In plain English: HSF.BF.B.4 is the Common Core functions standard that asks students to find inverse functions: solve f(x) = c for a simple invertible function and write an expression for the inverse. Its advanced (+) parts add verifying inverses by composition, reading inverse values from graphs and tables, and restricting a domain to make a function invertible. Part a is usually taught in Algebra II, the (+) parts in Precalculus.
Find inverse functions.
a.Solve an equation of the form f(x) = c for a simple function f that has an inverse and write an expression for the inverse. For example, f(x) = 2x3 or f(x) = (x+1)/(x-1) for x ≠ 1.
b.(+) Verify by composition that one function is the inverse of another.
c.(+) Read values of an inverse function from a graph or a table, given that the function has an inverse.
d.(+) Produce an invertible function from a non-invertible function by restricting the domain.
Common Core State Standards for Mathematics · Domain: Building Functions (BF) · Cluster: Build new functions from existing functions Also written as HSF-BF.B.4 or F-BF.4 · Official standard
An inverse function runs a function backward: if f takes an input to an output, f⁻¹ takes that output back to the input. Students find inverses the way part a of the standard describes, by solving f(x) = c for x. The answer, written in terms of c, is a formula for f⁻¹(c). They practice this with linear, cubic and rational functions, including the two official examples, f(x) = 2x³ and f(x) = (x + 1)/(x - 1) for x ≠ 1.
The (+) parts of the standard then ask for more than a formula. Students verify inverses by computing both compositions, read inverse values straight from a table or graph without any formula, and turn a function that fails the horizontal line test, such as a quadratic, into an invertible one by restricting its domain.
Learning Objectives
By the end of this lesson, students will be able to:
Solve f(x) = c for x when f is a simple invertible function, and use the result to write an expression for f⁻¹
Verify by composition that f(g(x)) = x and g(f(x)) = x when f and g are inverses
Read values of an inverse function from a table or graph of the original function
Restrict the domain of a function that is not one-to-one so that it has an inverse, and state the inverse's domain and range
Prior Knowledge Required
Students should already be comfortable with:
Function notation, domain and range HSF.IF.A.1
Evaluating functions and interpreting f(a) = b in context HSF.IF.A.2
Rearranging formulas to solve for a chosen variable HSA.CED.A.4
Composing two functions HSF.BF.A.1
Cube roots and square roots of real numbers 8.EE.A.2
Post a taxi fare rule, C(m) = 3 + 2m, where C is the fare in dollars for a ride of m miles. Ask the question below and give students three minutes with a partner.
Warm-Up Prompt
"Your ride cost $19. How many miles did you travel? Now write one rule that turns any fare c into the number of miles."
Most pairs undo the steps: subtract 3, then divide by 2, so 19 dollars means 8 miles. Record the general rule m = (c - 3)/2 next to the original. Ask which rule answers "how much?" and which answers "how far?" Name the second rule the inverse of the first, and point out that it came from solving C(m) = c for m, which is exactly what part a of the standard asks for.
Direct Instruction25 minutes
Part 1: Finding an inverse by solving f(x) = c (standard a). Write the routine on the board:
Set the function equal to a constant: write f(x) = c, where c stands for any output.
Solve for x: undo the operations of f in reverse order. The result is x in terms of c.
Name the result: that expression is f⁻¹(c). Rename the variable to write f⁻¹(x) if you like.
State any restrictions: note inputs of f⁻¹ that would divide by zero or take an even root of a negative number.
Check: pick one input a, compute b = f(a), and confirm that f⁻¹(b) = a.
Stress that f⁻¹ is notation for the inverse, not 1/f. Work the two official examples first, then the others below.
Official example: a cubic
f(x) = 2x³. Solve 2x³ = c for x. As a check, solve f(x) = 54.
Equation: x³ = c/2, so f⁻¹(x) = ∛(x/2); f(x) = 54 gives x = ∛27 = 3
Official example: a rational function
f(x) = (x + 1)/(x - 1) for x ≠ 1. Solve (x + 1)/(x - 1) = c: multiply by x - 1 to get x + 1 = cx - c, then collect the x terms.
