Inverse Functions in Precalculus: How to Find Inverses, Domains, Ranges and Graphs
Inverse functions are an important topic in precalculus, Algebra 2 and UK A-level Mathematics. They appear in algebraic equations, graph transformations, logarithms, exponential functions and real-world calculations involving money and measurement.
Many students learn a procedure for finding an inverse function: exchange x and y, solve for y, and write the answer as f⁻¹(x).
That procedure is useful, but it does not explain everything.
Why does x² need a domain restriction before it has an inverse? Why do the domain and range exchange places? Why does a square-root equation sometimes have two possible algebraic answers but only one valid inverse? And why can you not simply subtract the same percentage that was added to a price?
The key is to understand what an inverse function actually does.
An inverse reverses the operation of the original function. Once you understand that idea, you can find inverse functions systematically, check your answers and avoid the mistakes that commonly appear in precalculus examinations.
Formulas and Main Ideas to Remember
Before working through the examples, learn these essential rules.
1. Inverse-function notation
f⁻¹(x) represents the inverse function of f(x). It does not mean 1 ÷ f(x).
2. The composition rules
f(f⁻¹(x)) = x
f⁻¹(f(x)) = x
These identities hold on the appropriate domains.
3. Domain and range
Domain of f = Range of f⁻¹
Range of f = Domain of f⁻¹
4. The horizontal line test
A function has an inverse function on its stated domain if it is one-to-one. Every horizontal line must intersect its graph at most once.
5. The inverse-function procedure
Write y = f(x), exchange x and y, solve for y, and rename y as f⁻¹(x).
6. Graph of an inverse
The graph of f⁻¹ is the reflection of the graph of f across the line y = x.
7. Domain restriction for a quadratic
For f(x) = x², restricting the domain to x ≥ 0 gives f⁻¹(x) = √x. Restricting it to x ≤ 0 gives f⁻¹(x) = −√x.
These formulas form the foundation for the worked examples below.
1. What Is an Inverse Function?
A function is a rule that assigns exactly one output to each permitted input.
An inverse function reverses that rule.
Suppose a function takes an input of 3 and produces an output of 11.
The inverse takes 11 and returns 3.
We can represent this relationship as:
f(3) = 11
f⁻¹(11) = 3
The original function moves from input to output. The inverse moves from output back to input.
Think of a function as a machine that performs a sequence of operations.
If the machine adds 7, the inverse subtracts 7.
If the machine multiplies by 4, the inverse divides by 4.
If the machine first adds 3 and then squares the result, the inverse must first take the appropriate square root and then subtract 3.
The order matters because inverse functions undo operations in reverse order.
Example 1: Finding the inverse of a linear function
Find the inverse of:
f(x) = 4x − 7
Write the function using y:
y = 4x − 7
Exchange x and y:
x = 4y − 7
Add 7 to both sides:
x + 7 = 4y
Divide both sides by 4:
y = (x + 7) ÷ 4
Therefore,
f⁻¹(x) = (x + 7) ÷ 4
The inverse first adds 7 and then divides by 4, reversing the original operations.
How can you check the answer?
Substitute the inverse into the original function:
f(f⁻¹(x)) = 4((x + 7) ÷ 4) − 7
= x + 7 − 7
= x
The original function cancels the inverse and returns the input.
That is the central property of inverse functions.
2. What Is the Difference Between an Inverse and a Reciprocal?
This is a common source of confusion in precalculus.
The notation f⁻¹(x) does not mean the reciprocal of f(x).
The reciprocal is:
1 ÷ f(x)
The inverse is the function that reverses the original function.
For example, let:
f(x) = 2x
Its inverse is:
f⁻¹(x) = x ÷ 2
Its reciprocal is:
1 ÷ f(x) = 1 ÷ (2x)
These are different expressions.
The inverse divides an input by 2 to reverse multiplication by 2. The reciprocal takes the reciprocal of the function's output.
Remember this distinction when working with inverse trigonometric functions, logarithms and function composition.
3. Which Functions Have Inverse Functions?
A function has an inverse function on a given domain when every output corresponds to exactly one input.
This property is called being one-to-one.
Consider:
f(x) = x²
If x = 3, then f(3) = 9.
If x = −3, then f(−3) = 9.
Two different inputs produce the same output.
Consequently, the original function cannot be reversed uniquely over all real numbers.
If someone tells you the output is 9, you cannot determine whether the input was 3 or −3.
The function has lost information about the sign of its input.
The horizontal line test
The horizontal line test provides a graphical method for identifying one-to-one functions.
Imagine drawing horizontal lines across the graph.
