Question

In Exercises \(9-12,\) show that composing the functions in either order gets us back to where we started. $$ y=7 x-5 \text { and } x=\frac{y+5}{7} $$

Short answer

Question: Compose the given functions in both possible orders and verify that we end up with the same input variable we started with. Functions: 1. \(y = 7x - 5\) 2. \(x = \frac{y+5}{7}\) Answer: Composing the given functions in both orders results in \(x = x\) and \(y = y\), confirming that we end up with the same input variable we started with in both cases.

Step-by-step solution

Compose the functions in the first order (First Function -> Second Function)

To do this, we have to substitute the expression of \(y\) in terms of \(x\) (from the first function) into the second function: $$x = \frac{(7x - 5) + 5}{7}$$

Simplify and solve for \(x\)

Let's simplify the above equation: $$x = \frac{7x}{7}$$ $$x = x$$ The result we get is \(x = x\), which means that composing the functions in this order gives us back the original input variable \(x\).

Compose the functions in the second order (Second Function -> First Function)

Now, let's substitute the expression of \(x\) in terms of \(y\) (from the second function) into the first function: $$y = 7\left(\frac{y + 5}{7}\right) - 5$$

Simplify and solve for \(y\)

Let's simplify this equation: $$y = (y + 5) - 5$$ $$y = y$$ The result we obtained is \(y = y\), meaning that composing the functions in this order also gets us back to the original input variable \(y\). In conclusion, composing the given functions in either order gets us back to where we started, as shown by the results \(x = x\) and \(y = y\).

Key concepts

Inverse Functions

An inverse function reverses the operation of a particular function. Suppose we have a function that takes an input and modifies it to produce an output. The inverse function would take that output and revert it back to the original input. For example, if a function is given by \( y = 7x - 5 \), its inverse would resolve \( y \) back into \( x \). In the exercise, one function was provided as \( y = 7x - 5 \). To find the inverse, it was necessary to solve for \( x \) in terms of \( y \), which results in the inverse function \( x = \frac{y + 5}{7} \). The process of finding an inverse involves algebraic manipulation, where you essentially "undo" the operations applied in the forward function. Knowing how to find inverse functions is critical in mathematics because it allows us to determine original values from transformed ones. In essence, it enables reversibility within many processes. This concept is heavily utilized in solving equations and understanding function behaviors across domains.

Function Operations

Function operations involve performing various tasks such as addition, subtraction, multiplication, division, and composition using functions. A particularly critical operation is the composition of functions, where one function is applied to the result of another function. This is denoted as \((f \circ g)(x)\), meaning that function \( g \) is applied to \( x \) first, and then function \( f \) is applied to the result of \( g(x) \). In the exercise, we see this concept with the expressions \( y = 7x - 5 \) and \( x = \frac{y + 5}{7} \). Composing these functions shows that when applied in either order, you achieve the identity, getting back your original variable. This is a classic example where the composition yields the original input value \( x \) or \( y \), affirming that the functions are inverses of each other. By understanding how function operations work, especially composition, one can effectively combine multiple processes into a single function, which simplifies complex calculations and analysis.

Algebraic Manipulation

Algebraic manipulation involves using algebraic methods to simplify expressions or solve equations. It plays a key role in understanding and working with functions, including finding inverses and simplifying composed functions. In solving or composing functions, you often need to rearrange equations to isolate a particular variable. In the given exercise, after substituting one function into the inverse, the equation \( x = \frac{7x}{7} \) reduces to \( x = x \). Likewise, substituting in the other direction leads to \( y = (y+5)-5 \), which reduces neatly to \( y = y \). These processes illustrate the power of carefully applying operations such as factoring, expanding, adding, and subtracting to transform expressions into simpler forms.By mastering algebraic manipulations, you not only solve equations efficiently, but also enhance your ability to comprehend complex mathematical relationships and perform with ease on sophisticated algebraic tasks.