A fractional linear function has the form f(x) = (ax + b)/(cx + d). To find its inverse, write y = f(x), multiply through by the original denominator, collect the terms involving x and solve for x. The inverse’s domain is the original function’s range, so forbidden output values must be identified before stating the final formula.
For example, f(x) = (2x + 3)/(x − 1) is defined for x ≠ 1 and never produces y = 2. Solving y(x − 1) = 2x + 3 gives x = (y + 3)/(y − 2). Therefore f⁻¹(x) = (x + 3)/(x − 2), with domain x ≠ 2. This guide explains why the restrictions exchange, how to verify both compositions, and when a rational expression cannot have an inverse function.
An inverse reverses a function’s mapping
A function maps each allowed input x to one output f(x). Its inverse, when it exists, reverses this mapping: an output of the original becomes an input of the inverse.
For a fractional linear expression, finding the inverse is an algebraic rearrangement, but it is not merely swapping letters. A valid inverse also needs an appropriate domain and range.
The inverse must be single-valued: one original output cannot come from two distinct permitted inputs. Non-degenerate fractional linear functions are one-to-one on their natural domains, which makes a rational inverse possible.
The right question is therefore not only “what formula results after solving for x?” but also “which y-values does the original function actually produce?” Those values determine the inverse’s permitted inputs.
Start with the original denominator restriction
Let f(x) = (2x + 3)/(x − 1). The denominator is zero at x = 1, so the original function is defined for all real x except 1.
Write dom f = {x ∈ ℝ : x ≠ 1}. Do this before cancelling factors or performing other algebra.
At x = 1, the expression would involve division by zero and is undefined. A numerical output cannot be assigned there by ordinary function evaluation.
This restriction will eventually reappear as an excluded output of the inverse, because the inverse returns the original function’s input values. Tracking this exchange is one of the most important parts of any inverse-function problem involving rational expressions.
Find which output the original function can never attain
For f(x) = (2x + 3)/(x − 1), ask whether any allowed input could produce y = 2.
Set (2x + 3)/(x − 1) = 2. Multiplying by x − 1 gives 2x + 3 = 2x − 2, which simplifies to 3 = −2.
This contradiction means the original function never outputs 2. Its range is all real y except 2.
The result also follows by division: f(x) = 2 + 5/(x − 1). The reciprocal term is never zero at any allowed input, so f(x) cannot equal 2.
This missing output is exactly the forbidden input of the inverse. It is a range condition, not another forbidden x-value of the original function.
Derive the inverse in four algebraic lines
Write y = (2x + 3)/(x − 1). Multiply by x − 1, which is nonzero on the original domain: y(x − 1) = 2x + 3.
Expand: xy − y = 2x + 3. Collect terms involving x on one side: xy − 2x = y + 3.
Factor x: x(y − 2) = y + 3. Since the original function never gives y = 2, division is valid and x = (y + 3)/(y − 2).
Swap the variable labels to write f⁻¹(x) = (x + 3)/(x − 2). The inverse’s domain is x ≠ 2, matching the original range. The algebra and restrictions fit together without adding an arbitrary condition.
State both the domain and range of the inverse
The original function has domain x ≠ 1 and range y ≠ 2. An inverse exchanges those two sets.
Therefore f⁻¹(x) = (x + 3)/(x − 2) has domain x ≠ 2 and range y ≠ 1.
The inverse output cannot be 1 because that was not an allowed input of f. To check algebraically, suppose (x + 3)/(x − 2) = 1. This would give x + 3 = x − 2, a contradiction.
A complete school answer should state the inverse’s domain when the question asks for it or when restrictions are essential to its meaning. The rational formula alone may conceal an excluded value if read carelessly.
Check a specific input–output pair
Choose x = 0, which is allowed for the original function. Then f(0) = 3/(−1) = −3.
The inverse must send −3 back to 0. Evaluate f⁻¹(−3) = (−3 + 3)/(−3 − 2) = 0/(−5) = 0.
