A composite domain has two gates

Explain it like I am five

Picture a trip with two ticket checks. The first gate checks your starting ticket. The second checks the ticket you receive after the first stage. Passing the first gate does not promise that you can pass the second. A composite domain is the list of starts that pass both checks. First, the inner function must accept your starting number x. Then the outer function must accept the answer g(x). These are the starts for which neither calculation stops. In the heating model, a day outside the calendar fails first; a temperature outside the cost chart fails second. Keep both checks before simplifying, because a shorter formula can hide a failed first stage.

Gate 1 checks x. Gate 2 checks the answer g(x).

Reminder

Domain of a fraction. $4 over 3 x − 2$ requires 3x − 2 ≠ 0, so x ≠ $2 over 3$ .
Solving a linear equation. 4 = 3x − 2 becomes 6 = 3x, then x = 2.
Fraction equation. $4 over 3 x − 2$ = 1 permits multiplying by 3x − 2 only after requiring it nonzero.
Interval notation. Removing 2 gives (−∞, 2) ∪ (2, ∞); parentheses exclude the point.
Dividing fractions. 1 ÷ $1 over 4$ = 4, because division multiplies by the reciprocal.
Composition order. f(g(x)) checks g at x first and f at the resulting number second.

Why it works. You cannot evaluate f(g(x)) if g gives no answer. You also cannot finish when g gives an answer that f cannot accept. These are separate ways the path can stop. The outer restriction applies to the in-between answer g(x), not directly to x. Solving for starts that produce forbidden answers converts the second gate into a restriction on x. Simplifying afterward does not create an answer at an earlier failed stage. This is why a composition can accept an x that f alone rejects, provided g changes it into an input f accepts.

RuleThe domain of f ∘ g consists of the x values allowed by g for which g(x) is allowed by f.
Keep original restrictions when simplifying the composite formula.

The same idea, five ways

Say it

x must get through g, then g of x must get through f.

Write it

The domain of f after g is every starting input that both stages can use in order.

In math

(f ∘ g)(x) = f(g(x))
{x | x is in the domain of g and g(x) is in the domain of f}

Two ticket gates check different tickets.

See it

Gate 1 checks x. Gate 2 checks the answer g(x).

The same idea, other ways

As two ticket gates

A valid starting ticket gets you through gate 1. Gate 2 checks a new ticket made by gate 1. You must check both, because they examine different things.

The second gate inspects the intermediate ticket.

As two questions

Ask: can I calculate g(x)? Then ask: can I calculate f at that answer? If either answer is no, the starting input is outside the composite domain.

Does g accept x?

Does f accept g(x)?

Keep x only if both answers are yes

The second question concerns g(x), not x.

With the two excluded numbers

At $2 over 3$ , the inner fraction has bottom zero. At 2, the inner fraction works and returns 1, but the outer fraction has bottom zero. These failures occur at different gates.

x = : g fails

x = 2: g(2) = 1, then f fails

Each excluded input has a specific failed stage.

Why a canceled restriction survives

If g(x) = $1 over x$ and f(x) = $1 over x$ , the composite simplifies to x. But the first machine still cannot start at zero, so the composite is x with x ≠ 0.

Two reciprocals undo each other only after the first one exists.

start x	g(x)	f(g(x))	what happens
0	−2	− $5 over 3$	both stages work
1	4	$5 over 3$	both stages work
$2 over 3$	undefined	undefined	inner bottom is 0
2	1	undefined	outer bottom is 0

.1Watch each gate before solving

Try a few starting tickets first. Input 0 passes through −2 to − $5 over 3$ . Input 1 passes through 4 to $5 over 3$ . Input $2 over 3$ stops in the first machine. Input 2 produces 1, then stops in the second. Solving finds every start that stops, beyond the ones you happen to try.

These pictures use f(x) = $5 over x − 1$ and g(x) = $4 over 3 x − 2$ .
Undefined means a stage has no permitted answer.

Read the g answer below each start. The column under stops at gate 1. The table label 2 ÷ 3 means ; −5 ÷ 3 and 5 ÷ 3 mean − and .

Worked exampleFour starting tickets

Trace starts 0, 1, $2 over 3$ , and 2 through the two functions. You want to see which stage succeeds or fails. Plan: calculate g first and calculate f only if g gives an answer.