Equation: x(c - 1) = c + 1, so f⁻¹(x) = (x + 1)/(x - 1) for x ≠ 1: f is its own inverse
Verifying by composition (standard b)
Show that f(x) = 4x - 7 and g(x) = (x + 7)/4 are inverses.
Equation: f(g(x)) = 4·(x + 7)/4 - 7 = x and g(f(x)) = (4x - 7 + 7)/4 = x
Reading an inverse from a table (standard c)
An increasing function h has the values h(-2) = -9, h(-1) = -2, h(0) = -1, h(1) = 0, h(2) = 7, h(3) = 26. Find h⁻¹(7) and h⁻¹(-2).
Equation: Find each value in the output row: h⁻¹(7) = 2 and h⁻¹(-2) = -1
Restricting the domain (standard d)
p(x) = x² + 4 has p(-1) = p(1) = 5, so it is not one-to-one. Restrict it to x ≥ 0 and solve x² + 4 = c.
Equation: x = √(c - 4), so p⁻¹(x) = √(x - 4) with domain x ≥ 4 and range y ≥ 0
Part 2: Graphs, compositions and restrictions (standards b, c, d). Show Diagram 1. Each point (a, b) on the graph of f reflects across y = x to the point (b, a) on the graph of f⁻¹, so students can read f⁻¹ from the graph of f by finding the output first. Explain that composition checks the same idea with algebra: f⁻¹ undoes f, so f⁻¹(f(x)) = x on the domain of f, and f(f⁻¹(x)) = x on the domain of f⁻¹. Both must hold. Then show Diagram 2: a horizontal line that meets a graph twice means two inputs share one output, so no inverse function exists. Keeping only one side of the vertex removes the problem, and the restricted domain becomes the range of the inverse.
Guided Practice15 minutes
Pairs work four tasks on whiteboards and show them after each one. (1) For f(x) = 5x + 2, solve f(x) = 37 (x = 7), then write f⁻¹(x) = (x - 2)/5. (2) Find the inverse of g(x) = x³ - 4 (g⁻¹(x) = ∛(x + 4)). (3) For h(x) = 6/(x + 2), x ≠ -2, solve h(x) = c (x = 6/c - 2, so h⁻¹(x) = 6/x - 2 for x ≠ 0). (4) Verify by composition that p(x) = (x - 5)/3 and q(x) = 3x + 5 are inverses. Listen for these errors: writing 1/f(x) for f⁻¹(x), undoing steps in the wrong order, and checking only one composition.
Independent Practice15 minutes
Students work alone on four problems. (1) Find the inverse of k(x) = 8 - 3x (k⁻¹(x) = (8 - x)/3). (2) Solve m(x) = c for m(x) = (2x + 3)/(x - 4), x ≠ 4, and write m⁻¹(x) = (4x + 3)/(x - 2), x ≠ 2. (3) Verify by composition that r(x) = ∛x + 1 and s(x) = (x - 1)³ are inverses. (4) Restrict w(x) = (x + 1)² to x ≥ -1 and find w⁻¹(x) = √x - 1 for x ≥ 0. Circulate and ask each student to check one answer by choosing an input, applying the function and then the inverse.
Closure5-10 minutes
Exit ticket: (1) For f(x) = x/2 + 6, find f⁻¹(10). (Answer: 8.) (2) A table shows g(1) = 6, g(2) = 11, g(3) = 18, g(4) = 27. What is g⁻¹(18)? (Answer: 3.) (3) Give a domain restriction that makes v(x) = (x + 3)² invertible, and explain in one sentence why it works. (Answer: x ≥ -3 or x ≤ -3, because each side of the vertex passes the horizontal line test.)