Choose another input x = 2. The original output is f(2) = (4 + 3)/(2 − 1) = 7. The inverse gives f⁻¹(7) = (7 + 3)/(7 − 2) = 10/5 = 2.
Both checks support the derived formula. But a few matching numbers do not prove the inverse works for every allowed input; composition provides that stronger algebraic verification.
Verify f(f⁻¹(x)) exactly
Let g(x) = (x + 3)/(x − 2), defined for x ≠ 2. Substitute it into the original f.
We have g(x) − 1 = [(x + 3) − (x − 2)]/(x − 2) = 5/(x − 2).
Also 2g(x) + 3 = [2(x + 3) + 3(x − 2)]/(x − 2) = 5x/(x − 2).
Therefore f(g(x)) = [5x/(x − 2)]/[5/(x − 2)] = x for every x ≠ 2. The nested expression is defined throughout the inverse’s domain because g(x) never equals the original excluded input 1.
This establishes one direction of the inverse relationship symbolically rather than relying on the numerical examples alone.
Verify f⁻¹(f(x)) in the reverse direction
Start with f(x) = (2x + 3)/(x − 1), for x ≠ 1, and substitute it into g(y) = (y + 3)/(y − 2).
The numerator becomes f(x) + 3 = [2x + 3 + 3(x − 1)]/(x − 1) = 5x/(x − 1).
The denominator becomes f(x) − 2 = [2x + 3 − 2(x − 1)]/(x − 1) = 5/(x − 1).
Thus g(f(x)) = [5x/(x − 1)]/[5/(x − 1)] = x for every x ≠ 1.
Both compositions return the appropriate input on their respective domains. This is a complete functional check and shows why the excluded values cannot be discarded during the algebra.
A second worked example with different coefficients
Let h(x) = (3x − 1)/(2x + 1). The original denominator excludes x = −1/2.
Write y = (3x − 1)/(2x + 1). Multiply through: y(2x + 1) = 3x − 1, so 2xy + y = 3x − 1.
Collect x-terms: x(2y − 3) = −1 − y. Therefore x = (−1 − y)/(2y − 3) = (y + 1)/(3 − 2y).
Hence h⁻¹(x) = (x + 1)/(3 − 2x), with x ≠ 3/2. Its output range excludes −1/2. A quick check is h(0) = −1 and h⁻¹(−1) = 0.
Why the second example excludes the output 3/2
Suppose h(x) = (3x − 1)/(2x + 1) could equal 3/2. Multiplying by the allowed denominator gives 2(3x − 1) = 3(2x + 1).
This reduces to 6x − 2 = 6x + 3, an impossibility. Therefore 3/2 is not in h’s range.
The inverse expression (x + 1)/(3 − 2x) is undefined at x = 3/2, exactly as the range analysis predicts.
This is the same structure seen in the opening example. The output approached for very large |x| is often the ratio of leading coefficients, and that value is excluded for a non-degenerate fractional linear function with a nonzero coefficient of x in the denominator.
The general inverse formula
Consider f(x) = (ax + b)/(cx + d), with real constants a, b, c and d. Suppose ad − bc ≠ 0, and assume the denominator is not identically zero.
Set y = (ax + b)/(cx + d). Then y(cx + d) = ax + b. Rearranging gives x(cy − a) = b − dy.
Therefore x = (dy − b)/(a − cy), and f⁻¹(x) = (dx − b)/(a − cx), on its appropriate domain.
When c ≠ 0, the original domain excludes −d/c and its range excludes a/c. The inverse domain excludes a/c and its range excludes −d/c. The condition ad − bc ≠ 0 prevents the expression from degenerating into a constant.
What the determinant condition really protects
The expression ad − bc is not an arbitrary decorative formula. It determines whether the numerator and denominator define a nonconstant fractional linear map.
If ad − bc = 0 and the denominator is a genuine nonzero linear expression, the numerator is proportional to it. On the allowed domain, the ratio is constant and cannot have an inverse function over a domain containing multiple inputs.
For example, f(x) = (2x + 4)/(x + 2) equals 2 for every x ≠ −2. Many different inputs produce the same output 2.
