Lesson 20: Precalculus Capstone: Readiness for Calculus

This capstone gathers the four pillars the course was built on and stress-tests them together, the way calculus will on its first day. The pillars: algebraic fluency (factoring, rationalising, equation-solving without hesitation), functions (graphs, transformations, inverses, domains), trigonometry (unit circle, identities, equations), and limits (the language of both derivatives and integrals). The problems below are deliberately mixed — calculus problems never announce which pillar they are using.

Synthesis 1: the difference quotient, one more time

For f(x) = x^2 − 5x + 2, the difference quotient is ((x + h)^2 − 5(x + h) + 2 − (x^2 − 5x + 2))/h = (2xh + h^2 − 5h)/h = 2x + h − 5, so f'(x) = 2x − 5. At x = 3: f'(3) = 1, and f(3) = 9 − 15 + 2 = −4, so the tangent line at (3, −4) is y + 4 = 1(x − 3), i.e. y = x − 7. In one short computation you used algebra (expansion), functions (evaluation) and limits (letting h → 0) — Lesson 11's workflow, now expected to be fluent. The readiness benchmark: any quadratic's difference quotient should take you under two minutes, error-free.

Synthesis 2: limits that previews calculus results

Some limits are famous because calculus needs them immediately. lim(x→0) (e^x − 1)/x: the table x = 0.1 → 1.0517, x = 0.01 → 1.00502, x = −0.01 → 0.99502 closes on 1 — this is why the derivative of e^x is e^x itself. lim(x→0) sin(4x)/(3x) = (4/3)·[sin(4x)/(4x)] → 4/3 by Lesson 9's special limit. And lim(x→0) (1 − cos x)/x^2 → 1/2 (multiply by the conjugate 1 + cos x to expose sin^2 x). A calculus course proves these; a precalculus course that has evaluated them numerically and algebraically has removed the mystery in advance.

Synthesis 3: area, estimated and exact

For f(x) = 4 − x^2 on [0, 2] with n = 4 (Δx = 0.5): the left sum is 0.5 × (4 + 3.75 + 3 + 1.75) = 6.25; the right sum is 0.5 × (3.75 + 3 + 1.75 + 0) = 4.25; the trapezoid average is 5.25. The exact integral, by antiderivative: ∫ = [4x − x^3/3] from 0 to 2 = 8 − 8/3 = 16/3 ≈ 5.333. The trapezoid underestimates here because the curve is concave down (its chords lie below the graph — compare Lesson 12, where concave-up x^2 made the trapezoid overshoot). Bracketing an exact answer between two computable bounds, then closing the bracket with an antiderivative, *is* the Fundamental Theorem in working clothes.

Synthesis 4: domains and trig, under time pressure

Domain drill: f(x) = ln(x − 2) + sqrt(5 − x) needs x − 2 > 0 (logarithm) and 5 − x ≥ 0 (root): domain (2, 5]. Note the asymmetry — the log endpoint is excluded, the root endpoint included — a distinction worth marks. Trig drill: 2 sin x − 1 = 0 on [0, 2π) gives sin x = 1/2, x = π/6 and 5π/6 (Lesson 5's two families). Identity drill: from sin θ = 3/5 (quadrant II), cos 2θ = 1 − 2(9/25) = 7/25 (Lesson 4). None of these is hard; the capstone skill is doing all of them, mixed, without warming up on kindred problems first.

Readiness checklist

You are ready for calculus if you can, without notes:

• Expand, factor and rationalise fluently, including difference quotients for quadratics and roots
• State domains combining roots, denominators and logarithms, and sketch transformed graphs from a parent
• Produce exact trig values from the unit circle, prove a short identity, and solve a trig equation on [0, 2π) completely
• Evaluate limits by factoring, rationalising, and the special trig limits — and say what a limit means
• Estimate an area with left/right/trapezoid sums and check it against an antiderivative for a polynomial
• Fit a line or an exponential (via logs) to small data and interpret slope, intercept and residuals

Common mistakes (the capstone edition)

• Letting h = 0 too early. In difference quotients, cancelling h is the step that makes the limit computable; substituting first yields 0/0 and a stuck problem. The order is: expand, cancel, *then* let h → 0.
• Reaching for the calculator on exact-value questions. sin(4x)/(3x) → 4/3 is an exact statement about a theorem, not 1.333 from a table. Tables support; theorems conclude.
• One-family trig solutions. On [0, 2π), sin x = 1/2 has two answers. Under time pressure the second family is the most commonly dropped item in this entire course.
• Endpoint inclusion errors in domains. (2, 5] versus [2, 5] versus (2, 5): logarithms exclude their boundary, even roots include theirs. Write the inequality for each feature separately, then intersect.
• Treating estimates as exact. 5.25 (trapezoid) and 16/3 ≈ 5.333 (exact) are different objects. Label estimates with ≈ and say which method produced them.

Exam Experience

1. lim(x→0) sin(4x)/(3x) =

• (A) 3/4
• (B) 1
• (C) 4/3
• (D) 4

Answer: (C). sin(4x)/(3x) = (4/3)·[sin(4x)/(4x)] → (4/3)(1) = 4/3. (A) is the inverted ratio — from dividing by 4x incorrectly; (D) forgets the denominator's 3.

