Composition and inverse functions

Use the inside function first

A composition sends an input through two function rules in sequence. In f(g(x)), start with x, use g to obtain g(x), then use that result as the input to f. Write this new function as f ∘ g, read f composed with g: (f ∘ g)(x) = f(g(x)). The function written nearest to x acts first. Its output must be a permitted input to the outside function.

For f(x) = 3x − 2 and g(x) = x + 4, both on all real inputs, f(g(x)) = 3(x + 4) − 2 = 3x + 10. Replace every input occurrence in f with the whole expression x + 4. At x = 2, g first gives 6 and f then gives 16. Multiplying the separate outputs gives f(2)g(2) = 4 × 6 = 24, a different operation. Composition is not the product f(x)g(x).

Check the order, domain and meaning of each stage

With those same rules, g(f(x)) = (3x − 2) + 4 = 3x + 2, so reversing the order gives a different function. Composition is not commutative in general: the two orders need not agree. They can agree for particular rules, however. If a(x) = x + 3 and b(x) = x − 1 on all real inputs, both a(b(x)) and b(a(x)) equal x + 2 for every real x. Compare full formulas and domains; agreement at one input alone does not establish equality of functions.

For a constant-speed model, let D take elapsed time t ≥ 0 in hours to distance D(t) = 4t in kilometres, and let F take distance d ≥ 0 in kilometres to fuel F(d) = d/20 in litres. The stated rates are 4 kilometres per hour and 1/20 litre per kilometre. Then F(D(t)) = t/5 gives litres used in that model. Reversing the named stages would feed litres into a rule that requires hours, so D(F(d)) has no meaning for these quantities even when its numbers can be substituted into a formula. A stated function model describes the assumed relationship; notation alone does not establish a physical cause.

Require both the first input and the intermediate output to be allowed

The domain of f ∘ g contains exactly those x for which x belongs to g's stated domain and g(x) belongs to f's stated domain. The starting x need not belong to f's domain: f receives g(x), not x. First apply the inside function's restrictions, then check its output against the outside function's restrictions. If a problem has already narrowed either domain, keep that chosen restriction as well as the formula's natural restrictions.

Let s(x) = √x on x ≥ 0 and h(x) = x − 2 on all real inputs. In s(h(x)) = √(x − 2), h allows every starting input, but its output must satisfy x − 2 ≥ 0, giving domain x ≥ 2. In h(s(x)) = √x − 2, the inside root requires x ≥ 0 and h accepts every resulting output, giving domain x ≥ 0. At starting input 1 the first order is undefined over the reals, while the second produces −1.

Keep original restrictions when a formula simplifies

Let r(x) = 1/x on x ≠ 0. The composition r(r(x)) = 1/(1/x) requires the starting x to be nonzero. Its intermediate output 1/x is then also nonzero, so the outside stage is allowed. The result simplifies to x, but its domain remains x ≠ 0. It is the identity on the nonzero real numbers, not the identity on every real number. A newly defined identity function on all real inputs would be a different function.

For p(t) = (t − 1)/(t − 1) on t ≠ 1 and g(x) = 1/x on x ≠ 0, p(g(x)) equals 1 only when x ≠ 0 and 1/x ≠ 1, so its domain excludes both 0 and 1. In the other order, g(p(x)) first requires x ≠ 1; p(x) then equals 1, which g accepts. This order has domain x ≠ 1 and also gives constant output 1. The simplified constant formulas agree, but their domains differ, so the two composite functions are not equal.

A reversible function must not merge different inputs

A function is one-to-one, also called injective, when different permitted inputs always produce different outputs. Equivalently, for any permitted a and b, f(a) = f(b) implies a = b. The ordinary function condition gives one output per input; the one-to-one condition additionally gives at most one input per output. The square function on all real inputs is a function but is not one-to-one: inputs −2 and 2 both produce 4.

A horizontal line fixes an output. A function is one-to-one exactly when every horizontal line meets its complete graph at most once. This is the horizontal line test, distinct from the vertical line test for being a function. A cropped graph or a few samples may miss a second intersection. For a complete finite table, inspect every output for sharing between distinct inputs; for an infinite rule, give a general argument or find a counterexample. For example, 3a − 2 = 3b − 2 implies a = b, proving that 3x − 2 is one-to-one on all real inputs.