Differentiation Strategies
For Struggling Students
Use a two-column "do and undo" chart: list the steps of f in order on the left, then write the opposite steps in reverse order on the right
Start with input-output tables and arrow diagrams before any algebra, so students see the inverse as the reversed arrows
Give a checklist for composition: substitute, simplify, and confirm that exactly x is left, then do the other order
For Advanced Students
Ask for all linear functions that are their own inverse, and explain the answer with the graph and the mirror line y = x
Ask students to find the inverse of f(x) = (ax + b)/(cx + d) in general and state when f is its own inverse
Ask students to restrict q(x) = x² - 4x in two different ways and compare the two inverse formulas
Assessment Guidance
What to Look For
Check that students state an inverse as a function with its own domain, not only as a formula: √(x - 4) is only the inverse of x² + 4 when the original domain is restricted to x ≥ 0. In composition work, look for both orders written out and simplified to exactly x. When students read inverse values from a table or graph, listen for the phrase "find the output first": a common error is to look up the given number in the input row.
02
Classroom Activities
3 Activities
1
Inverse Match-Up
20 minPairs
Pairs match 8 function cards with 8 inverse cards and must prove each match by composition. Two extra decoy cards look like inverses but are not, so students cannot finish by elimination alone.
The Cards
Function cards: x + 9, 7x, 2x - 10, x³ + 2, 1/x (x ≠ 0), (x - 1)/4, √(x - 3) for x ≥ 3, and 3/(x - 5) for x ≠ 5
Inverse cards: x - 9, x/7, x/2 + 5, ∛(x - 2), 1/x (x ≠ 0), 4x + 1, x² + 3 for x ≥ 0, and 3/x + 5 for x ≠ 0
Decoy cards: x/2 - 5 and ∛x - 2
Procedure
Partner A proposes a match; Partner B computes f(g(x)) and Partner A computes g(f(x)); the match counts only when both equal x
For each decoy, pairs write which function it seems to belong to and show the composition that fails, for example 2(x/2 - 5) - 10 = x - 20
Pairs note the one function that is its own inverse and explain why its graph is symmetric about y = x
Discussion Questions
Why does √(x - 3) need the restriction x ≥ 0 on its inverse card?
Which matches could you predict by "undoing" the steps before doing any composition?
Modification for Distance Learning
Put the cards on a shared slide as draggable text boxes. Pairs type both compositions in the speaker notes before they move a card into place.
2
Charging Table Detectives
20 minGroups of 3-4
Groups read inverse values from a table and a graph with no formula at all. The context is a phone charging from 20% battery; B(t) is the charge, in percent, after t minutes on the charger.
The Data
t (minutes)
0
10
20
30
40
50
60
B(t) (%)
20
38
54
68
80
90
98
Procedure
Confirm that B is increasing, so every charge level in the table comes from exactly one time
Answer in words and in notation: B⁻¹(80) = 40 means the phone reaches 80% after 40 minutes; also find B⁻¹(54) and B⁻¹(98)
Build the table of B⁻¹ by swapping the rows, then plot both B and B⁻¹ on one grid with the line y = x
Estimate B⁻¹(95) from the graph and explain why the table only shows that it is between 50 and 60
Discussion Questions
What are the input and the output of B⁻¹, with units?
Why would B⁻¹ fail to exist if the charge stayed at 98% from minute 60 to minute 70?
3
Restrict and Rescue Stations
20 minGroups of 3
Each station has a function that is not one-to-one. Groups sketch it, show a horizontal line that meets it twice, choose a domain restriction, and find the inverse with its domain and range.
Station Cards
Station A: f(x) = x² - 6x + 10. Complete the square: (x - 3)² + 1. On x ≥ 3 the inverse is 3 + √(x - 1), with domain x ≥ 1
Station B: g(x) = |x - 4|. On x ≥ 4, g(x) = x - 4 and the inverse is x + 4, with domain x ≥ 0
Station C: h(x) = 9 - x². On x ≥ 0 the inverse is √(9 - x), with domain x ≤ 9
Procedure
Spend 6 minutes per station; the recorder writes the restriction, the inverse, and one check point such as f(5) = 5, so f⁻¹(5) = 5
At Station A, groups also find the inverse for the other choice, x ≤ 3, which is 3 - √(x - 1)
Groups post their answers and compare restrictions with the next group
Challenge Variation
Ask groups to find the largest interval containing x = 1 on which k(x) = x² - 4 is one-to-one, and to write the inverse on that interval.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: A Function and Its Inverse Reflected Across y = x
Graphs of f(x) = x³/8 for -4 ≤ x ≤ 4 and f⁻¹(x) = 2∛x, drawn to scale. Each point (a, b) on f matches the point (b, a) on f⁻¹, so the value of f⁻¹ at a number is found by locating that number as an output of f.