If every horizontal line intersects the graph at most once, the function is one-to-one.
If any horizontal line intersects the graph more than once, the function is not one-to-one on that domain.
For example, the graph of y = x³ passes the horizontal line test.
The graph of y = x² fails because many horizontal lines above the x-axis intersect the parabola twice.
This is why the graph of a cubic function such as y = x³ has an inverse over the real numbers, whereas the full quadratic function y = x² does not.
Which familiar functions are one-to-one?
The following functions are one-to-one over their usual real domains:
Linear functions with nonzero slope.
Cubic functions of the form f(x) = x³.
Exponential functions such as f(x) = 2ˣ.
Logarithmic functions such as f(x) = log₂(x).
The following functions are not one-to-one over their full usual domains:
Quadratic functions such as f(x) = x².
Absolute-value functions such as f(x) = |x|.
Sine and cosine over the entire real number line.
Some of these functions can have inverses after their domains are restricted appropriately.
4. How to Find the Inverse of a Function Step by Step
The exchange-and-solve method is one of the most reliable methods for finding an inverse function.
Follow these steps.
Write the function as y = f(x).
Replace f(x) with y.
Exchange x and y.
Solve the resulting equation for y.
Write the answer as f⁻¹(x).
State any necessary domain restrictions.
Let's apply the method to several types of functions.
Example 2: Inverse of a linear function
Find the inverse of:
f(x) = 5x + 12
Write:
y = 5x + 12
Exchange x and y:
x = 5y + 12
Subtract 12:
x − 12 = 5y
Divide by 5:
y = (x − 12) ÷ 5
Therefore,
f⁻¹(x) = (x − 12) ÷ 5
The original function multiplies by 5 and adds 12. The inverse subtracts 12 and divides by 5.
Example 3: Inverse of a cubic function
Find the inverse of:
f(x) = x³ + 4
Write:
y = x³ + 4
Exchange x and y:
x = y³ + 4
Subtract 4:
x − 4 = y³
Take the cube root:
y = ∛(x − 4)
Therefore,
f⁻¹(x) = ∛(x − 4)
Unlike a quadratic function, a cubic function is one-to-one over all real numbers. No additional domain restriction is needed.
Example 4: Inverse of a function containing a square root
Find the inverse of:
f(x) = √(x − 2) + 5
First, identify the original domain.
The expression inside the square root must be nonnegative:
x − 2 ≥ 0
Therefore, x ≥ 2.
Now write:
y = √(x − 2) + 5
Exchange x and y:
x = √(y − 2) + 5
Subtract 5:
x − 5 = √(y − 2)
Square both sides:
(x − 5)² = y − 2
Add 2:
y = (x − 5)² + 2
Therefore,
f⁻¹(x) = (x − 5)² + 2
The original square-root function has a minimum output of 5.
Its range is y ≥ 5.
Consequently, the inverse has domain x ≥ 5 and range y ≥ 2.
The correct inverse is therefore:
f⁻¹(x) = (x − 5)² + 2, where x ≥ 5.
The restriction is essential because the inverse must reverse the original function's outputs without introducing extra values.
5. How to Find the Inverse of a Rational Function
Rational functions contain a variable in a denominator, so finding their inverses often requires collecting terms and factoring.
Consider:
f(x) = (2x + 1) ÷ (x − 3)
Find its inverse.
Write:
y = (2x + 1) ÷ (x − 3)
Exchange x and y:
x = (2y + 1) ÷ (y − 3)
Multiply both sides by y − 3:
x(y − 3) = 2y + 1
Expand:
xy − 3x = 2y + 1
Collect the terms containing y:
xy − 2y = 3x + 1
Factor out y:
y(x − 2) = 3x + 1
Divide by x − 2:
y = (3x + 1) ÷ (x − 2)
Therefore,
f⁻¹(x) = (3x + 1) ÷ (x − 2)
The original function is undefined at x = 3.
It also never produces the output 2, because solving
(2x + 1) ÷ (x − 3) = 2
would require 1 = −6, which is impossible.
Therefore, the original function has domain x ≠ 3 and range y ≠ 2.
The inverse has domain x ≠ 2 and range y ≠ 3.
Notice how the excluded values exchange places.
How to avoid mistakes with rational inverses
When finding the inverse of a rational function:
Multiply through by the denominator carefully.
Collect all terms containing the variable you want to isolate.
Factor before dividing.
Check the values that make the original denominator zero.
Compare the original range with the inverse domain.
Never assume that exchanging x and y automatically resolves the domain restrictions.