Reverse the function on its actual range

For a one-to-one function f with domain D and actual range R, its inverse takes each output in R back to its unique original input in D. If f(a) = b, then f⁻¹(b) = a. Thus the inverse's domain is R and its range is D. An unused value in a larger declared output target, or codomain, has no original input to recover. If the codomain is larger than R, f is not invertible onto that entire target; regard f as mapping to R to define its inverse there. A function that reaches every member of its declared target is called onto, or surjective.

Read f⁻¹ as f inverse. The raised −1 is part of function notation and does not mean the reciprocal 1/f(x). For f(x) = 3x − 2 on all real inputs, the inverse is f⁻¹(y) = (y + 2)/3 on all real y; the reciprocal rule is 1/f(x) = 1/(3x − 2) and excludes x = 2/3. At input 4, the inverse gives 2, whereas the reciprocal gives 1/10. The inverse reverses the input–output pairing; the reciprocal divides 1 by a function's output.

Solve for the original input and check both identities

To find an inverse formula, write y = f(x) with the original domain conditions. Solve for the original input x in terms of output y, carrying every restriction and branch choice. The resulting rule takes y as its input. You may then rename that input letter x, but renaming does not restore the original domain: the inverse's inputs still belong to the original range. If solving gives multiple permitted original inputs for an output, there is no single inverse function on the stated domain.

For f(x) = ax + b on all real inputs with fixed a ≠ 0, solving y = ax + b gives x = (y − b)/a. Every real y is attained by this allowed input, so the inverse domain is all real numbers. Put k(y) = (y − b)/a. Then k(f(x)) = (ax + b − b)/a = x for every real original input, and f(k(y)) = a((y − b)/a) + b = y for every real inverse input. These two identities prove the reversal on the complete domains. If a = 0, the rule is constant and cannot recover a unique input on all real numbers.

Restrict a square before choosing a square-root inverse

The square rule on all real inputs has no inverse function because each positive output comes from two inputs. Restrict it instead to x ≥ 0. The restricted rule q₊(x) = x² has range y ≥ 0 and inverse q₊⁻¹(y) = √y. The root denotes the principal, nonnegative root. For every x ≥ 0, √(x²) = x; for every y ≥ 0, (√y)² = y. The inverse inputs are all nonnegative real numbers, including irrational values, rather than only the perfect squares of whole numbers.

A separate restriction to x ≤ 0 gives q₋(x) = x² with the same range y ≥ 0 but inverse q₋⁻¹(y) = −√y. For every x ≤ 0, −√(x²) = −|x| = x; for every y ≥ 0, (−√y)² = y. Here |x| is the nonnegative magnitude of x, and in general √(x²) = |x|. Choose the branch from the stated original domain. Writing ±√y gives two answers for positive y and therefore a relation, not a single inverse function. Checking only that squaring √y returns y would not prove an inverse for the unrestricted square, because √(x²) fails to return negative x.

Exchange graph coordinates and their labels

If the graph of a one-to-one real function contains the point (a, b), its inverse graph contains (b, a). For f(x) = 3x − 2, the point (2, 4) becomes (4, 2) because f⁻¹(4) = 2. On a mathematical coordinate plane with equal scales on the axes, exchanging coordinates reflects the full graph across the line y = x. Exchange included and excluded endpoints in the same way: reversing a pairing does not turn an omitted point into an included one.

Reflecting the unrestricted square graph still exchanges its coordinates, but the reflected relation assigns both −√y and √y to every positive y and fails to be a function. Reflecting either stated half of the square graph gives its corresponding single inverse branch. For functions between physical quantities, also exchange the axis labels and units: a cost-to-count inverse takes a cost as input and returns a count. A graph window shows only the portion plotted; the domain and range statements determine the full inverse.