Diagram 2: Restricting a Domain to Make a Function Invertible
The parabola f(x) = (x - 2)², drawn to scale, fails the horizontal line test: y = 9 meets it at x = -1 and x = 5. Keeping only x ≥ 2 makes f one-to-one, and its inverse is f⁻¹(x) = 2 + √x for x ≥ 0.
04
Homework Assignment
~30 min
HSF.BF.B.4 Homework: Finding Inverse Functions
Directions: Show all work. When you write an inverse, state its domain if it is not all real numbers. Check every inverse you find with at least one input-output pair.
Part 1: Solving f(x) = c and Writing the Inverse (Problems 1-2)
Let f(x) = 4x³ - 1. (a) Solve f(x) = 31. (b) Solve f(x) = c for x, and use your result to write f⁻¹(x). (c) Check your formula with the pair you found in part (a).
Let f(x) = (2x - 1)/(x + 3) for x ≠ -3. (a) Solve f(x) = c for x and write f⁻¹(x). (b) For which value of c does f(x) = c have no solution? Explain what this means for the domain of f⁻¹.
Part 2: Composition, Tables and Graphs (Problems 3-4)
(a) Verify by composition that f(x) = (x³ + 4)/2 and g(x) = ∛(2x - 4) are inverses: compute f(g(x)) and g(f(x)). (b) A student claims that p(x) = 3x - 4 and q(x) = x/3 + 4 are inverses. Compute p(q(x)) and explain why the claim is false.
A print shop's table shows the cost C(n), in dollars, of printing n T-shirts: C(10) = 180, C(20) = 310, C(30) = 440, C(40) = 570, C(50) = 700. (a) Find C⁻¹(440) and explain what it means. (b) Find C⁻¹(700). (c) The costs fit C(n) = 50 + 13n. Write an expression for C⁻¹ and use it to find how many shirts cost $843.
Part 3: Restricting the Domain (Problems 5-6)
Let f(x) = x² + 2x - 3. (a) Show that f is not one-to-one by finding two inputs with the same output. (b) Complete the square and restrict the domain to x ≥ -1. (c) Find f⁻¹ on that domain and state its domain and range. (d) Check that f⁻¹(5) = 2.
A rule of thumb for a car's braking distance on dry pavement is d(v) = v²/20, where v is the speed in miles per hour and d is the distance in feet. (a) Explain why d has no inverse if v can be any real number. (b) What restriction fits the context? (c) Find d⁻¹ and use it to estimate the speed of a car that needed 125 feet to stop. (d) Find d⁻¹(80) and interpret it.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Solving f(x) = c
Correct x in terms of c, inverse written with restrictions
Correct method with one algebra error
Not solved or inverse missing
Composition
Both orders computed and simplified to x
One order only, or a simplification error
No composition shown
Tables and Context
Inverse values read correctly and interpreted with units
Values correct, interpretation missing
Inputs and outputs confused
Domain Restriction
Valid restriction, inverse with domain and range, check shown
Restriction or inverse correct, not both
Missing or incorrect
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Let f(x) = 6x - 11. Solve f(x) = 25.
Answer: C
Add 11 to both sides to get 6x = 36, then divide by 6: x = 6. Check: 6(6) - 11 = 25. Choice A evaluates f(25) instead of solving f(x) = 25. Choice B subtracts 11 instead of adding it, and choice D forgets to divide by 6.
Question 2 of 20 · Multiple Choice
What is the inverse of f(x) = (x - 8)/3?
Answer: A
f subtracts 8 and then divides by 3, so the inverse multiplies by 3 and then adds 8: f⁻¹(x) = 3x + 8. Check: f(3x + 8) = (3x + 8 - 8)/3 = x. Choice C undoes the subtraction but divides by 3 again instead of multiplying, so f((x + 8)/3) = (x - 16)/9, not x, and choice D is the reciprocal 1/f(x), which is not the inverse.
