Problem 12 - Contemporary Precalculus

Question

Use algebra to find the inverse of the given one-to-one function. $$f(x)=-3 x+5$$

Step-by-Step Solution

Answer

Answer: The inverse function of \(f(x) = -3x + 5\) is \(f^{-1}(x) = \frac{-x + 5}{3}\).

Step 1: Replace f(x) with y

First, replace the function notation f(x) with y: $$y = -3x + 5$$

Step 2: Swap x and y

In the next step, we swap the roles of x and y: $$x = -3y + 5$$

Step 3: Solve for y

To find the inverse function, we'll need to solve for y. To do this, follow these steps: 1. Add 3y to both sides of the equation to isolate the y term on the left side: $$x + 3y = 5$$ 2. Subtract x from both sides of the equation to move it to the right side: $$3y = -x + 5$$ 3. Finally, divide both sides by 3 to completely isolate y: $$y = \frac{-x + 5}{3}$$

Step 4: Write the inverse function

Now that we have solved for y, we can write the inverse function, \(f^{-1}(x)\), as follows: $$f^{-1}(x) = \frac{-x + 5}{3}$$ The inverse function of the given one-to-one function \(f(x) = -3x + 5\) is: $$f^{-1}(x) = \frac{-x + 5}{3}$$

Key Concepts

Function Notation

Function notation is a way to represent functions in a clear and concise manner. Typically, functions are written in the form \( f(x) \), where \( f \) denotes the function, and \( x \) is the variable input. The notation helps to distinguish between different functions and their respective variables, making it easier to work with them in equations and mathematical operations.

In our exercise, we started with the function \( f(x) = -3x + 5 \). In order to find the inverse, we converted this functional notation to a more straightforward equation by replacing \( f(x) \) with \( y \), resulting in \( y = -3x + 5 \). This simple step allows us to treat the function as an equation, making it easier to perform algebraic manipulations to find the inverse.

Function notation is not just a tool for writing equations; it also helps to communicate the idea that a function assigns each input exactly one output. Being comfortable with this notation is essential, as it opens the door to understanding more complex mathematical concepts.

One-to-One Functions

A one-to-one function is a special type of function where each output value is paired with exactly one input value, and vice versa. This means that every element in the function's range corresponds to a unique element in the domain. One-to-one functions are crucial when finding inverses because only these types of functions have true inverses.

An important characteristic of one-to-one functions is that they pass the Horizontal Line Test. If a horizontal line cuts the graph of the function more than once, it is not a one-to-one function. Since \( f(x) = -3x + 5 \) is linear with a non-zero slope, it is one-to-one, ensuring that its inverse can be found.

To re-emphasize, for a function to have an inverse, it must be one-to-one. Therefore, identifying whether a function is one-to-one is an important preliminary step when dealing with inverse functions. This step ensures that every aspect of the domain and range pair up correctly with every flip of inputs and outputs.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to find unknown variables. When finding the inverse of a function, we use algebraic manipulation to exchange the roles of variables, often by swapping \( x \) and \( y \), and then solve for the new 'output' variable.

In our exercise, after swapping \( x \) and \( y \) to form the equation \( x = -3y + 5 \), we needed to isolate \( y \). This involved:

• Adding \( 3y \) to both sides: \( x + 3y = 5 \)
• Subtracting \( x \) from both sides: \( 3y = -x + 5 \)
• Dividing every term by 3: \( y = \frac{-x + 5}{3} \)

Through these steps, we arrived at the inverse: \( f^{-1}(x) = \frac{-x + 5}{3} \).

Algebraic manipulation is all about careful logical steps. Each operation helps to transform the equation correctly, leading us to our desired form. Practicing this manipulation will improve your skills in solving various mathematical problems, including inverses.