Recover only inputs allowed by the original context

Suppose a stated order model charges C(n) = 4n + 3 euros for n items, with n restricted to 0, 1, 2, 3 or 4. Its exact range is {3, 7, 11, 15, 19} euros, and these five costs are distinct. The inverse takes a cost c from this attained collection to (c − 3)/4 items. Cost 11 euros therefore recovers 2 items. A larger declared target such as all nonnegative euro amounts does not make every target amount an allowed inverse input.

At 9 euros, the algebraic expression gives 3/2, but fractional item counts are excluded and no permitted order produces 9 euros. The original context therefore has no inverse value there. On a discrete domain, checking both inverse identities at every permitted member is a complete proof because the finite lists are the entire definition. On an infinite real domain, a few successful evaluations remain samples; use identities that hold for every permitted input.

Inspect both stages and distinguish preimages from inverse functions

The explorer's first composition case uses f(t) = 1/t on t ≠ 0 and g(t) = t − 2 on all real inputs. Compare f(g(x)) = 1/(x − 2) on x ≠ 2 with g(f(x)) = 1/x − 2 on x ≠ 0. Its second case uses f(t) = (t − 1)/(t − 1) on t ≠ 1 and g(t) = 1/t on t ≠ 0: both orders give 1 when allowed, but f ∘ g excludes 0 and 1 whereas g ∘ f excludes only 1. Read the inside-domain and outside-domain checks before interpreting any simplified output.

The inverse cases compare 3x − 2 on all real inputs, the square on all real inputs, the square on x ≥ 0, the square on x ≤ 0, and 2n + 1 on integer inputs n = 0, 1, 2, 3, 4, 5. The discrete case declares the integer output target 0 through 12 but attains only {1, 3, 5, 7, 9, 11}. Choose an original input and a proposed output independently. An output check asks which permitted original inputs produce that output; these are its preimages. A unique preimage at one chosen output does not establish that the whole function is one-to-one. Use the full definitions and proofs to decide whether an inverse function exists, and keep any negative square-root branch and unattained output restrictions visible. The next lesson studies polynomial functions.

Worked examples

Try it yourself

1. Let f(x) = 2x + 3 and g(x) = x − 4, both on all real inputs. Find f(g(x)) and g(f(x)), evaluate both at x = 1, and compare with f(1)g(1).

   Show the answer and reasoning

   f(g(x)) = 2x − 5 and g(f(x)) = 2x − 1, giving −3 and 1 at x = 1. The product f(1)g(1) is −15.

   Substitute the whole inside expression: 2(x − 4) + 3 = 2x − 5, whereas (2x + 3) − 4 = 2x − 1. Separately, f(1) = 5 and g(1) = −3, so their product is −15. Both compositions have all real inputs, but their formulas differ.

2. Let s(x) = √(x + 1) on x ≥ −1 and r(x) = 1/x on x ≠ 0. Find the formulas and exact real domains of r(s(x)) and s(r(x)). State whether starting input −1 is permitted for each order.

   Show the answer and reasoning

   r(s(x)) = 1/√(x + 1) on x > −1. s(r(x)) = √(1/x + 1) on x ≤ −1 or x > 0. Input −1 is excluded in the first order and included in the second.

   In r ∘ s, x + 1 must be nonnegative and its root must be nonzero, giving x > −1. In s ∘ r, first require x ≠ 0, then 1/x + 1 = (x + 1)/x ≥ 0. For x > 0 this holds; for x < 0 it requires x + 1 ≤ 0, giving x ≤ −1. At −1, the second order gives √0 = 0, while the first would divide by zero.

3. Let p(t) = (t² − 1)/(t − 1) on t ≠ 1 and g(x) = x − 2 on all real inputs. Simplify p(g(x)) and find its exact domain. Can its simplified formula be used to include x = 3?

   Show the answer and reasoning

   p(g(x)) = x − 1 on x ≠ 3. Input 3 remains excluded.

   The intermediate input x − 2 must differ from 1, so x ≠ 3. Factoring gives p(t) = t + 1 only where the original denominator is nonzero; substituting t = x − 2 then gives x − 1. At x = 3, the original outside expression has denominator zero. The unrestricted rule x − 1 would be a different function.

