7.1: Inverses of Functions
This activity is intended to supplement Calculus, Chapter 6, Lesson 1.
ID: 11405
Time Required: 45 minutes
Activity Overview
In this activity, students will explore three ways to find the inverse of a function. First, students graph two scatter plots and find the line of reflection. Then, they will graph a line and use the \begin{align*}x\end{align*}
Topic: Sequences, Series & Functions
 Reflection
 Inverse of data points
 Inverse of functions
Teacher Preparation and Notes

Before beginning this activity, students should clear lists \begin{align*}L_1\end{align*}
L1 and \begin{align*}L_2\end{align*}L2 and turn off all functions and plots.  Students will need to know how to find the midpoint of two points. The formula is given later in this document.
Associated Materials
 Student Worksheet: Inverses of Functions http://www.ck12.org/flexr/chapter/9731
Exploring the Problem
Students are given data for a wind tunnel experiment. They are to use the data to create a scatter plot, then answer questions about the data and associated graph.
Students begin by constructing a scatter plot and drawing their graph on their worksheet. To create a scatter plot, students first enter the data into their lists by pressing STAT and selecting Edit.
Once the data is entered, students should press \begin{align*}2^{nd}\end{align*}
Students will switch \begin{align*}L1\end{align*}
The points obtained by switching the domain and range appear to be a reflection of the original points across a line.
Students are instructed to find the midpoint between the first point on each of the scatter plots and the midpoint between the last points for each of the scatter plots on the graph.
\begin{align*}M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\end{align*}
Then, to find the equation of the line, they need to find the slope using the two midpoints and the pointslope form.
4. \begin{align*} \left( \frac{10 + 6.4}{2}, \frac{6.4 + 10}{2} \right) & = \left( \frac{16.4}{2}, \frac{16.4}{2} \right) = (8.2, 8.2) \\
\left( \frac{55 + 59.2}{2}, \frac{55 + 59.2}{2} \right) & = \left( \frac{114.2}{2}, \frac{114.2}{2} \right) = (57.1, 57.1)\end{align*}
5. \begin{align*}y = x\end{align*}; This line represents the reflection line.
Developing the Pattern Further
At this point, students should clear out lists \begin{align*}L1\end{align*} and \begin{align*}L2\end{align*} and turn off all functions and plots.
Students are to graph \begin{align*}y = 2x + 3\end{align*} in a standard viewing window. To find the \begin{align*}x\end{align*}intercept, they will use the zero function. When selected, it asks for the lower bound, upper bound, and a guess. There must be a sign change in order for this command to work. The \begin{align*}y\end{align*}intercept can be found by just pressing TRACE.
Then, students will switch the locations of the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}coordinates of these points to \begin{align*}(y, x)\end{align*}, input these as \begin{align*}L1\end{align*} and \begin{align*}L2\end{align*}, and graph using Plot1. They can find the equation of the line connecting these two points by finding the slope and then using the slopeintercept form. They will graph it in \begin{align*}Y2\end{align*}.
Students and teachers are encouraged to explore the concept of inverses further. The necessity of a function being onetoone in order to have an inverse should be addressed by the teacher.
6. (0, 3) and (1.5, 0)
7. \begin{align*}y = \frac{(x  3)}{2}\end{align*} or \begin{align*}y = \frac{x}{2}  \frac{3}{2}\end{align*} or \begin{align*}y = 0.5x  1.5\end{align*}
8. It is the same.
9. Switch the domain and range, switch \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in the equation, reflect the graph of the function across the line \begin{align*}y = x\end{align*}.
10. Agree; Sample Answer: In order for a function to have an inverse, it must be onetoone. In other words, it passes both the vertical and horizontal line tests, which will result in a reflection across \begin{align*}y = x\end{align*} that will also pass the vertical line test, and is therefore a function.
11. a. \begin{align*}f^{1}(x) = \frac{x + 2}{6}\end{align*}
b. \begin{align*}f^{1}(x) = 2x + \frac{3}{2}\end{align*}
c. \begin{align*}f^{1}(x)\end{align*} does not exist
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