The final table records success only when both stages work. The table label 2 ÷ 3 means ; −5 ÷ 3 and 5 ÷ 3 mean − and .

g(0) = $4 over −2$ = −2; f(−2) = $5 over −3$ = − $5 over 3$ .Both bottoms are nonzero.
g(1) = $4 over 1$ = 4; f(4) = $5 over 3$ .The in-between input 4 is allowed by f.
At $2 over 3$ , g has bottom 3 × $2 over 3$ − 2 = 0, so stop.A failed first stage gives no answer to hand over.
g(2) = $4 over 4$ = 1, then f(1) has bottom 0, so stop.An existing in-between answer can still be forbidden by the outer function.

Answer

0 works.
1 works.
$2 over 3$ fails at gate 1.
2 fails at gate 2.

Check For input 1, the final multiplied by its outer bottom 3 returns the top 5.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: f accepts every start except 1, so the composite should exclude 1.

The composite sends g(1) = 4 into f, so start 1 works.

✓ Instead: Exclude start 2 because g(2) = 1.

Tips and tricks

The pictures show the handoff. Solving finds every starting input that creates a failed handoff.

.2Gate 1: the inner function accepts x

Begin with the original inner formula. Its accepted inputs are the starting territory for the composition. A failure here means there is no intermediate answer to hand over.

Every composite input must be in the inner domain.
An unrestricted outer function cannot rescue a failed inner stage.

The inner denominator excludes input 5 before any outer operation.

Worked exampleAn unrestricted outer rule still needs an inner answer

Let g(x) = $1 over x − 5$ and f(x) = x + 2. Find the domain of f(g(x)). You want starting inputs that make the inner fraction exist. Plan: inspect the original inside rule, then inspect the outside rule at the inside answer, keeping both restrictions.

The outer addition cannot begin until the fraction produces an answer.

Set x − 5 = 0 to locate the failed inner start. Add 5: x = 5. Plug back in: 5 − 5 = 0. Therefore require x ≠ 5.The equation finds the input that would make the inner bottom zero, so we exclude it.
There are no further restrictions from f.Adding 2 accepts every real intermediate answer.

Answer

(−∞, 5) ∪ (5, ∞).

Check At 6, g(6) = 1 and f(1) = 3; at 5, the first division has no value.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: f(x) = x + 2 accepts everything, so f(g(x)) accepts everything.

At start 5, g(5) = has no value.

✓ Instead: The domain excludes 5 before f can act.

Tips and tricks

Begin with the original inner formula, even if the outer operation looks unrestricted.

.3Gate 2: the outer function accepts g(x)

The outer function's forbidden numbers are intermediate outputs. Find which starting inputs would produce those numbers. The starting number itself may be allowed by the outer formula or forbidden by it; that alone does not decide the composition.

For f(x) = $1 over x − 1$ , the second gate requires g(x) ≠ 1.
Do not merely intersect the two original domains as if both received x.

Starting at causes an outer failure even though the inner rule works.

Worked exampleA forbidden outer input changes its starting location

Let f(x) = $1 over x − 1$ and g(x) = 2x. Find the domain of f(g(x)). You want to avoid starting inputs whose doubled value is 1. Plan: inspect the original inside rule, then inspect the outside rule at the inside answer, keeping both restrictions.

The open point 0.5 is exactly . Start 1 is allowed because its handoff is 2.

g accepts every real starting input.Doubling has no denominator or square root restriction.
Find starts that hand f the forbidden input 1: solve 2x = 1. Divide by 2: x = $1 over 2$ . Plug back in: g( $1 over 2$ ) = 1 and f(1) = $1 over 0$ . Exclude $1 over 2$ .The solve step translates the outer forbidden input into the failed starting input.

Answer

(−∞, ) ∪ (, ∞).

Check Starting at 1 works: g(1) = 2 and f(2) = 1. Starting at fails: g() = 1 and f(1) is undefined.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Exclude x = 1 because f forbids 1.

Start 1 gives g(1) = 2 and f(2) = 1, so it passes both gates.

✓ Instead: Exclude , the start whose doubled value is 1.

Tips and tricks

Write g(x) beside the outer input requirement before solving.

