Problem 33 - Algebra 2

Question

For Exercises \(31-34, f(x)=10 x-10 .\) Find each value. $$ \left(f^{-1} \circ f\right)(0.2) $$

Step-by-Step Solution

Answer

The value of \((f^{-1} \circ f)(0.2)\) is 0.2.

Step 1: Identify the function

From the problem statement, we have the function \(f(x)=10 x-10\). Let's write that down, this is going to be our base function for the exercise.

Step 2: Determine the inverse function

The inverse function \(f^{-1}(x)\) is obtained by replacing \(f(x)\) with \(y\), swapping \(x\) and \(y\), and then solving for \(y\). So the equation becomes \(x = 10y - 10\). Solving this for \(y\) to get the inverse function, we find that \(f^{-1}(x) = \frac{x+10}{10}\).

Step 3: Apply the composite function concept

We know that \((f^{-1} \circ f)(x)\) is equivalent to applying the function \(f(x)\) to \(x\) first and then applying its inverse. This essentially brings the input back to its original value. Hence, \((f^{-1} \circ f)(x)\) simply results in \(x\). Therefore, \((f^{-1} \circ f)(0.2) = 0.2\).

Key Concepts

Understanding Composite Functions

Composite functions combine two or more functions to create a new function. Imagine it like a machine where you place an input, and it goes through a series of processes.
In the end, you get an output. For example, if you have two functions, say \( g(x) \) and \( h(x) \), the composite function \( (g \circ h)(x) \) means we first apply \( h(x) \) to \( x \) and then use the output of \( h(x) \) as the input for \( g(x) \).
This can seem confusing, but breaking it down helps:

• Start with the inside function (apply \( h(x) \)).
• Use this result as an input for the outside function (then \( g \)).
• The final result is your composite function output.

In our example, understanding that \( (f^{-1} \circ f)(x) \) simplifies to just \( x \) helps since applying both a function and its inverse cancels out their effects, leading back to the original value.

Demystifying Function Notation

Function notation is a way to represent functions that clarify how inputs are converted into outputs. When using function notation, you typically see expressions such as \( f(x) \), where \( f \) is the function name, and \( x \) is the input value.
It's like a label for what the function does to any given number you plug in. In example, if \( f(x) = 10x - 10\), then \( f(2) \) is found by substituting \( x \) with \( 2 \). This gives us \( 10(2) - 10 = 10 \).

• \( f(x) \) indicates the function process.
• Substitute in function to get an outcome.
• Functions can have inverses, which switch the input and output roles.

The notation helps differentiate different functions and clearly communicate how they operate, especially when performing operations like composition or finding inverses.

Solving Equations with Inverse Functions

Solving equations can be quite straightforward if you understand how to work with inverse functions. In mathematics, the inverse of a function \( f(x) \) is a function \( f^{-1}(x) \) that reverses the operation of \( f \). This means if you start with a value, apply \( f \), and then apply \( f^{-1} \), you end up back where you started.
Here’s how to determine and use inverse functions:

• Rewrite \( f(x) \) in the form \( y = \ldots \).
• Swap \( x \) and \( y \) in the equation to set up for finding the inverse.
• Solve for \( y \), which becomes \( f^{-1}(x) \).

For instance, starting with \( f(x) = 10x - 10 \), we find its inverse by letting \( y =10x - 10 \), swapping variables to get \( x = 10y - 10 \), and then solving for \( y \) to get \( f^{-1}(x) = \frac{x + 10}{10} \).
Inverse functions simplify solving complicated equations by turning them into more straightforward processes that lead you back to your starting point.