6. How to Find the Inverse of a Quadratic Function
A quadratic function needs special attention because its graph is a parabola.
Consider:
f(x) = x²
Over all real numbers, this function is not one-to-one.
However, restricting its domain to x ≥ 0 makes it one-to-one.
Find its inverse under this restriction.
Write:
y = x², where x ≥ 0
Exchange x and y:
x = y²
Take the square root:
y = √x
Therefore,
f⁻¹(x) = √x, where x ≥ 0.
If the original domain is instead restricted to x ≤ 0, the inverse becomes:
f⁻¹(x) = −√x, where x ≥ 0.
The choice of domain restriction determines which inverse you obtain.
Example 5: A quadratic with a shifted vertex
Find the inverse of:
f(x) = (x − 3)² + 4
Restrict the original domain to x ≥ 3.
Write:
y = (x − 3)² + 4
Exchange x and y:
x = (y − 3)² + 4
Subtract 4:
x − 4 = (y − 3)²
Take the square root.
Because the original domain is restricted to x ≥ 3, choose the positive square root:
√(x − 4) = y − 3
Add 3:
y = 3 + √(x − 4)
Therefore,
f⁻¹(x) = 3 + √(x − 4), where x ≥ 4.
The original range is y ≥ 4, so the inverse domain is x ≥ 4.
If the original domain had instead been restricted to x ≤ 3, the inverse would be:
f⁻¹(x) = 3 − √(x − 4), where x ≥ 4.
Both results are valid for their respective restrictions.
7. How Do the Domain and Range Change in an Inverse Function?
The domain of a function contains its permitted inputs.
The range contains its possible outputs.
An inverse reverses the relationship between inputs and outputs. Therefore, the domain and range exchange places.
Suppose the original function has:
Domain: x ≥ 2
Range: y ≥ 5
Its inverse must have:
Domain: x ≥ 5
Range: y ≥ 2
This relationship is one of the most important inverse-function rules in precalculus.
Example 6: Identifying the inverse domain and range
Suppose:
f(x) = √(x + 1) + 2
The square root requires:
x + 1 ≥ 0
Therefore, the original domain is x ≥ −1.
Since the square root is always nonnegative, the smallest output is 2.
The original range is y ≥ 2.
After finding the inverse, these restrictions exchange places.
The inverse domain is x ≥ 2.
The inverse range is y ≥ −1.
Always determine the original domain and range before finalising the inverse.
This is particularly important for square-root functions, quadratic functions and rational functions.
8. Why Is the Graph of an Inverse Reflected Across y = x?
The graph of an inverse function is the reflection of the original graph across the line y = x.
The reason follows directly from the meaning of an inverse.
If the point (a, b) lies on the graph of f, then:
f(a) = b
The inverse reverses that relationship:
f⁻¹(b) = a
Therefore, the corresponding point on the inverse graph is (b, a).
The coordinates have exchanged places.
For example, if the original graph contains the point (2, 7), the inverse graph contains the point (7, 2).
This coordinate exchange produces reflection across the line y = x.
How to sketch an inverse function
Draw the original graph.
Draw the line y = x.
Reflect important points across that line.
Exchange the original domain and range.
Check that the reflected graph satisfies the inverse formula.
This graphical method is useful for understanding inverse functions even when the algebra becomes complicated.
9. How to Prove That Two Functions Are Inverses
Finding a possible inverse is only part of the process. You can verify it using function composition.
Two functions f and g are inverses when:
f(g(x)) = x
and
g(f(x)) = x
These identities must hold for all inputs in the appropriate domains.
Example 7: Proving two linear functions are inverses
Let:
f(x) = 3x − 8
g(x) = (x + 8) ÷ 3
First, calculate f(g(x)):
f(g(x)) = 3((x + 8) ÷ 3) − 8
= x + 8 − 8
= x
Now calculate g(f(x)):
g(f(x)) = ((3x − 8) + 8) ÷ 3
= 3x ÷ 3
= x
Both compositions return x.
Therefore, f and g are inverse functions.
A common proof mistake
When dealing with restricted functions, do not ignore the domain.
For example:
√(x²) = |x|
It is not always equal to x.
It equals x when x ≥ 0.
A proof involving square roots and squared expressions must respect the relevant restrictions.
10. Real-World Applications of Inverse Functions
Inverse functions are useful whenever you need to work backwards from a final result to an original value.
They appear in temperature conversion, taxation, measurement, financial calculations and scientific models.
Application 1: Converting Fahrenheit to Celsius
The formula for converting Celsius to Fahrenheit is:
F = (9 ÷ 5)C + 32
To reverse the conversion, subtract 32 and multiply by 5 ÷ 9.
