Last updated .
4.6 — Bernie’s Bikes
In Section 4.5, you learned that inverse relationships reverse inputs and outputs. Today you will write the rules yourself, use them in a practical situation, and decide when an inverse is actually a function.
1. In class: A Solidify Understanding Task
Reversing an input–output relationship
Imagine a machine that takes a number, doubles it, and adds 5. An input of 3 produces 11. To run the relationship backward, start with 11, subtract 5, and divide by 2. The result is 3.
| Original input | Original output | Inverse input | Inverse output |
|---|---|---|---|
| 1 | 7 | 7 | 1 |
| 3 | 11 | 11 | 3 |
| 5 | 15 | 15 | 5 |
A function’s domain (allowed inputs) becomes its inverse’s range (outputs), and vice versa. If both are functions, their graphs mirror each other across \(y=x\).
Bernie’s signs: a table, a formula, and its inverse
Ink required by sign length — task problems 1–3
The packet records sign lengths of 3, 4, 2, and 15 feet with corresponding ink quantities of 9, 16, 4, and 225 ounces. Every output is the square of the input, suggesting the model \(I(L)=L^2\).
| Sign length \(L\) (feet) | 2 | 3 | 4 | 15 | \(L\) |
|---|---|---|---|---|---|
| Ink \(I(L)\) (ounces) | 4 | 9 | 16 | 225 | \(L^2\) |
Because sign lengths are nonnegative, the physically meaningful domain is \([0,\infty)\), and the range is \([0,\infty)\). The equation \(I(L)=L^2\) models the supplied table; the numerical coefficient carries the necessary units of ounces per square foot.
If Bernie uses 49 ounces of ink, what sign length does the model predict?
Start with \(I=L^2\). Since a sign’s length is nonnegative, \(L=\sqrt{I}\). Thus \(L=\sqrt{49}=7\) feet.
The inverse model is \(I^{-1}(z)=\sqrt z\), where \(z\) represents ounces of ink. Its domain and range are both \([0,\infty)\). Notice how the units and roles switch: ink in, length out.
Two operations: height from length — task problems 4–7
Bernie specifies that a sign’s vertical height is 5 inches more than twice its horizontal length. Both columns of the packet’s table, however, are labeled in feet. We will first make the units consistent.
Five inches equals \(\frac5{12}\) feet. For horizontal length \(L\) feet, the height in feet is therefore
| Length, feet | 2 | 3 | 4 | 15 |
|---|---|---|---|---|
| Height, feet | \(4\frac5{12}\) | \(6\frac5{12}\) | \(8\frac5{12}\) | \(30\frac5{12}\) |
For nonnegative length, the domain is \([0,\infty)\), and the height range is \([\frac5{12},\infty)\). The inverse takes height in feet as its input. Undo the addition first and doubling second:
Thus \(H^{-1}(y)=\frac{y-5/12}{2}\). Its domain is \([\frac5{12},\infty)\), and its range is \([0,\infty)\). For a height of \(8\frac5{12}\) feet, the inverse gives a 4-foot horizontal length.
The printed rule says “5 inches more,” while the table measures length and height in feet. The mathematically consistent feet-to-feet rule is \(H(L)=2L+5/12\), not \(2L+5\). The wording of problem 6 (“adding 5”) suggests that the exercise may have intended the table to use inches or the verbal rule to say 5 feet. If your teacher directs the latter interpretation, use \(H(L)=2L+5\), with inverse \((y-5)/2\), and adjust the table accordingly.
Finding inverse rules algebraically
For an inverse, interchange which quantity is known and which is sought. A reliable procedure is to write \(y=f(x)\), solve for \(x\), then express the result using the input letter of the inverse. The inverse may require a domain restriction.
Given \(g(x)=3x+2\), reverse the two operations:
\[y=3x+2\quad\Rightarrow\quad y-2=3x\quad\Rightarrow\quad x=\frac{y-2}{3}.\]Therefore \(g^{-1}(x)=(x-2)/3\). Both functions have domain and range all real numbers. Check: \(g(4)=14\) and \(g^{-1}(14)=4\).
Consider \(f(x)=x^2-1\), with all real inputs. Solving \(y=x^2-1\) gives \(x=\pm\sqrt{y+1}\). Each input above \(-1\) has two inverse outputs, so this inverse relation is not a function on the original domain.
To make an inverse function, restrict the original domain to \(x\ge0\). Now \(f^{-1}(x)=\sqrt{x+1}\), with inverse domain \([-1,\infty)\) and range \([0,\infty)\). Or restrict the original to \(x\le0\), obtaining \(-\sqrt{x+1}\) instead.
