Problem 21 - Algebra 2 and Trigonometry

Question

If \(\mathrm{f}=\\{(x, y) : y=5 x\\}\) is a direct variation function, find \(\mathrm{f}^{-1}\)

Step-by-Step Solution

Answer

The inverse function is \\( ext{f}^{-1}(x) = \frac{x}{5}\\).

Step 1: Understanding Direct Variation

The given function is \((x, y): y = 5x\). This means that for every change in \(x\), \(y\) is a direct multiple of 5 times that change. The function establishes the relationship \(y = 5x\).

Step 2: What is an Inverse Function?

The inverse of a function \( ext{f}\), denoted as \( ext{f}^{-1}\), reverses the inputs and outputs of \( ext{f}\). For the function \(y = ext{f}(x)\), the inverse function will give us \(x = ext{f}^{-1}(y)\).

Step 3: Setting Up the Inverse Function

To find the inverse, set \(y = 5x\) and solve for \(x\) in terms of \(y\). Start by swapping \(x\) and \(y\), which gives us \(x = 5y\).

Step 4: Solving for the Inverse Variable

With \(x = 5y\) set up, solve for \(y\) to express \(y\) in terms of \(x\). Divide both sides by 5 to isolate \(y\), resulting in \(y = \frac{x}{5}\).

Step 5: Writing the Inverse Function

The expression \(y = \frac{x}{5}\) gives us the formula for the inverse function. Thus, \( ext{f}^{-1}(x) = \frac{x}{5}\).

Key Concepts

Direct Variation

In mathematics, direct variation describes a simple relationship between two variables where one variable is a constant multiple of the other. For our given function,

• we had \(y = 5x\).
• This means that as \(x\) changes, \(y\) changes proportionally by a factor of 5.

Understanding direct variation can be quite simple when you break it down:

• The equation always takes the form \(y = kx\), where \(k\) is the constant of variation.
• In our exercise, the constant \(k\) is 5, showing that any increase or decrease in \(x\) by a certain amount results in a change in \(y\) by 5 times that amount.
• This type of relationship is linear, meaning the graph of a direct variation function is a straight line that passes through the origin (0,0).

Recognizing direct variation is key to solving problems related to proportional relationships in algebra.

Inverse Operation

The concept of an inverse operation is fundamental in algebra. It refers to operations that "undo" each other. For example,

• addition and subtraction are inverse operations, as are multiplication and division.

In the context of the inverse function problem:

• we start with a direct variation function \(y = 5x\).
• The task is to find the inverse, which essentially means finding an operation that reverses the effect of the original function.

To achieve this, we need to

• exchange the roles of \(x\) and \(y\): taking \(x = 5y\)
• and then solve for \(y\), giving us the inverse function, \(y = \frac{x}{5}\).

Here, division by 5 is the inverse operation of multiplying by 5. Understanding inverse operations helps us to reverse processes and solve equations, making it a crucial part of algebra.

Algebraic Functions

An algebraic function is any function defined by a finite series of algebraic operations (addition, subtraction, multiplication, division, and exponents).

• In our example, both the original function \(y = 5x\) and its inverse \(y = \frac{x}{5}\) are algebraic functions.
• Such functions are common in algebra problems because they express relationships in simple terms.

Key characteristics of algebraic functions include:

• They often represent linear, quadratic, polynomial, or rational relationships.
• The solutions, like in our task, involve straightforward algebraic manipulation—solving for one variable in terms of another.

Algebraic functions allow us to model real-world situations mathematically, understanding how one quantity changes in response to another. They are the backbone of many equations and operations you'll encounter in algebra.