Question 3 of 20 · Multiple Choice
What is the inverse of f(x) = 5x³?
Answer: B
Solve 5x³ = c: x³ = c/5, so x = ∛(c/5) and f⁻¹(x) = ∛(x/5). Check: f(∛(x/5)) = 5 · x/5 = x. Choice A divides after taking the cube root, which undoes the steps in the wrong order: f(∛x/5) = x/25. Choice D is the reciprocal of f.
Question 4 of 20 · Multiple Choice
Which function g is the inverse of f(x) = 2x + 6?
Answer: D
Check by composition: g(f(x)) = (2x + 6)/2 - 3 = x + 3 - 3 = x and f(g(x)) = 2(x/2 - 3) + 6 = x - 6 + 6 = x. For choice A, g(f(x)) = x + 3 - 6 = x - 3, so it fails: the 6 must be divided by 2 too. Choice B is the reciprocal and choice C is the opposite of f.
Question 5 of 20 · Multiple Choice
A student shows that f(g(x)) = x for two functions f and g. To verify by composition that f and g are inverses, what else should the student show?
Answer: A
Verifying inverses by composition means checking both orders: f(g(x)) = x on the domain of g and g(f(x)) = x on the domain of f. One order can hold while the other fails, as with f(x) = x² and g(x) = √x, where g(f(x)) = |x|. Choice B describes reciprocals, not inverses.
Question 6 of 20 · Multiple Choice
A table for an invertible function f shows f(1) = 3, f(2) = 5, f(3) = 9, f(4) = 17 and f(5) = 33. What is f⁻¹(9)?
Answer: B
f⁻¹(9) is the input whose output is 9. The table shows f(3) = 9, so f⁻¹(9) = 3. Choice A treats f⁻¹ as a reciprocal. Choice D looks for 9 in the input row instead of the output row, a common error when reading an inverse from a table.
Question 7 of 20 · Multiple Choice
The graph of an invertible function f passes through (-2, 5) and (3, -1). Which point must be on the graph of f⁻¹?
Answer: C
Every point (a, b) on f gives the point (b, a) on f⁻¹, so (-2, 5) becomes (5, -2). The point (3, -1) would give (-1, 3), which is not offered. Choice A negates both coordinates, which reflects through the origin, and choice B swaps and negates them, which reflects across y = -x instead of y = x.
Question 8 of 20 · Multiple Choice
T(h) is the temperature in °F h hours after sunrise, from sunrise to early afternoon, and it is increasing on that interval. What does T⁻¹(62) = 4 mean?
Answer: D
T⁻¹ takes a temperature back to a time, so T⁻¹(62) = 4 means T(4) = 62: four hours after sunrise it is 62°F. Choice A swaps the roles of input and output, and choice B reads the statement as a rate of change.
Question 9 of 20 · Multiple Choice
Which domain restriction makes f(x) = (x + 5)² one-to-one while keeping every output value y ≥ 0?
Answer: B
The vertex is at x = -5, and f increases on x ≥ -5, taking every value y ≥ 0 exactly once. Choices A and C are also one-to-one but lose outputs: on x ≥ 0 the smallest output is 25. Choice D fails the horizontal line test, since f(-7) = f(-3) = 4.
Question 10 of 20 · Multiple Choice
Find the inverse of f(x) = x² - 7 for x ≥ 0.
Answer: C
Solve x² - 7 = c: x² = c + 7 and, because x ≥ 0, x = √(c + 7). So f⁻¹(x) = √(x + 7) for x ≥ -7. Choice B is not a function: the restriction x ≥ 0 is what removes the minus sign. Choice A adds 7 on the wrong side.
Question 11 of 20 · Multiple Choice
What is the inverse of f(x) = (x + 4)/(x - 2) for x ≠ 2?
Answer: B
Solve (x + 4)/(x - 2) = c: x + 4 = cx - 2c, so x - cx = -2c - 4 and x = (2c + 4)/(c - 1). Thus f⁻¹(x) = (2x + 4)/(x - 1) for x ≠ 1. Choice A is the reciprocal of f. Choice C assumes f is its own inverse, which is true for (x + 1)/(x - 1) but not here: f(0) = -2, while f(-2) = -1/2.
