01 · Linear inverse
Find the inverse of f(x) = 4x + 9 on the real numbers.
Hint
Undo addition, then multiplication.
Worked solution
f⁻¹(x) = (x − 9)/4. Domain and range are both all real numbers.
Understand · explore · practise
Inverse functions
Find inverse functions, choose the correct quadratic branch and exchange domain and range. Explore reflection in y = x, verify compositions and practise with solutions.
Before you startOne-to-one functions, domain and range, rearranging formulas and composite functions
01 / Reverse a function
An inverse takes an output back to its unique input.
If f maps a to b, then its inverse maps b back to a. This requires f to be one-to-one on its domain: otherwise a shared output would have two possible original inputs.
f(a) = b ⇔ f⁻¹(b) = a
The inverse is defined on the range actually reached by f. For example, f(x) = x² on all real numbers has no inverse function because 9 came from both −3 and 3. Restricting the domain to x ≥ 0 gives a unique inverse √x.
f⁻¹(x) means the inverse function. It does not mean 1/f(x).
02 / Rearrange and relabel
Solve y = f(x) for the original input.
Find the inverse of f(x) = 3x − 7Worked example
y = 3x − 7 ⇒ x = (y + 7)/3
Undo subtracting 7, then undo multiplying by 3.
f⁻¹(x) = (x + 7)/3
Relabel the new input as x after rearranging.
Both domain and range are all real numbers
The linear rule is one-to-one and reaches every real output.
For comparison, the reciprocal of f(x) is 1/(3x − 7), which is a completely different function.
03 / Choose the correct branch
The original domain decides the sign of the square root.
f(x) = (x − 1)² + 2, with x ≥ 1Worked example
y − 2 = (x − 1)²
The range of f is y ≥ 2.
x − 1 = √(y − 2)
Since x ≥ 1, the original x − 1 is nonnegative. Choose the positive branch.
f⁻¹(x) = 1 + √(x − 2), with x ≥ 2
The inverse has range y ≥ 1.
If instead the original domain were x ≤ 1, its inverse would be 1 − √(x − 2), still with domain x ≥ 2 but range y ≤ 1. Never keep an unexplained ± in the answer for an inverse function.
Swap the coordinatesExplore
Blue: f(x) = (x − 1)² + 2, x ≥ 1.
Gold: f⁻¹(x) = 1 + √(x − 2), x ≥ 2.
Only the part within the displayed axes is shown.
(2, 3) on f ↔ (3, 2) on f⁻¹.
The dashed connector is perpendicular to y = x and its midpoint lies on that line.
Watch an input-output pair reflect
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Exchange domain and range
The allowed inputs and achieved outputs swap roles.
If f has domain D and range R, its inverse has domain R and range D. These restrictions are part of the answer, even when a rearranged formula appears to work more widely.
A quadratic on a shorter intervalWorked example
f(x) = (x − 1)² + 2 for 1 < x ≤ 3
The function increases throughout its domain.
Range of f: 2 < y ≤ 6
The lower value is approached, not attained.
f⁻¹(x) = 1 + √(x − 2)
Domain: 2 < x ≤ 6
Range: 1 < y ≤ 3
Keep the open and closed endpoints when exchanging the sets.
05 / Rational inverses
Collect the terms containing the original input.
f(x) = (2x + 1)/(x − 3), x ≠ 3Worked example
y(x − 3) = 2x + 1
Multiply by the nonzero denominator.
x(y − 2) = 3y + 1
Collect x terms on one side.
x = (3y + 1)/(y − 2)
Here y cannot equal 2. The original expression is 2 + 7/(x − 3), so its range excludes 2.
f⁻¹(x) = (3x + 1)/(x − 2)
Domain x ≠ 2; range y ≠ 3
Each allowed output has exactly one original input.
06 / Roots and exponentials
Retain restrictions created by nonnegative roots and logarithms.
p(x) = √(3x − 2) + 1, x ≥ 2/3Worked example
y − 1 = √(3x − 2), so y ≥ 1
The sign restriction must survive squaring.
(y − 1)² = 3x − 2
Now rearrange for x.
p⁻¹(x) = ((x − 1)² + 2)/3
Domain x ≥ 1; range y ≥ 2/3
The polynomial expression alone does not justify using x < 1.
q(x) = e^(2x) − 4, x realWorked example
y + 4 = e^(2x) > 0
The range of q is y > −4.
ln(y + 4) = 2x
Take logarithms of positive quantities.
q⁻¹(x) = ½ ln(x + 4)
Domain x > −4; range all real y
The inverse domain comes from the original range.
07 / Verify by composition
Both round trips return the original input on their domains.
For f⁻¹ to undo f, check f⁻¹(f(x)) = x on the domain of f and f(f⁻¹(x)) = x on the domain of f⁻¹.
