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4.5 — What is Your Pace?
This lesson starts with the familiar idea of speed, then explains inverses using tables, equations, and coordinates. The radical sections near the end support the separate Ready and Go homework.
1. In class: A Develop Understanding Task
Rate: miles per hour and minutes per mile
A rate compares two changing quantities. When the units are reversed, the numerical rate also changes. Chandler bikes 12 miles per hour. This is the same as one mile every five minutes, because \(60\div12=5\).
If \(t\) is elapsed time in minutes and \(d\) is distance in miles, Chandler’s distance rule is
\[d=\frac{12}{60}t=\frac{t}{5}.\]For \(t=0,5,15,30,60\) minutes, multiply by \(\frac15\).
| Time \(t\), minutes | 0 | 5 | 15 | 30 | 60 |
|---|---|---|---|---|---|
| Distance \(d\), miles | 0 | 1 | 3 | 6 | 12 |
After 30 minutes, \(d(30)=30/5=6\) miles. On a graph of distance against time, the slope is \(1/5\) mile per minute.
Isaac says he takes five minutes to bike each mile. If the input is distance \(d\) and the output is time \(t\), Isaac’s rule is \(t=5d\). That means 6 miles take \(5(6)=30\) minutes. There is no speed difference: Chandler and Isaac both travel 12 mph.
In the first rule, \(t=15\) minutes gives \(d=3\) miles. In the reversed rule, \(d=3\) miles gives \(t=15\) minutes. The input and output trade roles:
| Distance from time | Time from distance |
|---|---|
| Input 15 min → Output 3 miles | Input 3 miles → Output 15 min |
| \((15,3)\) | \((3,15)\) |
What makes two functions inverses?
A function takes an input and produces an output. Its inverse reverses that correspondence: it takes the old output and returns the old input. The inverse of \(f\) is written \(f^{-1}\), pronounced “f inverse.” The \(-1\) is not an exponent here; it does not mean \(1/f(x)\).
| Input to \(f\) | −4 | 0 | 3 |
|---|---|---|---|
| Output of \(f\) | 7 | 2 | −1 |
| Input to \(f^{-1}\) | 7 | 2 | −1 |
| Output of \(f^{-1}\) | −4 | 0 | 3 |
Since \(f(3)=-1\), we know \(f^{-1}(-1)=3\).
For our biking rules, \(f(t)=t/5\) returns distance in miles. Its inverse \(f^{-1}(d)=5d\) returns minutes. An input of 6 now means six miles, not six minutes. Interchanging the axes’ units is an essential part of interpreting the inverse correctly.
Suppose \(f(x)=2x+3\). Begin with \(y=2x+3\), exchange the input and output letters, and then isolate the new output:
\[x=2y+3\quad\Longrightarrow\quad x-3=2y\quad\Longrightarrow\quad y=\frac{x-3}{2}.\]Thus \(f^{-1}(x)=(x-3)/2\). Check a pair: \(f(4)=11\), and \(f^{-1}(11)=4\). We are simply undoing “multiply by two, then add three” in reverse order.
Graphing an inverse and the function test
On a coordinate plane, reversing the coordinates turns \((a,b)\) into \((b,a)\). This is a reflection across \(y=x\). To graph an inverse, identify important points, switch each point’s coordinates, and connect the reflected points in the matching order and shape.
Does the inverse always pass the function test? No. An ordinary function assigns exactly one output to each input. For its inverse also to be a function, the original must never assign the same output to two different inputs. Such a function is called one-to-one. Graphically, its horizontal lines meet the original graph at most once: the horizontal line test.
Imagine \(f\) contains \((1,4)\) and \((3,4)\). It is allowed for two different inputs to have the same output in an ordinary function. But the inverse relation contains \((4,1)\) and \((4,3)\). Now input 4 produces two outputs, so the inverse is not a function.
The vertical line test checks whether the original graph is a function. The horizontal line test checks whether its inverse will be a function. The inverse relation can always be drawn by exchanging coordinates, even if it cannot be represented by one output for every input.
2. Ready (problems 1–7): square roots and square areas
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 square with side length \(s\) has area \(A=s^2\). Since physical lengths are nonnegative, the side length is \(s=\sqrt A\), the principal (nonnegative) square root. For example, 49 square centimeters of area means a side length of 7 cm.
If the area is \((x-8)^2\) square feet, then the side length is
\[\sqrt{(x-8)^2}=|x-8|\text{ ft}.\]The absolute value is necessary: \(\sqrt{z^2}=|z|\), not always \(z\). If the problem additionally says that the side length is exactly \(x-8\) feet, then \(x-8\ge0\), so \(x\ge8\).