The original graph has a hole at x = −2 and is otherwise a horizontal line. It is not one-to-one, so the attempted inverse would require the output 2 to return many inputs. That is not a function.
Do not confuse an inverse with a reciprocal
The notation f⁻¹(x) means inverse function, not 1/f(x). These are different operations.
For f(x) = (2x + 3)/(x − 1), the reciprocal is (x − 1)/(2x + 3), where defined. The inverse is (x + 3)/(x − 2). They are clearly different rational expressions.
The reciprocal reverses multiplication of numerical outputs; the inverse reverses the mapping from input to output.
A reliable check is composition. If g is genuinely the inverse of f, then f(g(x)) and g(f(x)) return x on the appropriate domains. A reciprocal expression generally will not have that property.
A fractional linear function can be its own inverse
Consider p(x) = (x + 1)/(x − 1), for x ≠ 1. To find its inverse, write y(x − 1) = x + 1.
Rearrange: xy − y = x + 1, so x(y − 1) = y + 1 and x = (y + 1)/(y − 1). Therefore p⁻¹(x) = (x + 1)/(x − 1) = p(x).
This is called a self-inverse or involution on its domain. For instance p(0) = −1 and p(−1) = 0.
It is an instructive special case, not a general rule for fractional linear functions. Always derive an inverse rather than assume its formula resembles the original with coefficients rearranged by guesswork.
A domain restriction can make the inverse’s domain smaller
Suppose f(x) = (2x + 3)/(x − 1) is restricted to x > 1 rather than its full natural domain.
Rewrite f(x) = 2 + 5/(x − 1). For x > 1, the reciprocal term is positive, so f(x) > 2. As x approaches 1 from the right, f grows without bound; as x increases without bound, f approaches 2 from above.
Thus the range of the restricted f is (2, ∞). Its inverse formula is still g(x) = (x + 3)/(x − 2), but the inverse domain is now x > 2.
Writing merely x ≠ 2 would be too broad for the inverse of this particular restricted function. Domain and range belong to the defined function, not just to its algebraic formula.
The left branch gives another restricted inverse
Now restrict the same original formula to x < 1. Then x − 1 is negative, so f(x) = 2 + 5/(x − 1) is strictly less than 2.
Its range is (−∞, 2). The inverse formula remains (x + 3)/(x − 2), but its domain must be x < 2.
These two branches cover the natural domain on either side of the vertical asymptote. Together their ranges cover every real value except 2, consistent with the full inverse.
A student who treats a specified interval as irrelevant because the algebraic inverse formula did not change has missed the functional question. Restrictions can change the inverse domain even when the final expression looks identical.
Horizontal and vertical asymptotes exchange under inversion
For f(x) = 2 + 5/(x − 1), the graph has vertical asymptote x = 1 and horizontal asymptote y = 2.
The inverse g(x) = (x + 3)/(x − 2) can be rewritten as 1 + 5/(x − 2). Its vertical asymptote is x = 2 and horizontal asymptote y = 1.
This exchange is exactly what reflection in the line y = x predicts. Original output exclusions become inverse input exclusions, and vice versa.
The observation offers a strong graphical check on an inverse calculation. If the derived inverse has the wrong forbidden denominator value or asymptote level, revisit the cross-multiplication and coefficient signs.
Why the graph reflects in y = x
The inverse relationship swaps each coordinate pair. If (u, v) is on y = f(x), then (v, u) is on y = f⁻¹(x), provided the values belong to the corresponding domains.
In the main example, (0, −3) on f becomes (−3, 0) on the inverse. Likewise, (2, 7) on f becomes (7, 2) on the inverse.
This coordinate swap is the geometric meaning of reflection across the diagonal line y = x. It explains why the asymptotes x = 1 and y = 2 exchange roles.
A sketch can support understanding, but exact algebra remains the clearest method for determining the inverse formula and all excluded values. A drawing alone may not identify a small numerical error in the constants.