.4Record the domain before simplifying

A shorter formula can conceal the first gate. Keep the original restrictions next to the simplified answer. The formula together with its domain describes the function.

Simplification keeps outputs equal on the original domain.
It does not add starting inputs that the original composition rejected.

The formula x returns its input, but the original reciprocal chain still excludes zero.

Worked exampleTwo reciprocals hide a hole

Let f(x) = $1 over x$ and g(x) = $1 over x$ . Find f(g(x)) and its domain. You want the result of taking the reciprocal twice, while respecting the first division. Plan: inspect the original inside rule, then inspect the outside rule at the inside answer, keeping both restrictions.

The two stages return the original input only when the first stage exists.

Require x ≠ 0 for g(x).The inner reciprocal cannot divide by zero.
The outer function requires g(x) ≠ 0; this holds for every x ≠ 0. $1 over x$ cannot equal zero when its denominator is nonzero.
f(g(x)) = 1 ÷ $1 over x$ = x, with x ≠ 0.Dividing by a nonzero fraction multiplies by its reciprocal.

Answer

f(g(x)) = x with x ≠ 0.
Domain: (−∞, 0) ∪ (0, ∞).

Check At x = 4, g(4) = and f() = 4. At 0 the first reciprocal fails even though the simplified expression x would have a value.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Two reciprocals give x, so zero is now allowed.

The first reciprocal at zero is still .

✓ Instead: Keep x ≠ 0 beside the formula x.

Tips and tricks

A function answer includes its formula and the original allowed inputs.

.5The inner range can settle the second gate

The range is the list of answers a machine can produce. If every answer on that list is accepted by the next machine, gate 2 needs no extra exclusions. If some answers fail, keep only starts that avoid them.

For g(x) = $x^{2}$ , its outputs are [0, ∞). The root f(x) = $\sqrt{x}$ accepts every one of them.
If g(x) = x − 5, its outputs include negatives, so an outer root requires x ≥ 5.

The square only produces heights zero and above, which the next root accepts.

Worked exampleAll square outputs fit the root

Let g(x) = $x^{2}$ and f(x) = $\sqrt{x}$ . Find f(g(x)) and its domain. You want to know whether every square answer can enter the root. Plan: compare g’s outputs with f’s accepted inputs.

Every square output is an accepted root input.

Every real x can be squared, and $x^{2}$ ≥ 0.A square of a real number never gives a negative answer.
Every such output can enter f, so f(g(x)) = $\sqrt{x^{2}}$ = |x| for every real x.The root returns the nonnegative size of x.

Answer

f(g(x)) = |x|.
Domain: (−∞, ∞).

Check At x = −3, the path is −3 to 9 to 3. The answer is |−3| = 3, so the final output need not equal the signed start.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Every g can be followed by a root on its whole domain.

For g(x) = x − 5, start 4 gives −1, which a real root rejects.

✓ Instead: For this g, solve x − 5 ≥ 0 to find allowed starts x ≥ 5; the domain is [5, ∞).

Tips and tricks

Compare the inner outputs to the outer accepted inputs; the two lists describe different stages.

Strategy: step by step

1. Find the inner domain. Record inputs making an inner denominator zero or violating any other inner requirement.
2. Find what the outer function cannot accept. For an outer denominator, determine its forbidden input values.
3. Set g(x) equal to each forbidden outer input and solve. Exclude the solutions that are in the inner domain.
4. Keep only starting inputs passing both gates. For roots, use inequalities rather than only equations, as the next lesson explains.
5. Write the result in interval notation and check a permitted input and each excluded input using the original stages.

Strategy

Find the domain of f(g(x))

Does g have a fraction bottom that depends on x?

YesSet that bottom = 0, solve, plug back in, and exclude those starts at gate 1.

NoContinue to the root check.

↓

Does g have an even root?

YesRequire the inside ≥ 0, or > 0 when that root is a bottom. Solve for the allowed starts.

NoContinue to gate 2.

↓

Does f have a fraction bottom that depends on its input?

YesReplace its input with g(x). Set the resulting bottom = 0 and solve to locate starts to exclude at gate 2.

NoContinue to the outer root check.

↓

Does f have an even root?

YesReplace its input with g(x). Require its inside ≥ 0, or > 0 when that root is a bottom. Keep the starts satisfying this requirement.