Find the side length when \(A=9x^2+30x+25\) square inches:
\[9x^2+30x+25=(3x+5)^2,\qquad s=|3x+5|\text{ in}.\]First and last terms are squares, and the middle term equals \(2(3x)(5)=30x\).
In the tables of \(f(x)=\sqrt{x}\), distinguish these situations:
| Input form | How to simplify | Example |
|---|---|---|
| Perfect-square number | Take its nonnegative square root | \(\sqrt{81}=9\) |
| Number with a square factor | Pull out the largest square factor | \(\sqrt{72}=\sqrt{36\cdot2}=6\sqrt2\) |
| Perfect-square expression | Recognize \((x-a)^2\) | \(\sqrt{x^2-12x+36}=|x-6|\) |
The area expressions contain unknown variables. A square's side is the nonnegative root of its area, so \(\sqrt{(x-11)^2}=|x-11|\), \(\sqrt{(5a+6)^2}=|5a+6|\), and \(\sqrt{x^2-2kx+k^2}=|x-k|\). Replacing these by \(x-11\), \(5a+6\), or \(x-k\) requires knowing those expressions are nonnegative.
Check your work — Ready 1–7 answer key
- 4 inches.
- \(|x-11|\) feet.
- \(|5a+6|\) centimeters.
- \(x>24\) because a square has positive side length.
- Outputs in row order: \(1,2,3,4,5,6,7,8,9,10\).
- Outputs in row order: \(5,5\sqrt2,5\sqrt3,10,5\sqrt5,5\sqrt6,5\sqrt7,10\sqrt2,15,5\sqrt{10}\).
- Outputs in row order: \(|x-1|,|x-2|,|x-3|,|x-4|,|x-5|,|x-6|,|x-7|,|x-8|,|x-9|,|x-10|\).
3. Set (problems 8–14): apply inverse ideas
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.
For the shared ideas used in these exercises, revisit the In class explanation.
Problem 8 — Ordered pairs. If a relation contains \((-13,5)\), its inverse contains \((5,-13)\). Reverse every pair, including pairs containing zero or negative numbers. To determine whether the reversed relation is a function, check whether the original list has duplicate output values.
Problems 9–10 — A piecewise graph. The packet’s graph descends and then rises; some horizontal lines intersect it twice. The reflected relation will have a section that doubles back sideways. Therefore its inverse graph does not pass the vertical line test.
Problem 11 — The suitcase story. Packing something last means it is near the top and available to remove first. The order of operations reverses, just as inverse functions undo steps in the reverse order. This analogy helps explain inverses but does not by itself establish whether an inverse is a function.
Problems 12–14 — Table and reflection. The packet table shows \((-10,-2),(-6,-1),(-2,0),(2,1),(6,2)\). Its inverse points are \((-2,-10),(-1,-6),(0,-2),(1,2),(2,6)\). Each original output appears only once, so the inverse of this finite table is also a function. Reflect these points over \(y=x\) on the provided grid.
In problems 9–10, graph the inverse relation even though it is not itself a function. Problem 14 asks about the inverse of the tabulated points in problem 12; that finite inverse does pass the function test. A table alone does not determine a unique rule for values between its listed inputs.
Check your work — Set 8–14 answer key
- \(f^{-1}=\{(5,-13),(-9,-9),(-2,-5),(-5,-1),(-4,0),(6,4),(10,9),(32,14)\}\)
- Plot/construction not shown. Reflect both segments across \(y=x\) by swapping the coordinates of their endpoints. If each unlabeled grid square represents one unit, the reflected segments join \((2,-7),(-2,-1),(4,1)\).
- No. The original graph has repeated output values (fails the horizontal line test), so its inverse has multiple outputs for an input (fails the vertical line test).
- Pack the pajamas last, so they can be removed first; unpacking reverses the packing order.
- \(g^{-1}=\{(-2,-10),(-1,-6),(0,-2),(1,2),(2,6)\}\)
- Original points: \((-10,-2),(-6,-1),(-2,0),(2,1),(6,2)\). Inverse points are those in 12. Reflection line: \(y=x\). Plot not shown.
- Yes. The original outputs are all distinct, so each inverse input has exactly one output.
4. Go (problems 15–20): multiplying radical expressions
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.
Use the distributive property and the product rule for nonnegative radicands: \(\sqrt a\sqrt b=\sqrt{ab}\). In particular, \((\sqrt a)^2=a\). Combine like radicals only after simplifying their radicands.