Question 12 of 20 · Multiple Choice
Let h(x) = 2√x + 1. Solve h(x) = 9.
Answer: A
Subtract 1 and divide by 2: √x = 4, so x = 16. Check: 2√16 + 1 = 9. Choice B stops at √x = 4 without squaring. Choice C adds 1 instead of subtracting (√x = 5), and choice D is h(9), which evaluates instead of solving.
Question 13 of 20 · Multiple Choice
Let f(x) = 3x - 2 and g(x) = (x + 2)/3. What is g(f(5))?
Answer: C
f(5) = 13 and g(13) = (13 + 2)/3 = 5. Because g is the inverse of f, g(f(5)) returns the original input 5. Choice A stops after f(5), and choice B computes g(5) instead of g(f(5)).
Question 14 of 20 · Multiple Choice
Why does f(x) = x², with domain all real numbers, have no inverse function?
Answer: D
An inverse must send each output back to one input, but f(-3) = f(3) = 9, so f⁻¹(9) would have to be both -3 and 3. The graph fails the horizontal line test. Choice A is not the reason: f(x) = x³ is not a line but has an inverse. Choice C is false, since the graph passes through (0, 0).
Question 15 of 20 · Short Answer
Let f(x) = 7 - 2x³. Solve f(x) = -47, then write an expression for f⁻¹(x).
7 - 2x³ = -47 gives -2x³ = -54, so x³ = 27 and x = 3. In general 7 - 2x³ = c gives x³ = (7 - c)/2, so f⁻¹(x) = ∛((7 - x)/2). Check: f⁻¹(-47) = ∛(54/2) = ∛27 = 3.
Question 16 of 20 · Short Answer
Verify by composition that f(x) = 4/(x - 1), x ≠ 1, and g(x) = 4/x + 1, x ≠ 0, are inverses.
f(g(x)) = 4/((4/x + 1) - 1) = 4/(4/x) = x for x ≠ 0. g(f(x)) = 4/(4/(x - 1)) + 1 = (x - 1) + 1 = x for x ≠ 1. Both compositions give x, so f and g are inverses.
Question 17 of 20 · Short Answer
An object is dropped from a tall tower. The distance it has fallen after t seconds is d(t), in meters: d(0) = 0, d(1) = 4.9, d(2) = 19.6, d(3) = 44.1, d(4) = 78.4. Find d⁻¹(44.1) and say what it means. Then write the table for d⁻¹.
d⁻¹(44.1) = 3: the object has fallen 44.1 meters after 3 seconds. The table for d⁻¹ swaps the rows: inputs 0, 4.9, 19.6, 44.1, 78.4 meters give outputs 0, 1, 2, 3, 4 seconds. The inverse exists because the distance increases for t ≥ 0.
Question 18 of 20 · Short Answer
Let f(x) = (x - 6)² + 2. Choose a domain restriction that makes f invertible, find f⁻¹, and state its domain and range.
Restrict to x ≥ 6. Solve (x - 6)² + 2 = c: x - 6 = √(c - 2), so f⁻¹(x) = 6 + √(x - 2) with domain x ≥ 2 and range y ≥ 6. The other choice, x ≤ 6, gives f⁻¹(x) = 6 - √(x - 2) with range y ≤ 6. Check: f(7) = 3 and 6 + √(3 - 2) = 7.
Question 19 of 20 · Short Answer
The Fahrenheit temperature for a Celsius temperature C is F(C) = 9C/5 + 32. Write an expression for the inverse, which gives C from F, and use it to convert 104°F.
Solve 9C/5 + 32 = F: 9C/5 = F - 32, so C = 5(F - 32)/9. For F = 104: C = 5(72)/9 = 40°C. Check: 9(40)/5 + 32 = 104.