NoGate 2 has no root requirement.

↓

Does a square-root output have to equal a negative forbidden input?

YesThere is no such start because a square root cannot give a negative answer. Do not exclude a candidate obtained by squaring that impossible equation.

NoSolve any remaining equation, then check its candidates in the original expression.

Gate 1: begin with any stated input list or interval for the inner function, then write every requirement in its original formula. A bottom must be nonzero; an even root must have a nonnegative inside; a root in a bottom needs a positive inside.
Gate 2: keep any stated outer input list or interval, and write the outer formula’s requirements with g(x) in every input slot. These requirements check the in-between answer.
To locate a zero bottom, set it equal to 0 and solve. This finds starts that break the division. Substitute each candidate into the original stages to confirm which gate fails.
For a root, solve an inequality for its inside. When removing a root by squaring, first check that both compared quantities are nonnegative.
Keep only starting inputs that pass both gates. Write intervals with included endpoints closed and excluded endpoints open. Retain every original restriction after simplification.

Worked exampleTwo different failures in a fraction composition

Let f(x) = $5 over x − 1$ and g(x) = $4 over 3 x − 2$ . Find the domain of (f ∘ g)(x). You want every starting number that can get through g and then f. Plan: find where the first bottom is zero, then find where g hands f the number 1; exclude both starts.

The open points exclude and 2. The label 0.67 approximates ; use the exact fraction in your answer.

Gate 1: set 3x − 2 = 0 to find the starting input that makes g divide by zero.A zero inner bottom prevents any in-between answer.
Add 2: 3x = 2. Divide by 3: x = $2 over 3$ . Plug back in: 3 × $2 over 3$ − 2 = 2 − 2 = 0. Exclude $2 over 3$ .Balanced operations find the failed start, and substitution confirms the zero bottom.
Gate 2: f cannot accept 1, since 1 − 1 = 0. Find every start whose g answer is 1.The outer function receives g(x), so the forbidden number is a handoff value.
Solve $4 over 3 x − 2$ = 1. On the inner domain multiply by the nonzero 3x − 2: 4 = 3x − 2.This locates starts that reach the forbidden outer input. The first exclusion makes the multiplication legal.
Add 2: 6 = 3x. Divide by 3: x = 2.These operations find the start that causes the second failure.
Plug back in: g(2) = $4 over 6 − 2$ = 1, and f(1) = $5 over 0$ . Exclude 2.The first stage works at 2, but its answer breaks the second division.
Keep x ≠ $2 over 3$ and x ≠ 2. Split the line at these two excluded points.All remaining starts pass both original bottoms.

Answer

Domain: (−∞, $2 over 3$ ) ∪ ( $2 over 3$ , 2) ∪ (2, ∞).
Set-builder: {x | x ≠ $2 over 3$ and x ≠ 2}.

Check At x = 0, g(0) = −2 and f(−2) = −, so a permitted input works. At x = , g fails. At x = 2, g(2) = 1 exists but f(1) fails. The two exclusions have different causes.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: both gates accept every real number

Let g(x) = x + 3 and f(x) = 2x. Find the domain of f(g(x)). You only need to decide which starting inputs can pass both rules. Plan: inspect the original inside rule, then inspect the outside rule at the inside answer, keeping both restrictions.

No starting inputs are excluded.

g accepts every real x.Adding 3 has no restriction.
f accepts every real intermediate output.Multiplying by 2 has no restriction.

Answer

(−∞, ∞).

Check Even x = −10 works: g(−10) = −7 and f(−7) = −14; the same two arithmetic actions work for every real input.

Rung 2Rung 2: one forbidden outer value

Let g(x) = x + 2 and f(x) = $1 over x$ . Find the domain of f(g(x)). You must prevent the intermediate answer from becoming zero. Plan: inspect the original inside rule, then inspect the outside rule at the inside answer, keeping both restrictions.

The forbidden intermediate zero comes from starting input −2.

g accepts all real inputs, but f requires g(x) ≠ 0.The reciprocal's denominator is its input.
Set x + 2 = 0 to locate the forbidden handoff zero. Subtract 2: x = −2. Plug back in: g(−2) = 0, so f(g(−2)) = $1 over 0$ . Exclude −2.This solves for the starting input that breaks the outer division.

