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Lesson 10 — Trim 1 Consolidation: integrating workshop

Integration workshop for the 9 previous lessons. Problems combining functions, rate of change, exponential, and modeling. ENEM/EJU/Abitur style.

Used in: 1.º ano EM

\frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}

The average rate of change is the guiding thread of Trim 1: it links the linear function (constant ARC), the quadratic (linear ARC), the exponential (ARC proportional to value), and paves the way for the derivative.

Choose your door

Rigorous notation, full derivation, hypotheses

Trimester roadmap

This lesson introduces no new content. It is an integrating workshop with problems requiring you to combine:

Lesson 1: set notation, intervals, set operations
Lesson 2: domain, range, composition, injectivity
Lessons 3–4: linear and quadratic functions
Lesson 5: formal composition and inverse
Lessons 6–8: exponential, logarithm, growth/decay models
Lesson 9: average rate of change

Conceptual arc of the trimester

Trim 1 builds a single idea from the ground up: how to describe change.

\underbrace{\text{Sets}}_{\text{L1}} \to \underbrace{\text{Functions}}_{\text{L2}} \to \underbrace{f(x) = ax+b}_{\text{L3: constant ARC}} \to \underbrace{f(x) = ax^2+bx+c}_{\text{L4: linear ARC}} \to \underbrace{f \circ g,\ f^{-1}}_{\text{L5}} \to \underbrace{a^x,\ \ln x}_{\text{L6-7: ARC proportional to value}} \to \underbrace{N_0 e^{kt}}_{\text{L8}} \to \underbrace{\Delta y/\Delta x}_{\text{L9: gateway to calculus}}

Each step answers the question "what happens to $y$ when $x$ changes a little?": linear (always the same), quadratic (grows linearly), exponential (grows proportionally).

Prerequisite map

Concept	Lesson	Used for here
Sets and intervals	1	Domain of exponential/log; intersection of conditions
Function and composition	2, 5	$(f \circ g)(x)$ , inverse
Linear and quadratic	3, 4	Linear/parabolic modeling
Exponential and log	6, 7, 8	Interest, decay, half-life
ARC	9	Average speed, marginal cost

Suggested self-assessment

Set aside 4 h without references to solve. Check with the answer key (25% have inline answers). If you get less than 50% correct, re-read the corresponding lessons; if you get 70–90%, you are ready for Trim 2; above 90%, additional reading is suggested.

Worked examples

Example— 1· Domain of a composition with log (application)

Problem: Find the maximal domain of $f(x) = \log_2(x^2 - 4)$ .

Strategy: The argument of the logarithm must be strictly positive. Solve the inequality $x^2 - 4 > 0$ by factoring and sign analysis.

Solution:

Condition: $x^2 - 4 > 0$ , i.e., $(x - 2)(x + 2) > 0$ .
Roots: $x = 2$ and $x = -2$ . Parabola opening upward.
The inequality holds "outside the roots": $x < -2$ or $x > 2$ .
Domain: $\text{Dom}(f) = (-\infty, -2) \cup (2, +\infty)$ .

Verification: At $x = 3$ : $f(3) = \log_2(5) \approx 2{,}32$ , defined. At $x = 0$ : $0^2 - 4 = -4 < 0$ — undefined, consistent with $0 \notin \text{Dom}(f)$ . At $x = 2$ : $\log_2(0) \to -\infty$ — not in the domain.

Source. OpenStax College Algebra 2e §6.3 — CC-BY 4.0 license.

Example— 2· Exponential equation via substitution (intermediate)

Problem: Solve $4^x - 5 \cdot 2^x + 4 = 0$ .

Strategy: Note that $4^x = (2^x)^2$ . Substitute $u = 2^x$ (with $u > 0$ ): the equation becomes quadratic in $u$ .

Solution:

Substitute $u = 2^x$ : $u^2 - 5u + 4 = 0$ .
Factor: $(u - 1)(u - 4) = 0 \Rightarrow u = 1$ or $u = 4$ .
Back to $x$ : $2^x = 1 \Rightarrow x = 0$ . $2^x = 4 \Rightarrow x = 2$ .
Verify in the original: $4^0 - 5 \cdot 2^0 + 4 = 1 - 5 + 4 = 0$ ✓. $4^2 - 5 \cdot 2^2 + 4 = 16 - 20 + 4 = 0$ ✓.
Answer: $x = 0$ or $x = 2$ .

Verification: Both solutions satisfy the original equation. The substitution $u = 2^x$ is canonical: whenever $a^{2x}$ appears together with $a^x$ , substitute $u = a^x$ .

Source. Stitz–Zeager Precalculus §6.3 — CC-BY-NC-SA license.

Example— 3· Modeling with a piecewise function (intermediate)

Problem: A pool is filled in two stages: during the first 2 hours, at a constant flow rate of 500 L/h; after that, at a constant flow rate of 800 L/h. Model the volume $V(t)$ . How long does it take to fill a 6,000 L pool?