Multiply \(\sqrt5(4+3\sqrt5)\).
\[\sqrt5(4+3\sqrt5)=4\sqrt5+3(\sqrt5)^2=4\sqrt5+15.\]We distributed first, then replaced \((\sqrt5)^2\) by 5.
Multiply \((2+\sqrt7)(3-2\sqrt7)\).
\[\begin{aligned}(2+\sqrt7)(3-2\sqrt7)&=6-4\sqrt7+3\sqrt7-2(7)\\&=-8-\sqrt7.\end{aligned}\]Every term in the first binomial must multiply every term in the second. Then combine like radical terms.
Multiply \((5+3\sqrt2)(5-3\sqrt2)\).
\[\begin{aligned}(5+3\sqrt2)(5-3\sqrt2)&=25-15\sqrt2+15\sqrt2-9(2)\\&=25-18=7.\end{aligned}\]The radical terms cancel. More generally, \((a+b)(a-b)=a^2-b^2\); these are called conjugate binomials.
Multiply \((3-2\sqrt3)(4+5\sqrt3)\).
\[\begin{aligned}(3-2\sqrt3)(4+5\sqrt3)&=12+15\sqrt3-8\sqrt3-10(3)\\&=-18+7\sqrt3.\end{aligned}\]Remember: \(\sqrt3\cdot\sqrt3=3\), but \(\sqrt3\) and a plain number are not like terms.
Check your work — Go 15–20 answer key
- \(15+4\sqrt3\)
- \(-66+12\sqrt{11}\)
- \(15-8\sqrt2\)
- \(-43\)
- \(-29\)
- \(41-17\sqrt6\)
5. Additional learning resources
| Resource | What to use it for | Scope / What to skip |
|---|---|---|
| Paul’s Online Math Notes — Inverse Functions | One-to-one functions, input/output reversal, worked algebraic examples. | Some extra material. Begin with the meaning of inverses and one-to-one; save more complicated inverse-finding or verification for later. |
| Khan Academy — Reading inverse values from a graph | Matched practice for reversing coordinates and drawing a reflected inverse graph. | Close match. Focus on reading and plotting inverse points. |
| Professor Leonard — One to One Functions (Precalculus / College Algebra 50) | Optional longer explanation of one-to-one functions and why they matter for inverses. | Some extra material. Use the one-to-one and function-test discussion; timestamps not verified. |
| Professor Leonard — Finding Inverse Functions (Precalculus / College Algebra 51) | Optional extended lesson for algebraic inverse procedures. | Significantly broader. The algebraic linear examples are useful; skip domain restriction and more advanced examples for now. Timestamps not verified. |
6. Teacher’s review (for teachers and parents)
Main task, problems 5–8 — Critical distinction. Chandler and Isaac describe the same speed (12 mph = 5 min/mile), not two different travel behaviors. Their descriptions become inverse functions only if one chooses minutes → miles and the other chooses miles → minutes. If both are modeled as distance as a function of time, they are the same function. Clarify the choice of inputs and outputs before stating they are inverse.
Set, problems 9–10 — Inverse relation versus inverse function. The inverse of the graphed V-like piecewise function fails the vertical line test. Problem 9’s request to “graph \(f^{-1}(x)\)” may imply a function exists. Clearer wording: “Reflect the graph across \(y=x\) to graph the inverse relation. Determine whether that relation is a function.”
Ready, problems 2, 3, 7 — Principal square root. For variable inputs, the square root of a squared expression is its absolute value. State any needed nonnegativity restrictions rather than automatically dropping absolute value signs.
7. Packet coverage
| Packet problems | Mathematics practiced | Lesson location |
|---|---|---|
| Main task 1–7 | Equivalent rates, units, tables, graphs, equations, comparison at 30 minutes | In class |
| Main task 8 | Inverse relationships | In class and clarification |
| Ready 1–4 | Square area, side length, variable restrictions | Section 2 |
| Ready 5–7 | Square root tables, simplifying radicals, perfect-square polynomials | Section 2 |
| Set 8–14 | Reverse ordered pairs, inverse graphs, one-to-one/function tests, reversing order | In class and Section 3 |
| Go 15–20 | Distributive multiplication, radical products, conjugate binomials | Section 4 |
Companion to the MVP Secondary Math II Module 4 (2017), Section 4.5, pp. 22–26. Original curriculum © Mathematics Vision Project, CC BY 4.0. This independently written supplement is not affiliated with or endorsed by MVP.
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