4. A complete function has domain {0, 1, 2}, declared output target {2, 5, 8, 11}, and pairs (0, 2), (1, 5), (2, 8). Is it one-to-one? State its inverse pairs and inverse domain. What changes if the last pair is replaced by (2, 5)?

   Show the answer and reasoning

   The original function is one-to-one. Its inverse pairs are (2, 0), (5, 1), (8, 2), with domain {2, 5, 8}. Replacing the last pair by (2, 5) makes it fail to be one-to-one, so the changed function has no inverse function on its actual range.

   In the original complete table, each output is assigned to one input. Target value 11 is unattained and cannot be an inverse input. The inverse range is {0, 1, 2}; f is invertible when its output target is restricted to its actual range. In the changed table, output 5 comes from two different inputs, 1 and 2, so reversal would assign two outputs to input 5.

5. Let f(x) = 4x − 7 on all real inputs. Find f⁻¹, state its domain, prove both inverse identities and evaluate f⁻¹(1). Is it the same rule as 1/f(x)?

   Show the answer and reasoning

   f⁻¹(y) = (y + 7)/4 on all real y, and f⁻¹(1) = 2. It differs from the reciprocal 1/(4x − 7).

   Solving y = 4x − 7 gives x = (y + 7)/4, an allowed original input for every real y. Then f⁻¹(f(x)) = (4x − 7 + 7)/4 = x and f(f⁻¹(y)) = 4((y + 7)/4) − 7 = y. The inverse recovers an input, while the reciprocal divides by the output and excludes x = 7/4.

6. Let q(x) = x² + 3 only for x ≤ 0. Find its range and inverse function, including the inverse domain and range. Recover the input for output 5 and prove both inverse identities.

   Show the answer and reasoning

   q has range y ≥ 3. Its inverse is q⁻¹(y) = −√(y − 3) on y ≥ 3, with range x ≤ 0. Output 5 recovers input −√2.

   All outputs satisfy y ≥ 3, and every such y is attained uniquely at the permitted input −√(y − 3). For every x ≤ 0, q⁻¹(q(x)) = −√(x²) = −|x| = x. For every y ≥ 3, q(q⁻¹(y)) = (−√(y − 3))² + 3 = y. The positive root would recover an input outside the original domain except at zero.

7. A stated model assigns C(n) = 5n + 2 euros to integer item counts n = 0, 1, 2, 3. State the actual cost range and inverse rule. Recover the count for 12 euros, and decide whether 10 euros or 22 euros is an allowed inverse input.

   Show the answer and reasoning

   The range is {2, 7, 12, 17} euros. C⁻¹(c) = (c − 2)/5 items only for those costs; 12 euros recovers 2 items. Neither 10 nor 22 euros is an allowed inverse input.

   The four allowed counts produce four distinct costs, so the complete finite rule is one-to-one. At 10 euros the expression gives 8/5 items, which is not an allowed integer count. At 22 euros it gives 4 items, which is integer but outside the stated domain. Both costs are unattained; algebraic calculation alone cannot enlarge the context domain.

What to carry forward

• In f(g(x)), evaluate g first and feed its whole output into f. Composition is different from multiplying the two separate outputs.
• The two composition orders need not agree, although some functions do commute. Equality of functions requires the same domain and the same output at every permitted input.
• A composite input must belong to the inside domain, and the intermediate output must belong to the outside domain. Preserve original restrictions after simplification and check the units in a model.
• One-to-one means distinct permitted inputs have distinct outputs. Horizontal lines test that condition; a counterexample disproves it, while samples alone do not prove it on an infinite domain.
• An inverse reverses a one-to-one function on its actual range, exchanging domain and range. Unattained members of a larger output target are not inverse inputs.
• f⁻¹ denotes an inverse function, not the reciprocal 1/f. Carry constraints while solving and prove both inverse identities on their complete stated domains.
• The unrestricted square has no inverse function. The restriction x ≥ 0 gives inverse √y, while x ≤ 0 gives −√y; each has domain y ≥ 0, including irrational inverse inputs.