Strategy: Piecewise function (linear in each piece). Calculate $V(2)$ at the end of the first phase; then solve $V(t) = 6\,000$ in the second phase.

Solution:

Model: $V(t) = 500t$ for $0 \le t \le 2$ , and $V(t) = 1{,}000 + 800(t-2)$ for $t > 2$ .
At $t = 2$ : $V(2) = 1\,000$ L.
Remaining: $6\,000 - 1\,000 = 5\,000$ L.
Time in 2nd phase: $5\,000 / 800 = 6{,}25$ h.
Total time: $2 + 6{,}25 = 8{,}25$ h.

Verification: $V(8{,}25) = 1\,000 + 800 \cdot (8{,}25 - 2) = 1\,000 + 5\,000 = 6\,000$ L. ✓ Continuity at $t = 2$ : both expressions give 1,000 L.

Source. Yoshiwara — Modeling, Functions, and Graphs ch. 2 — open license.

Example— 4· Maximum profit with quadratic (advanced)

Problem: A company has cost $C(q) = 50 + 20 q + 0{,}5 q^2$ and revenue $R(q) = 60 q$ . Profit is $L(q) = R(q) - C(q)$ . (a) When is profit zero? (b) What quantity maximizes profit?

Strategy: Compute $L(q)$ , find roots for (a) and the vertex of the parabola for (b). Use the quadratic formula and $x_v = -b/(2a)$ .

Solution:

$L(q) = 60 q - (50 + 20 q + 0{,}5 q^2) = -0{,}5 q^2 + 40 q - 50$ .
(a) Zero profit: $-0{,}5 q^2 + 40 q - 50 = 0$ . Multiplying by $-2$ : $q^2 - 80 q + 100 = 0$ . Quadratic formula: $q = (80 \pm \sqrt{6\,000})/2 \approx (80 \pm 77{,}46)/2$ . Roots: $q \approx 1{,}27$ or $q \approx 78{,}73$ .
(b) Maximum profit: vertex at $q^* = -40/(2 \cdot (-0{,}5)) = 40$ . Maximum profit: $L(40) = -800 + 1\,600 - 50 = 750$ .
Answer: (a) break-even at $q \approx 1{,}27$ and $q \approx 78{,}73$ ; (b) maximum profit R $750 at$ q = 40$ units.

Verification: Parabola opening downward (coeff. $-0{,}5 < 0$ ) — vertex is a maximum ✓. Average of roots: $(1{,}27 + 78{,}73)/2 = 40 = q^*$ ✓.

Source. OpenStax College Algebra 2e §3.2 — CC-BY 4.0 license.

Example— 5· Logistic model — inflection point (real-world modeling)

Problem: $P(t) = 200/(1 + 9 e^{-0{,}5 t})$ (logistic model). (a) Compute $P(0)$ . (b) What is the carrying capacity? (c) At what time $t$ does the population reach half the carrying capacity?

Strategy: For (a) substitute $t = 0$ . For (b) examine $t \to \infty$ . For (c) solve $P(t) = 100$ .

Solution:

(a) $P(0) = 200/(1 + 9) = 20$ . Initial population = 20.
(b) As $t \to \infty$ : $e^{-0{,}5 t} \to 0$ , so $P(t) \to 200$ . Carrying capacity = 200.
(c) $200/(1 + 9 e^{-0{,}5 t}) = 100 \Rightarrow e^{-0{,}5 t} = 1/9 \Rightarrow t = 2 \ln 9 \approx 4{,}39$ .

Verification: At $t = 4{,}39$ : $P = 200/(1 + 9 \cdot 1/9) = 100$ ✓. This is the inflection point of the logistic curve — where the ARC is at its maximum.

Source. Lebl — Notes on Diffy Qs §1.5 — CC-BY-SA license.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 26Understanding 9Modeling 3Challenge 2

Sources

Only books that directly fed the text and exercises. Full catalog at /livros.

OpenStax — College Algebra 2e — Abramson et al. · 2022 · EN · CC-BY 4.0 · §3–6. Central source for this workshop.
Stitz–Zeager Precalculus — Stitz, Zeager · 2013, v3 · EN · CC-BY-NC-SA · ch. 1, 4–6.
Yoshiwara — Modeling, Functions, and Graphs — Katherine Yoshiwara · 2020 · EN · open · ch. 1–6. Source for the modeling block.
Active Calculus 2.0 — Matt Boelkins · 2024 · EN · CC-BY-NC-SA · §1.3 (ARC).
OpenStax — Calculus Volume 1 — OpenStax · 2016 · EN · CC-BY-NC-SA · §2.1.
Notes on Diffy Qs — Jiří Lebl · 2024, v6.6 · EN · CC-BY-SA · §1.4–1.5.
Hammack — Book of Proof — Richard Hammack · 2018 · EN · open · ch. 4, 12.

Full catalog (80+ books in 12 languages) at /livros.