price.

For example, if the final price was $50, the inverse function can be used to find out the original price:

$x(50) = \frac{1}{0.85}(50 + 10) \approx 70.59$

So, the original price of the item was approximately $70.59 before the discounts were applied to reach a final price of $50.

This is a simple example of how linear functions and their inverses can be used in real-world applications.

Application of Linear Functions and Their Inverses

To find the inverse of this function, we follow these steps:

Replace $f(x)$ with $y$, so the equation becomes $y = 0.85x – 10$.
Swap $x$ and $y$, resulting in $x = 0.85y – 10$.
Solve for $y$ to get the inverse function. In this case, the inverse function is $f^{-1}(x) = \frac{1}{0.85}(x + 10)$.

For example, if the final price was $50, the inverse function can be used to find out the original price:

$x(50) = \frac{1}{0.85}(50 + 10) \approx 70.59$

So, the original price of the item was approximately $70.59 before the discounts were applied to reach a final price of $50.

This is a simple example of how linear functions and their inverses can be used in real-world applications.