Therefore:
C = (5 ÷ 9)(F − 32)
Suppose a thermometer reads 86°F.
Substitute into the inverse formula:
C = (5 ÷ 9)(86 − 32)
C = (5 ÷ 9) × 54
C = 30°C
The inverse function converts the Fahrenheit measurement back into Celsius.
Application 2: Finding the original price before sales tax
Suppose a product costs $50 before tax and the sales tax rate is 8%.
The final price is:
Final price = 1.08 × Original price
For a $50 product:
Final price = 1.08 × 50
Final price = $54
Now suppose you know only the final price of $54.
To find the original price, divide by 1.08:
Original price = 54 ÷ 1.08
Original price = $50
The inverse operation is division by 1.08, not subtraction of 8% from the final price.
Application 3: Removing VAT from a price
Suppose a product costs £60 including VAT at a rate of 20%.
The relationship is:
Gross price = 1.20 × Net price
Therefore:
Net price = Gross price ÷ 1.20
Substitute the gross price:
Net price = 60 ÷ 1.20
Net price = £50
VAT = £60 − £50
VAT = £10
Subtracting 20% of £60 would give £48, which is incorrect.
The 20% VAT was calculated on the original net price of £50, not on the final gross price of £60.
This illustrates a broader principle: the inverse of multiplying by a factor is dividing by that factor.
11. How Inverse Functions Connect to Exponential and Logarithmic Functions
Inverse functions are particularly important in precalculus because exponential and logarithmic functions reverse one another.
Consider:
f(x) = 2ˣ
Its inverse is:
f⁻¹(x) = log₂(x)
The exponential function raises 2 to a power. The logarithm identifies the power needed to produce a given positive number.
For example:
2³ = 8
Therefore:
log₂(8) = 3
The exponential function converts 3 into 8, while the logarithm converts 8 back into 3.
This gives the identities:
log₂(2ˣ) = x
2^(log₂(x)) = x, for x > 0.
The domain of the exponential function is all real numbers, and its range is positive real numbers.
For the logarithmic inverse, the domain is x > 0 and the range is all real numbers.
The same domain-range exchange applies.
Understanding inverse functions first makes logarithmic equations and exponential equations easier to learn.
12. Inverse Trigonometric Functions and Domain Restrictions
Trigonometric functions provide another important application.
The sine function is not one-to-one over all real numbers because its values repeat periodically.
For example, different angles can have the same sine value.
To define an inverse sine function, the original sine function is restricted to the interval:
−π ÷ 2 ≤ x ≤ π ÷ 2
On this interval, sine is one-to-one.
Its inverse is written:
y = arcsin(x)
or
y = sin⁻¹(x)
The domain of arcsin is −1 ≤ x ≤ 1, and its range is:
−π ÷ 2 ≤ y ≤ π ÷ 2
The notation sin⁻¹(x) means inverse sine, not 1 ÷ sin(x).
Similar restrictions are used to define inverse cosine and inverse tangent.
These restrictions explain why inverse trigonometric functions return particular principal angles instead of every angle that has the same trigonometric value.
13. Six Common Inverse-Function Mistakes
Understanding common errors can improve accuracy in Algebra 2 and precalculus examinations.
Mistake 1: Confusing an inverse with a reciprocal
Remember that f⁻¹(x) reverses the function. It does not mean 1 ÷ f(x).
Mistake 2: Forgetting to exchange x and y
After writing y = f(x), exchange the variables before solving for y.
Mistake 3: Ignoring the domain restriction
A quadratic function generally needs an appropriate restriction before it has an inverse function.
Mistake 4: Choosing the wrong square-root sign
The correct sign depends on the branch of the original function.
Mistake 5: Forgetting to exchange the domain and range
The inverse's domain comes from the original range, and its range comes from the original domain.
Mistake 6: Failing to verify the answer
Substitute the inverse into the original function whenever possible. Function composition is an effective way to identify algebraic errors.
14. Inverse Functions Practice Problems With Answers
Try these questions before reading the solutions.
The problems progress from basic linear functions to rational functions and restricted quadratics.
Question 1: Linear function
Find the inverse of:
f(x) = 6x + 5
Question 2: Cubic function
Find the inverse of:
f(x) = x³ − 7
Question 3: Square-root function
Find the inverse of:
f(x) = √(x + 4) + 2
State the domain and range of the inverse.
Question 4: Rational function
Find the inverse of:
f(x) = (x + 2) ÷ (x − 5)
State the excluded value in the inverse's domain.