The derivative shows why non-degenerate functions are one-to-one
For f(x) = (ax + b)/(cx + d), differentiating by the quotient rule gives f′(x) = (ad − bc)/(cx + d)² wherever the denominator is nonzero.
If ad − bc ≠ 0, the derivative never vanishes on the natural domain. Its sign is fixed by ad − bc because the denominator is squared.
This shows that the function is strictly monotonic on each connected branch. More generally, direct algebra establishes that no two allowed inputs can have the same output, since the inverse rearrangement determines a unique original x for each permitted y.
The derivative route is optional enrichment but connects inverses, rational graphs and calculus. It also explains why the degenerate case ad − bc = 0 loses the nonzero gradient.
A composite inverse problem needs inherited restrictions
Suppose an expression uses f⁻¹(g(x)), where f is the main fractional linear function and g(x) = x². The inverse f⁻¹ accepts every real input except 2.
Therefore f⁻¹(g(x)) = (x² + 3)/(x² − 2) requires x² ≠ 2, so x ≠ ±√2.
This restriction comes from the inverse’s input domain applied to the inner output g(x). The inner square itself is defined for all real x, but it can produce the forbidden inverse input 2.
The result connects directly with composite-function domain tracking. Algebraic substitution into an inverse formula does not remove the need to check whether the inner value belongs to the inverse’s domain.
An incorrect cancellation can conceal an excluded input
Consider the degenerate ratio (2x + 4)/(x + 2). Cancelling its common factor gives 2, but the original is undefined at x = −2.
The simplified expression and original rational function agree only on x ≠ −2. They do not define the same unrestricted function.
This matters when deciding whether an inverse exists. On the original allowed domain, the result is constant; keeping the hole does not make it one-to-one.
Before searching for an inverse, factor and simplify only with explicit domain notes. A cancelled denominator restriction remains part of the function’s definition, even if it no longer appears in the simplest formula.
A quick five-stage inverse solution
First write the original domain by excluding zeros of its denominator. Second solve y = f(x) for x using legitimate cross-multiplication.
Third identify any denominator produced while isolating x. The excluded y-value is a candidate missing output of the original function; verify that it truly cannot be attained.
Fourth swap labels and state f⁻¹ with its domain. Fifth check f(f⁻¹(x)) = x and f⁻¹(f(x)) = x on their corresponding domains.
If the original function has an additional interval restriction or degenerates to a constant, handle that before claiming a full inverse. This routine is compact but protects both the algebraic and functional meaning of the answer.
Practice questions with checked answers
Question one: f(x) = (2x + 3)/(x − 1), x ≠ 1. Find f⁻¹. Answer: (x + 3)/(x − 2), x ≠ 2.
Question two: for the same function, state its range. Answer: every real output except 2.
Question three: h(x) = (3x − 1)/(2x + 1). Find h⁻¹. Answer: (x + 1)/(3 − 2x), x ≠ 3/2.
Question four: p(x) = (x + 1)/(x − 1). Show that p is self-inverse on x ≠ 1.
Question five: f(x) = (2x + 4)/(x + 2). Does it have an inverse on its natural real domain? No; it is constant 2 wherever defined.
Question six: restrict the main function to x > 1. What is its inverse domain? Answer: x > 2.
For each, check a specific input–output pair and the forbidden denominator value before finalising.
Syllabus guidance and connected study
Fractional linear inverses connect rational expressions, function domains and ranges, graph reflection, asymptotes and composition. Specific extensions such as determinant conditions and derivatives should follow the student’s actual syllabus and school sequence.
For Singapore, refer to the applicable SEAB resources for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2.
Related guides include How to Select the Correct Branch of a Quadratic Inverse, How to Track Domain Restrictions Through Composite Functions, How to Distinguish a Hole from a Vertical Asymptote and the Additional Mathematics archive.
Parents may enquire about suitable eduKate Singapore three-student tutorials near Sixth Avenue MRT. The objective is not merely rearranging a fraction, but showing that the reversed mapping is a correctly defined function. Properly taught kids shine a bright light into the future.