Check the restricted quadraticWorked example
f⁻¹(f(x)) = 1 + √((x − 1)²)
= 1 + |x − 1|
The square root of a square is an absolute value.
For x ≥ 1, this equals 1 + (x − 1) = x
The branch restriction is exactly what makes it work.
f(f⁻¹(x)) = (√(x − 2))² + 2 = x for x ≥ 2
The second round trip starts in the inverse domain.
08 / Reflection and intersections
Swapping coordinates reflects a graph across y = x.
A point (a, b) on f corresponds to (b, a) on f⁻¹. Equal axis scales make this a geometric reflection. A fixed point f(a) = a lies on both graphs and on y = x.
For a strictly increasing function, every intersection with its inverse lies on y = x. To see why, an intersection gives f(a) = b and f(b) = a. If a < b, increasing behaviour would require b < a, a contradiction. The case a > b is equally impossible. Therefore a = b.
Increasing exampleWorked example
f(x) = 2x + 1; f⁻¹(x) = (x − 1)/2
Both are increasing.
2x + 1 = x ⇒ x = −1
The only intersection is (−1, −1).
Do not use this shortcut for every inverse pair. The decreasing function f(x) = 5 − x is its own inverse. The two graphs coincide along the entire line, including (1, 4), which is not on y = x. Their only fixed point is (5/2, 5/2), but it is not their only intersection.
09 / Your turn
Give inverse domains as well as formulas.
State the branch and any excluded values explicitly.
02 · Negative quadratic branch
f(x) = (x + 2)² − 5, with x ≤ −2. Find f⁻¹ and its domain/range.
Hint
The original x + 2 is nonpositive.
Worked solution
f⁻¹(x) = −2 − √(x + 5). Its domain is x ≥ −5 and range y ≤ −2.
03 · Short interval
g(x) = x² for 2 ≤ x < 5. Find g⁻¹, including its domain and range.
Hint
The original range is [4, 25).
Worked solution
g⁻¹(x) = √x for 4 ≤ x < 25. Its range is 2 ≤ y < 5.
04 · Rational inverse
h(x) = (3x − 2)/(x + 1), x ≠ −1. Find h⁻¹ and its domain/range.
Hint
Collect x terms after multiplying by x + 1.
Worked solution
yx + y = 3x − 2 gives x(y − 3) = −y − 2. Thus h⁻¹(x) = (x + 2)/(3 − x), with domain x ≠ 3 and range y ≠ −1.
05 · Root inverse
p(x) = √(2x + 6), x ≥ −3. Find its inverse.
Hint
The original output is nonnegative.
Worked solution
p⁻¹(x) = (x² − 6)/2, domain x ≥ 0 and range y ≥ −3. Negative inverse inputs must be excluded.
06 · Exponential inverse
q(x) = 3eˣ + 2 for real x. Find q⁻¹ and its domain.
Hint
Isolate eˣ before taking logarithms.
Worked solution
q⁻¹(x) = ln((x − 2)/3), with domain x > 2 and range all real numbers.
07 · Why a branch matters
Why is √(x²) = x not true for all real x? Explain its relevance to inverting x².
Hint
Test x = −3.
Worked solution
√(x²) = |x|. At −3 it equals 3. The inverse √x undoes x² on x ≥ 0; on x ≤ 0 the correct inverse is −√x. Without a restriction, x² has no inverse function.
08 · Self-inverse
Show that r(x) = 7 − x is its own inverse. Are all intersections of r and r⁻¹ on y = x?
Hint
Compute r(r(x)).
Worked solution
r(r(x)) = 7 − (7 − x) = x for every real x. The graphs are the same whole line, so they have infinitely many intersections, most off y = x. For example, (0, 7) is common to both.
09 · Increasing intersection
Find the intersection of f(x) = 3x − 4 and f⁻¹(x).
Hint
The function is strictly increasing, so solve f(x) = x.
Worked solution
3x − 4 = x gives x = 2 and y = 2. The inverse is (x + 4)/3, which also gives 2 at x = 2.
10 · Inverse notation
If f(6) = −2 and f is invertible, what is f⁻¹(−2)? Can you deduce 1/f(−2)?
Hint
Reverse the given mapping. Do not confuse notation.
Worked solution
f⁻¹(−2) = 6. The given information does not determine f(−2), so it does not determine its reciprocal; that reciprocal might even be undefined.
10 / Recap
An inverse is a reversed mapping with an explicit domain.
- One-to-one behaviour is required on the chosen domain.
- Rearrange, then select the branch using the original restriction.
- Swap domain and range, including open endpoints.
- Check both compositions on the appropriate domains.
- Reflect points by swapping coordinates; the y = x intersection shortcut needs justification.
Section 1 of 10 · Reverse a function