Connect functions through intermediate values

Try

A fictional sensor turns a raw count r into a temperature T(r)=r/5−4 degrees Celsius. A display then converts Celsius to Fahrenheit using F(c)=9c/5+32. For r=70, calculate the intermediate Celsius temperature before the displayed result. T(70)=10 and F(10)=50.

Draw two boxes with the intermediate value on the arrow between them. Label raw counts, Celsius and Fahrenheit. This representation exposes an error that a long formula can hide: feeding a raw count directly into the conversion rule gives the wrong units as well as the wrong value.

Predict

Predict whether reversing the two formulas produces the same result. Arithmetic permits evaluating T(F(70)), but the intermediate Fahrenheit value is not a raw sensor count. A formally defined composition can still be meaningless as a model of the physical process.

Experiment

Set f(u)=u² and g(x)=x−3. Compute both orders.

x	g(x)	f(g(x))	f(x)	g(f(x))
0	−3	9	0	−3
2	−1	1	4	1
4	1	1	16	13

The two orders agree at x=2 but disagree elsewhere. One matching input does not prove the rules equivalent. Algebra gives f(g(x))=(x−3)² and g(f(x))=x²−3. Setting them equal yields −6x+9=−3, so their agreement occurs only at x=2.

Understand

The notation f∘g means “apply g first, then f.” In f(g(x)), the parentheses show the same nesting. Composition is not ordinary multiplication: f(g(x)) generally differs from f(x)g(x), and it generally differs from g(f(x)).

The domain has two requirements. First x must be in the domain of g. Second g(x) must be in the domain of f. If f(u)=√u and g(x)=x−3, then x≥3. If their order is reversed, g(f(x))=√x−3 only requires x≥0. The same two component rules produce different composite domains.

Simplification cannot replace this check. With f(u)=1/u and g(x)=1/(x−2), the composite formula simplifies to x−2. Nevertheless x=2 remains excluded because the inner operation cannot start. On allowed inputs the inner output is never zero, so it causes no additional exclusion for the outer reciprocal.

To build a composite rule, name what each stage measures. A price discount D(p)=0.9p followed by a fixed fee A(p)=p+4 becomes A(D(p))=0.9p+4. Applying the fee before the discount gives D(A(p))=0.9p+3.6. The difference arises because the fee is discounted in only one ordering. Whether either sequence is contractually correct is a question about the stated policy, not algebra alone.

Three stages can be grouped without changing their order when all intermediate values are permitted. Both f(g(h(x))) and (f∘g)(h(x)) describe the same pipeline. This grouping property does not authorize swapping stages. Always preserve the actual order and intermediate-domain checks.

Run the pipeline backwards: build the component functions

Evaluating a supplied composition and decomposing a new rule are different actions. For H(x)=4+3√(2x−6), first name the operations in the order an input encounters them: a(x)=2x−6; b(u)=√u, defined for u≥0; c(v)=4+3v. Then H=c∘b∘a. The condition a(x)≥0 gives x≥3; the final affine map adds no restriction. Verify by substitution, not by merely listing operations. Grouping b∘a as one function gives another valid two-stage decomposition. There is no uniqueness theorem for this choice of components.

Translations and scalings are themselves functions. If Th(x)=x+h and S_a(x)=ax, then g(x)=A f(B(x−h))+k is T_k∘S_A∘f∘S_B∘T(−h). A graph point (u,v) moves to (h+u/B,Av+k) when B≠0. This derives the horizontal reciprocal scale rather than guessing it. The transformed domain consists precisely of x for which B(x−h) belongs to the original domain. The identity I(x)=x satisfies I∘f=f∘I=f, including the same domain, not a larger one.

Changed practice. Decompose 5−2√(3x+6) into three nonidentity stages, then two stages. Record the intermediate domain and verify the final rule. Next decompose −3f(2(x−4))+5 when f is defined only on [−1,2]. A complete response must retain both the operation order and the restricted domain; the registered tasks below supply worked reasoning and separate rubric criteria.

Construct the entire composite graph

Reading one composite value does not construct a graph. Suppose the exact graph of f joins (0,1),(1,3),(3,−1), with domain [0,3], and g joins (0,0),(2,3),(4,1), with domain [0,4]. First g’s full range [0,3] lies in f’s domain, so f∘g is defined throughout [0,4]. Then include corners of g and preimages of corners of f. The outer corner at input 1 is reached at x=2/3 and at the endpoint x=4.

The full composite has vertices (0,1),(2/3,3),(2,−1),(4,3). Its successive affine rules are 1+3x on [0,2/3], 5−3x on [2/3,2], and 2x−5 on [2,4]. At shared endpoints the formulas agree, so either adjoining convention produces the same function. This is an exact complete coordinate-and-segment graph description, accessible without inspecting a rendered image. Connecting only values at g’s original corners would miss the peak at 2/3. The registered task asks you to derive the breakpoints and pieces, not merely copy sampled pairs.

Apply

The practice includes a sensor pipeline, a canceled reciprocal and two discount policies. Write intermediate values even when mental arithmetic seems quicker. For a domain question, explicitly state both checks. For a contextual comparison, identify the units and any constraints such as nonnegative charges.

A useful error-repair question is “Which input does this outer function actually receive?” Answering it often identifies the wrong branch, excluded value or reversed operation immediately.

Criterion-linked retry: the inner output is the next input

Let f(u)=√(u−1) and g(x)=(x+1)/(x−2), with real-valued functions. The composite f(g(x)) first requires x≠2. Its outer radicand is g(x)−1=3/(x−2). Because 3 is positive, a nonnegative radicand requires x>2. The denominator cannot be zero and the radicand can never equal zero, so the endpoint is not included. The domain is the open interval (2,∞).

Reverse the order. For g(f(x)), the inner square root requires x≥1. The outer function excludes input 2, meaning √(x−1)≠2, or x≠5. Thus the reversed domain is [1,5) together with (5,∞). The forbidden original input is five, not two. Naming the intermediate input prevents that common mistake.

Use the control to inspect x=1,2,3 and 5 in the first composition. The table stops at an undefined inner division instead of trying to evaluate an outer operation on it. A valid rational value may still fail the real-square-root condition. No finite selection of sample points proves the whole interval; the sign argument supplies that proof.

In the changed retry, write two lines before looking at feedback: “inner operation exists when …” and “its output is allowed by the outer operation when …”. Intersect these conditions, then test any boundary separately. This is public criterion-linked practice for composite domains, not a new AP score or evidence that all framework objectives have been assessed.

Repair a composite-domain claim

Review criterion: Check the inner domain first, then require its output to belong to the outer domain.

Predict, then change one input

Original input x:

2

Inner g(x)

Undefined in the real model

Radicand g(x)−1

Undefined in the real model

Real f(g(x))

Undefined in the real model

The inner division is undefined. No outer operation is evaluated.

Explain the changed case: A rational input to a square root

Let f(u)=√(u−1) and g(x)=(x+1)/(x−2), using real values. What is the domain of f(g(x))?

• g(x)−1 = 3/(x−2)
• The inner denominator cannot be zero.

Write a reason before selecting a claim. Feedback is public practice, not a proof grader.

x>2x≥2All real x except 2

Switch to changed retryReset this practice

Inputs and choices are temporary. Nothing is saved, no lesson completion or score is changed, and repeating a public case is not unseen assessment.

Continue

An inverse function reverses a pipeline where the original input can be recovered uniquely. The next lesson examines why a many-to-one function needs a domain restriction before it can be reversed by a function.