For \(f(x)=(x+3)^2\), first solve \(y=(x+3)^2\):
\[x+3=\pm\sqrt y\quad\Rightarrow\quad x=-3\pm\sqrt y.\]The full inverse relation has two branches; it is not a function if the original domain is all reals. Restrict the original domain to \([-3,\infty)\) to obtain \(f^{-1}(x)=-3+\sqrt x\), whose domain is \([0,\infty)\) and range is \([-3,\infty)\). The other restriction, \(( -\infty,-3]\), gives \(-3-\sqrt x\).
For \(f(x)=x^3\), every output comes from exactly one real input. Therefore:
\[y=x^3\quad\Rightarrow\quad x=\sqrt[3]{y}\quad\Rightarrow\quad f^{-1}(x)=\sqrt[3]{x}.\]Both original and inverse have domain and range \(( -\infty,\infty)\). A cube root can be negative, unlike the principal square root.
2. Ready — classify functions from their graphs (1–6)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Ready asks you to find each function’s domain and range and classify its graph. Read the actual endpoints, gaps, and direction of the graph; do not assume all curves extend forever.
| Feature | How to identify it |
|---|---|
| Domain / range | List all x-values / y-values shown; use brackets for included endpoints and parentheses for excluded endpoints or infinities. |
| Discrete | Only isolated points are defined, often at whole-number inputs. |
| Continuous | The curve can be traced without lifting your pencil over the stated interval. |
| Discontinuous | It has a gap, jump, hole, or break. |
| Increasing / decreasing / constant | As you move to the right, outputs go up / down / stay level. A graph can have different behaviors on different intervals. |
Suppose a graph consists of a filled straight segment from \((-2,1)\) to \((3,6)\). Its domain is \([-2,3]\), its range is \([1,6]\); it is continuous and increasing. If the segment were replaced by just those two isolated points, its domain would be \(\{-2,3\}\), its range \(\{1,6\}\), and it would be discrete.
Check your work — Ready 1–6 answer key
- Domain \([-5,5]\); range \([0,10]\); continuous and increasing.
- Domain \([-10,10]\); range \([-4,7)\); discontinuous at \(x=0,3\). Decreases on \((-10,0)\), increases on \((0,3)\), decreases on \((3,10)\).
- Domain \([-4,5]\); range \([1,7]\); continuous. Decreases on \((-4,-2)\), increases on \((-2,0)\), constant on \([0,5]\).
- Domain \([-11,9]\); range approximately \([-7,4]\); continuous. Decreases to the minimum near \(x=-6.5\), increases to the maximum near \(x=3.5\), then decreases. Turning coordinates are approximate readings from the sketch.
- Domain \([0,\infty)\); range \((-\infty,5]\); continuous and decreasing.
- Domain \([-3,8]\); range \([-4,2]\); continuous. Decreases on \((-3,2)\), increases on \((2,8)\).
3. Set — square-root speed and stopping distance (7–10)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
The packet gives a model relating stopping distance \(d\), in feet, to speed \(s\), in miles per hour:
Square roots use the nonnegative root here because speed is not negative. The inverse equation follows by squaring both sides and dividing by 30:
\[s^2=30d\quad\Rightarrow\quad d(s)=\frac{s^2}{30},\qquad s\ge0.\]This is a mathematical model for the exercise, not a universal stopping-distance law. Actual stopping depends on conditions and response time.
| Stopping distance \(d\), ft | Speed \(\sqrt{30d}\), mph, rounded |
|---|---|
| 25 | 27 |
| 50 | 39 |
| 100 | 55 |
| 200 | 77 |
| 300 | 95 |
Skid marks are 150 feet long. The model predicts \(s(150)=\sqrt{30\cdot150}=\sqrt{4500}=30\sqrt5\approx67.1\) mph. Under this model, the driver’s statement that the speed was below 60 mph is not supported.
To check: at 60 mph, the predicted distance would be \(d(60)=60^2/30=120\) feet, less than 150.
To complete problems 9–10, put speed on the horizontal axis and stopping distance on the vertical axis for the inverse graph. The two graphs use exchanged coordinates, and when drawn on identically scaled axes their points are reflections across \(y=x\).
Check your work — Set 7–10 answer key
- Rounded outputs in row order: 27, 39, 55, 77, 95 mph. Plot not shown; graph \(s(d)=\sqrt{30d}\) for \(d\ge0\), starting at \((0,0)\).
- No. \(s(150)=\sqrt{4500}=30\sqrt5\approx67.1\) mph, greater than 60 mph.