Question 5: Quadratic function
Find the inverse of:
f(x) = x² − 4
given the restriction x ≥ 0.
Question 6: Absolute-value function
Is the function
f(x) = |x − 3|
one-to-one over all real numbers?
If not, restrict its domain to x ≥ 3 and find the inverse.
Question 7: Function composition
Let:
f(x) = 2x − 9
g(x) = (x + 9) ÷ 2
Show that the two functions are inverses.
Question 8: Sales tax
A product costs $81 after an 8% sales tax has been added.
Find the original price before tax.
Answers and worked solutions
Answer 1
Start with:
y = 6x + 5
Exchange x and y:
x = 6y + 5
Subtract 5 and divide by 6:
f⁻¹(x) = (x − 5) ÷ 6
Answer 2
Start with:
y = x³ − 7
Exchange x and y:
x = y³ − 7
Add 7 and take the cube root:
f⁻¹(x) = ∛(x + 7)
Answer 3
Start with:
y = √(x + 4) + 2
Exchange x and y:
x = √(y + 4) + 2
Subtract 2:
x − 2 = √(y + 4)
Square and subtract 4:
y = (x − 2)² − 4
Therefore:
f⁻¹(x) = (x − 2)² − 4, where x ≥ 2.
The inverse domain is x ≥ 2, and the inverse range is y ≥ −4.
Answer 4
Start with:
y = (x + 2) ÷ (x − 5)
Exchange x and y:
x = (y + 2) ÷ (y − 5)
Multiply through:
xy − 5x = y + 2
Collect the y terms:
xy − y = 5x + 2
Factor and divide:
y(x − 1) = 5x + 2
Therefore:
f⁻¹(x) = (5x + 2) ÷ (x − 1)
The inverse domain excludes x = 1.
Answer 5
Start with:
y = x² − 4, where x ≥ 0.
Exchange x and y:
x = y² − 4
Add 4 and take the positive square root:
y = √(x + 4)
Therefore:
f⁻¹(x) = √(x + 4), where x ≥ −4.
Answer 6
The original function is not one-to-one over all real numbers because different inputs can produce the same absolute value.
Restrict the domain to x ≥ 3.
Then:
f(x) = x − 3
Exchange x and y:
x = y − 3
Add 3:
f⁻¹(x) = x + 3, where x ≥ 0.
Answer 7
Calculate the first composition:
f(g(x)) = 2((x + 9) ÷ 2) − 9
= x + 9 − 9
= x
Now calculate the second composition:
g(f(x)) = ((2x − 9) + 9) ÷ 2
= 2x ÷ 2
= x
Both compositions return x, so the functions are inverses.
Answer 8
The final price includes 8% tax.
Therefore:
Final price = 1.08 × Original price
Original price = 81 ÷ 1.08
Original price = $75
The original price was $75, and the tax was $6.
15. Frequently Asked Questions About Inverse Functions
How do you find the inverse of a function?
Write y = f(x), exchange x and y, solve for y and rename the result f⁻¹(x). Then check the domain and range and apply any necessary restrictions.
How do you know whether a function has an inverse?
A function has an inverse function if it is one-to-one on its domain. The horizontal line test is a convenient graphical check.
Why does x² not have an inverse over all real numbers?
Because x = 2 and x = −2 both produce an output of 4. The original function is not one-to-one over all real numbers.
How do you find the domain of an inverse function?
The domain of the inverse is the range of the original function. Determine the original range and transfer its restrictions to the inverse's domain.
Is the inverse of every function another function?
No. The inverse relationship may fail to be a function if the original function is not one-to-one. Restricting the original domain can sometimes resolve this problem.
What is the horizontal line test used for?
It determines whether a graph represents a one-to-one function on its stated domain. A horizontal line must not cross the graph more than once.
Are inverse functions important for calculus?
Yes. Inverse functions are used in logarithmic and exponential equations, inverse trigonometric functions and the differentiation of inverse functions.
Final Summary
An inverse function reverses the relationship between inputs and outputs.
The most reliable method is to exchange x and y, solve for the new y, and check the restrictions.
Always remember these principles:
A function must be one-to-one on its domain to have an inverse function.
The horizontal line test checks whether a function is one-to-one.
The domain and range exchange places when a function is inverted.
A quadratic function may need a domain restriction before it has an inverse.
The inverse of an operation must undo it in reverse order.
Function composition can verify whether two functions are inverses.
Once these ideas become familiar, inverse functions provide a useful foundation for more advanced precalculus topics, including logarithms, exponential equations and inverse trigonometric functions.