Question 20 of 20 · Short Answer
Let f(x) = x² - 9 and g(x) = √(x + 9). Compute f(g(x)) and g(f(x)). On what domain of f are f and g inverses?
f(g(x)) = (√(x + 9))² - 9 = x for x ≥ -9. g(f(x)) = √(x² - 9 + 9) = √(x²) = |x|, which equals x only when x ≥ 0. So f and g are inverses when the domain of f is restricted to x ≥ 0. On all real numbers, g(f(-2)) = 2, not -2.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSF.BF.B.4 mean?
HSF.BF.B.4 means students can find inverse functions. Part a asks them to solve f(x) = c for a simple invertible function and turn the answer into a formula for the inverse. The (+) parts b, c and d add verifying inverses by composition, reading inverse values from tables and graphs, and restricting a domain so that a function becomes invertible.
Is HSF.BF.B.4 Algebra 2 or Precalculus?
Both, depending on the part. Part a, finding the inverse of a simple function by solving f(x) = c, is usually taught in Algebra II. Parts b, c and d are marked (+), which Common Core uses for additional mathematics in advanced courses, so they usually appear in Precalculus.
Does f⁻¹(x) mean 1/f(x)?
No. The -1 in f⁻¹ is notation for the inverse function, not an exponent. For f(x) = 2x, the inverse is f⁻¹(x) = x/2, while 1/f(x) = 1/(2x). Mixing the two is a frequent error on quizzes, so ask students to read f⁻¹ aloud as "f inverse."
How do you find the inverse of a function step by step?
Set f(x) = c and solve for x. The expression you get is f⁻¹(c).
Undo the operations of f in reverse order: for f(x) = 3x + 1, subtract 1, then divide by 3
Write the result as f⁻¹(x) = (x - 1)/3
State any restrictions, such as x ≠ 0 or x ≥ 0
Check with one pair: f(2) = 7, so f⁻¹(7) should be 2
Many textbooks swap x and y first and then solve for y. That gives the same inverse.
Why do you have to check both f(g(x)) and g(f(x))?
Because one composition can equal x while the other does not. With f(x) = x² and g(x) = √x, f(g(x)) = x for x ≥ 0, but g(f(x)) = |x|, so g(f(-5)) = 5. Part b of the standard asks students to verify the full inverse relationship, which takes both orders.
How do you read an inverse function from a table or a graph?
Find the given number among the outputs of f, then read the matching input. If a table shows f(4) = 10, then f⁻¹(10) = 4. On a graph, locate the height 10 on the y-axis, move across to the curve, and read the x-coordinate. The graph of f⁻¹ itself is the reflection of the graph of f across the line y = x.
Why do some functions not have an inverse?
A function has an inverse only if each output comes from exactly one input. If two inputs share an output, as f(-2) = f(2) = 4 for f(x) = x², the inverse would have to send 4 to two places, and that is not a function. On a graph, this shows up as a horizontal line that meets the curve more than once.
How do you choose a domain restriction for an inverse?
Pick an interval on which the function is always increasing or always decreasing and still takes every output value once. For a parabola, that means one side of the vertex: for (x - 1)², either x ≥ 1 or x ≤ 1. The choice x ≥ 1 is common because it gives the principal square root in the inverse, 1 + √x. Context can also decide it: time and length cannot be negative.
Why is f(x) = (x + 1)/(x - 1) its own inverse?
Solving (x + 1)/(x - 1) = c gives x = (c + 1)/(c - 1), which is the same rule as f. So f(f(x)) = x for every x ≠ 1, and its graph is symmetric about the line y = x. This is one of the two official examples in part a of the standard. The other, f(x) = 2x³, has the inverse ∛(x/2).
What do inverse functions lead to in later math?
Inverse functions come next in HSF.BF.B.5, where logarithms are introduced as the inverses of exponential functions: log₂ x undoes 2ˣ. Restricting domains returns in trigonometry (HSF.TF.B.6), where sine, cosine and tangent are restricted so that arcsin, arccos and arctan exist. In calculus, students use inverse functions to find derivatives of logarithms and inverse trigonometric functions.
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Related Standards
5 standards
These standards connect to HSF.BF.B.4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSF.IF.A.1Prerequisite
Understand functions, domain and range, and f(x) as the output for input x
