Problem 74 - College Algebra with Modeling and Visualization

Question

Find a formula for \(f^{-1}(x) .\) Identify the domain and range of \(f^{-1}\). Verify that \(f\) and \(f^{-1}\) are inverses. $$ f(x)=6-7 x $$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = \frac{6 - x}{7} \), with domain and range \( \mathbb{R} \).

Step 1: Find Inverse Function

To find the inverse of the function, we switch the roles of the dependent and independent variables. Start by setting \( y = 6 - 7x \). Then, solve for \( x \):\[ \begin{align*} y & = 6 - 7x \ 7x & = 6 - y \ x & = \frac{6 - y}{7}. \end{align*}\]Now, express the inverse function as \( f^{-1}(x) = \frac{6 - x}{7} \).

Step 2: Determine the Domain and Range

The original function \( f(x) = 6 - 7x \) is a linear function with a domain of all real numbers \( \mathbb{R} \). The range of \( f(x) \) is also all real numbers \( \mathbb{R} \). For the inverse \( f^{-1}(x) \), the domain is the range of the original function, which is \( \mathbb{R} \), and the range is the domain of the original function, which is also \( \mathbb{R} \). Hence, the domain and range of \( f^{-1} \) are both \( \mathbb{R} \).

Step 3: Verify the Inverses

To verify that \( f \) and \( f^{-1} \) are inverses, check if \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \):\[ \begin{align*} f(f^{-1}(x)) & = f\left(\frac{6 - x}{7}\right) = 6 - 7\left(\frac{6 - x}{7}\right) = 6 - (6 - x) = x, \ f^{-1}(f(x)) & = f^{-1}(6 - 7x) = \frac{6 - (6 - 7x)}{7} = \frac{7x}{7} = x. \end{align*} \]Since both compositions result in \( x \), \( f \) and \( f^{-1} \) are confirmed as inverses.

Key Concepts

Domain and Range

When we talk about the domain and range of a function, we're essentially discussing the set of input values that a function can accept, and the set of output values that the function can produce, respectively.
In the case of the given linear function, which is described by the equation \( f(x) = 6 - 7x \), the domain is quite broad. In fact, it includes all real numbers \( \mathbb{R} \). This means you can plug any real number into \( x \) and still have a valid output.
Similarly, the range of \( f(x) \) is also all real numbers \( \mathbb{R} \). That's because a linear function like this one, when graphed, forms a straight line that extends infinitely in either direction on the y-axis.

Now, when we take the inverse of this function, which we've found to be \( f^{-1}(x) = \frac{6 - x}{7} \), its domain and range are determined by swapping the original domain and range. So, \( f^{-1} \) also has a domain and range of all real numbers \( \mathbb{R} \). This swap is because every output (or y-value) of the original function can now serve as an input (x-value) for the inverse function and vice-versa.

Linear Functions

Linear functions are among the simplest and most fundamental functions in mathematics. They're typically expressed in the form \( y = mx + b \), where \( m \) represents the slope, and \( b \) indicates the y-intercept. For the function \( f(x) = 6 - 7x \), the slope \( m \) is \(-7\), and the y-intercept \( b \) is \(6\).
These functions graph as a straight line on the coordinate plane. The slope tells us how "steep" the line is and the direction in which it moves. Since our slope is negative, the line will slant downward from left to right. The y-intercept of 6 indicates that this line will cross the y-axis at the point \( (0,6) \).

Linear functions are notable because they exhibit a constant rate of change, which means for every increase in x by 1 unit, \( y \) will decrease by -7 units in our specific example. Understanding this relationship is key to working with straight line equations and helps in predicting how altering one variable will affect the other.

Verification of Inverse

Verifying inverses involves checking that two functions truly "undo" each other. In other words, if you apply one function to a value and then apply its inverse to the result, you should get back your original value.

To confirm that \( f(x) = 6 - 7x \) and its inverse \( f^{-1}(x) = \frac{6 - x}{7} \) are really inverses, we check two conditions:

• \( f(f^{-1}(x)) = x \)
• \( f^{-1}(f(x)) = x \)

For the first function, plug \( f^{-1}(x) \) into \( f(x) \):

\[ f\left( \frac{6 - x}{7} \right) = 6 - 7 \left( \frac{6 - x}{7} \right) = 6 - (6 - x) = x \].

This verifies the first condition. Similarly, for the second:

\[ f^{-1}(6 - 7x) = \frac{6 - (6 - 7x)}{7} = \frac{7x}{7} = x \].
Both mappings leave us with the input \( x \), demonstrating that \( f \) and \( f^{-1} \) are indeed inverse functions.