Answer

(−∞, −2) ∪ (−2, ∞).

Check At 0 the path is 0 to 2 to . At −2, the path stops after producing 0.

Rung 3Rung 3: an inner restriction only

Let g(x) = $2 over x + 4$ and f(x) = $x^{2}$ . Find the composite domain. You want a defined inner fraction followed by a square. Plan: inspect the original inside rule, then inspect the outside rule at the inside answer, keeping both restrictions.

Only the inner denominator limits this path.

Set x + 4 = 0 to locate the inner zero bottom. Subtract 4: x = −4. Plug back in: −4 + 4 = 0. Exclude −4.The equation identifies the failed first-stage input.
No further exclusion is needed.The outer square accepts any real intermediate output.

Answer

(−∞, −4) ∪ (−4, ∞).

Check At −3, g(−3) = 2 and f(2) = 4. At −4 the first stage divides by zero.

Rung 4Rung 4: two fraction exclusions

Let g(x) = $2 over x + 1$ and f(x) = $1 over x − 3$ . Find the domain of f(g(x)). You must avoid an inner zero denominator and an intermediate value of 3. Plan: inspect the original inside rule, then inspect the outside rule at the inside answer, keeping both restrictions.

Exclude −1 and −. The plotted −0.33 approximates −; keep the exact fraction in your answer.

Gate 1: x + 1 = 0 gives x = −1. Plug back in: −1 + 1 = 0. Exclude −1.This finds and confirms the inner zero bottom.
At gate 2, solve $2 over x + 1$ = 3: 2 = 3(x + 1).f rejects input 3, and x + 1 is nonzero on the inner domain.
2 = 3x + 3, so −1 = 3x and x = − $1 over 3$ . Plug back in: x + 1 = $2 over 3$ ; g(x) = 2 ÷ $2 over 3$ = 3, so f(3) has bottom 0. Exclude this start.This start makes the outer denominator zero.

Answer

(−∞, −1) ∪ (−1, −) ∪ (−, ∞).

Check At 0, g(0) = 2 and f(2) = −1. At −1 g fails. At −, g returns 3 and f fails.

Rung 5Rung 5: the first restriction disappears from the formula

Let g(x) = $1 over x − 5$ and f(x) = $1 over x$ . Find the formula and domain of f(g(x)). Taking the reciprocal twice will shorten the formula, but both stages must first exist. Plan: inspect the original inside rule, then inspect the outside rule at the inside answer, keeping both restrictions.

The domain preserves the missing starting input after simplification.

Set x − 5 = 0 to find the failed inner bottom. Add 5: x = 5. Substitution gives 5 − 5 = 0. Exclude 5.The shorter final expression must retain this original failed input.
The inner output is never zero on that domain.A fraction with numerator 1 and a nonzero denominator cannot be zero.
f(g(x)) = 1 ÷ $1 over x − 5$ = x − 5, still with x ≠ 5.The reciprocal undoes the first reciprocal only at allowed inputs.

Answer

f(g(x)) = x − 5, x ≠ 5.
Domain: (−∞, 5) ∪ (5, ∞).

Check At 7 the path is 7 to to 2, matching 7 − 5. At 5 the first step is undefined, so the apparent simplified value 0 cannot be used.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Intersect the domains of f and g directly to find the composite domain.

The two functions receive different inputs: g receives x and f receives g(x).

✓ Instead: For f(x) = and g(x) = 2x, exclude x = , not x = 1.

✗ Not this: If the composition simplifies to x, its domain must be all real numbers.

The original inner reciprocal may fail at zero before simplification is possible.

✓ Instead: For two reciprocal functions, retain x ≠ 0.

Tips and tricks

Know cold: two gates, both must open. Write gate 1 and gate 2 as separate lines on the exam.
Put on the cheat sheet for study: x must be accepted by the inner function, and the inner output must be accepted by the outer function.
The whole inner range must fit the outer domain to compose on every inner input. If only some outputs fit, keep only the starting inputs that produce those outputs.
Set-builder conditions that remove two points use and: {x | x ≠ a and x ≠ b}. Interval pieces use ∪ because an input may belong to any one piece.
Gate 1 checks x. Gate 2 checks g(x).

Trap. Checking only the simplified final denominator. That can lose an inner restriction which disappeared during algebra.