- \(d(s)=\frac{s^2}{30},\quad s\ge0\); the accident point is approximately \((67.1,150)\). Plot not shown. Swap the coordinates of the points from 7 to obtain further plotting points (rounding gives approximate points).
- They represent inverse functions: inputs and outputs exchange roles. With equally scaled axes they are reflections across \(y=x\).
4. Go — solve literal equations (11–16)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
A literal equation uses several letters, with each letter standing for a quantity. “Solve for \(r\)” means isolate \(r\), leaving the other letters in the expression.
Solve \(P=2a+2b\) for \(a\).
\[P-2b=2a\quad\Rightarrow\quad \boxed{a=\frac{P-2b}{2}=\frac P2-b}.\]First subtract the term without \(a\), then divide by its coefficient.
Solve \(V=\pi r^2h\) for \(r\), assuming \(V\ge0\), \(h>0\), and \(r\ge0\).
\[\frac{V}{\pi h}=r^2\quad\Rightarrow\quad\boxed{r=\sqrt{\frac{V}{\pi h}}}.\]Divide by the factors multiplying \(r^2\), then take the nonnegative square root because a radius cannot be negative.
Given \(A=\frac{(b_1+b_2)h}{2}\), solve for \(b_2\), with \(h\ne0\).
\[2A=(b_1+b_2)h\]\[\frac{2A}{h}=b_1+b_2\]\[\boxed{b_2=\frac{2A}{h}-b_1}.\]Reversing operations is useful for literal equations just as it is for inverse functions.
For problem 11, \(C=2\pi r\) gives \(r=C/(2\pi)\). For problem 12, \(A=\pi r^2\) gives \(r=\sqrt{A/\pi}\). For problem 13, \(V=\pi r^2h\) gives \(h=V/(\pi r^2)\) when \(r\ne0\). For problem 15, \(V=e^3\) gives \(e=\sqrt[3]{V}\). For problem 16, \(A=\frac{b_1+b_2}{2}h\) gives \(h=\frac{2A}{b_1+b_2}\), provided \(b_1+b_2\ne0\).
Check your work — Go 11–16 answer key
- \(r=\frac{C}{2\pi}\)
- \(r=\sqrt{\frac A\pi}\)
- \(h=\frac{V}{\pi r^2}\)
- \(r=\sqrt{\frac{V}{\pi h}}\)
- \(e=\sqrt[3]{V}\)
- \(h=\frac{2A}{b_1+b_2}\)
5. Additional learning resources
| Resource | How to use it | Scope / what to skip |
|---|---|---|
| Khan Academy — Inverse functions | Review inverse values, inverse graphs, and equations. | Some extra material; concentrate on introductory inverse rules and graph reflections. |
| Math Is Fun — Inverse Functions | Step-by-step examples of reversing operations and checking inverses. | Close match for the main task; read the sections on inverse existence carefully. |
| Paul’s Online Math Notes — Solving Equations | Optional equation-isolation review for Go. | Some extra material; use linear-equation manipulation, skip unrelated advanced cases. |
6. Teacher’s review
Significant issue — main task, problems 4–7: The verbal rule adds five inches to twice a length given in feet, while the table labels the output in feet. Problem 6 then says “adding 5,” without a unit conversion. Recommend making both units inches, changing the addition to five feet, or explicitly instructing students to add \(5/12\) feet.
Important missing restriction — main task, problems 8 and 10: The inverse of an unrestricted quadratic is a two-branch relation, not a function. The task asks for \(f^{-1}(x)\) without specifying a branch/domain restriction. Recommend adding \(x\ge0\) to problem 8 and \(x\ge-3\) to problem 10, or explicitly requesting the inverse relation instead.
Minor wording and labeling: The main task’s header reads “Secondary Math I” while the module and homework are Secondary Math II. The mathematics in this supplement follows the course and module designation.
7. Coverage checklist
| Packet | Instruction included here |
|---|---|
| Main task 1–3 | Square rule, ink quantities, domain/range, square-root inverse, reflection |
| Main task 4–7 | Two-operation height rule, inverse table, reversing operations, unit clarification |
| Main task 8–11 | Inverse of quadratic, line, shifted quadratic, cube; inverse existence and restrictions |
| Ready 1–6 | Domain, range, discrete/continuous/discontinuous; increasing/decreasing/constant |
| Set 7–10 | Stopping-distance square-root model, tables, graphs, swapped axes, context |
| Go 11–16 | Literal equations, linear, squares, cubes, cylinder, trapezoid; three original worked problems |
This independently authored supplement is aligned to MVP Secondary Math II, Module 4, Section 4.6. The MVP task supplies the instructional scope; worked examples and explanations are newly written. MathJax requires internet access for typeset equations.
Last updated .