2. For f(x) = x^2 − 5x + 2, the slope of the tangent line at x = 3 is

• (A) −4
• (B) 1
• (C) 6
• (D) −1

Answer: (B). f'(x) = 2x − 5 (from the difference quotient 2x + h − 5), so f'(3) = 1. (A) is f(3) = −4, the point's height rather than its slope — the last appearance of this course's most persistent confusion.

3. The domain of f(x) = ln(x − 2) + sqrt(5 − x) is

• (A) [2, 5]
• (B) (2, 5]
• (C) (2, 5)
• (D) [2, 5)

Answer: (B). The logarithm requires x > 2 (strict), the root requires x ≤ 5 (inclusive). Intersecting gives (2, 5] — the endpoints are decided by two different rules, which is the point of the question.

Constructed response. Let f(x) = 4 − x^2 on [0, 2]. (a) Compute the right-endpoint Riemann sum with n = 4. (b) Compute the left-endpoint Riemann sum with n = 4, and the trapezoidal estimate. (c) Evaluate ∫(0 to 2) (4 − x^2) dx exactly using an antiderivative. (d) Is the trapezoidal estimate an overestimate or underestimate? Justify using the shape of the graph.

Scoring guide. (a) 2 points: Δx = 0.5 and right endpoints 0.5, 1, 1.5, 2 with heights 3.75, 3, 1.75, 0 — 1 point; sum 0.5 × 8.5 = 4.25 — 1 point. (b) 2 points: left sum 0.5 × (4 + 3.75 + 3 + 1.75) = 6.25 — 1 point; trapezoid = (6.25 + 4.25)/2 = 5.25 — 1 point. (c) 1 point: [4x − x^3/3] from 0 to 2 = 8 − 8/3 = 16/3 ≈ 5.33. (d) 1 point: underestimate (5.25 < 5.33), because 4 − x^2 is concave down, so the trapezoid chords lie below the curve and shave off area.

Going deeper: a first-week-of-calculus rehearsal

Four problems, one per pillar, at the standard your first calculus week will assume. (1) Limits by rationalising: lim(x→0) (sqrt(x + 4) − 2)/x — multiply by the conjugate: x/(x(sqrt(x + 4) + 2)) = 1/(sqrt(x + 4) + 2) → 1/4. (2) Continuity repair: f(x) = (x^2 − 4)/(x − 2) for x ≠ 2 and f(2) = k is continuous at 2 for exactly one k: the limit is x + 2 → 4, so k = 4 — continuity means the value *equals* the limit (Lesson 9's checklist), and "find k" problems are that definition in disguise. (3) A derivative from the definition, completely: for f(x) = x^3, the difference quotient is ((x + h)^3 − x^3)/h = (3x^2 h + 3xh^2 + h^3)/h = 3x^2 + 3xh + h^2 → 3x^2. The binomial expansion from Lesson 10 is doing the work inside the limit — the course's tools were always one toolkit. (4) Area two ways: ∫(0 to 3) 2x dx = [x^2] = 9 by antiderivative, and the region is a triangle of base 3 and height 6 with area (1/2)(3)(6) = 9 by geometry. When two independent methods agree, confidence is earned; when they disagree, one of them has just taught you something.

Keep the capstone standard as a habit in calculus itself: estimate before you compute, verify after you compute, and know which pillar every problem stands on. The course ahead is new — its foundations are not.

Workshop: symmetry in integrals, and where the derivative vanishes

Symmetry halves the work. Odd functions integrate to zero over symmetric intervals: ∫(−2 to 2) x^3 dx = 0 without computing anything, because the signed areas cancel. Even functions double: ∫(−2 to 2) x^2 dx = 2∫(0 to 2) x^2 dx = 2 × 8/3 = 16/3. These are the integral analogues of Lesson 1's even/odd function tests — the same symmetry, harvested at the other end of the course. Where the tangent is horizontal: for f(x) = x^2 − 5x + 2, f'(x) = 2x − 5 vanishes at x = 5/2, and f(5/2) = −17/4 — the vertex, located by the derivative in one step instead of by completing the square. Optimisation in calculus is this observation industrialised: extrema live where the derivative is zero (or fails to exist), and precalculus has quietly built every tool that statement needs — functions, limits, and the difference quotient that ties them together.

Final drills: the reciprocal trick and equal bases

One limit that rewards a second look: lim(x→4) (x − 4)/(sqrt(x) − 2). Factor the numerator as a difference of squares in disguise: x − 4 = (sqrt(x) − 2)(sqrt(x) + 2), so the quotient is exactly sqrt(x) + 2 for x ≠ 4, and the limit is 4. This is Lesson 9's rationalisation read backwards — the same identity, deployed from the other side. Equal bases, one line: solve 3^(x − 1) = 9^(2x). Rewrite 9 = 3^2: 3^(x − 1) = 3^(4x), so x − 1 = 4x and x = −1/3. Equate-exponents problems are free marks when the bases align — and noticing that 9, 27, 1/3 and sqrt(3) are all powers of 3 is the pattern recognition the whole course has been training.