Week 3 — Drills

Exam format. Show all work. Core items are for everyone; stretch items for whoever finishes early. Solutions at the end, worked the way an examiner wants to see them.

The seven-point study, in the order the midterm expects it: domain → zeros → sign → stationary points → nature (max/min) → shape (convexity, inflection) → sketch.

A. The full study of a function

D1 (core — the midterm rehearsal). Study π(q) = −4q² + 1200q − 60,000 completely and sketch it. Give: domain, zeros, sign line, π' and the stationary point, π'' and its nature, behaviour as q → ∞, and a labelled sketch.

D2 (core). A cost function: C(q) = 0.4q² + 3q + 25 (€ thousand), for q ≥ 0. (a) Domain, and C(0). What is C(0) called? (b) Average cost AC(q) = C(q)/q. Find its domain and lim (q→0⁺) AC(q) and lim (q→∞) AC(q). (c) Minimise AC(q). At that quantity, compute AC and C'(q) and comment on what you notice.

D2b (core — asymptotes, in the resit’s own form). f(x) = x² − 3x + 7 ln(x + 3). (a) State the domain as an interval. (b) Evaluate lim (x→−3⁺) f(x) and say what it tells you about the graph. (c) Evaluate lim (x→+∞) f(x). (d) Sketch the shape, marking the asymptote.

D3 (stretch). f(q) = q³ − 12q² + 36q on [0, 10]. (a) Stationary points and their nature. (b) Inflection point. (c) Sketch.

D3b (core — step 6 on its own). f(q) = q³ − 6q² + 9q + 4, for q ≥ 0. (a) Compute f''(q) and solve f''(q) = 0. (b) Give the sign of f'' on each side of that point and name the intervals: convex where, concave where. (c) The inflection point, with both coordinates. (d) The stationary points, and the nature of each one — using f'', not a sketch. (e) One sentence a report would use for step 6.

B. Optimisation, and what it is for

D4 (core). A second firm: R(q) = 30 ln(q + 1), C(q) = 0.4q² + 3q + 25, both in € thousand, q ≥ 0. (a) Write π(q). (b) Write π'(q). (c) Solve π'(q) = 0. Leave it as an equation in q first, then solve the quadratic it becomes. (d) Confirm with π'' that it is a maximum. (e) Is the firm profitable at that quantity? Answer with a number.

D5 (core — the managerial reading). For the firm in D4, R'(q) is marginal revenue and C'(q) is marginal cost. (a) Write the condition π'(q) = 0 as R'(q) = C'(q), in words. (b) At q = 2 compute both. Which is larger, and what should the firm do?

D6 (stretch). A box with a square base and no lid must hold 32 m³. Minimise the material used. (Surface S = x² + 4xh, volume x²h = 32.)

C. Integration

D7 (core — exam form). Evaluate, splitting term by term: (a) ∫ (q⁵ − 3q³ + 2)/q² dq (b) ∫ (4q³ − 6q + 5) dq (c) ∫ (2/q + 3√q) dq

D8 (core — exam form). Find f(q) given f'(q) = 12q² − 6q + 1 and f(2) = 20.

D8b (core — exam form, and the one that catches people). Find f(x) given f'(x) = 12x² − 4x and f(−3) = 17. Watch every sign.

D8c (core — the third rule). Evaluate, by asking first whether the top is the derivative of the bottom: (a) ∫ 1/(q − 4) dq (b) ∫ q/(q² + 5) dq (c) ∫ (2q + 3)/(q² + 3q − 1) dq (d) ∫ 5/(2q + 1) dq

D9 (core). Evaluate the definite integrals and say what each measures: (a) ∫₀²⁰ (0.8q + 3) dq where 0.8q + 3 is marginal cost. (b) ∫₁⁴ (6/q) dq. Leave the answer in terms of ln.

D9b (core — exam form). Evaluate ∫₀^(1/2) [ 1/(x − 1) + x/(x² − 1) ] dx, leaving the answer in terms of ln. Both terms are logs; neither is 1/x.

D10 (stretch). Marginal profit is π'(q) = 1200 − 8q. The firm currently makes 100 bikes. (a) By how much does profit change if output rises from 100 to 150? Compute it as a definite integral. (b) Check your answer against π(150) − π(100) using π(q) = −4q² + 1200q − 60,000.

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Solutions

D1. - Domain. A polynomial: all of ℝ. For a quantity, [0, +∞). - Zeros. −4q² + 1200q − 60,000 = 0; divide by −4: q² − 300q + 15,000 = 0. Δ = 90,000 − 60,000 = 30,000, √Δ ≈ 173.2. q = (300 ± 173.2)/2 → q ≈ 63.4 and q ≈ 236.6. - Sign. a = −4 < 0, so positive between the roots: π > 0 on (63.4, 236.6), negative outside. - Stationary point. π'(q) = −8q + 1200 = 0 → q = 150. - Nature. π''(q) = −8 < 0 for every q, so the stationary point is a maximum; the curve is concave everywhere. π(150) = −90,000 + 180,000 − 60,000 = **€30,000**. - Behaviour. lim (q→∞) π(q) = −∞: the −4q² term wins. There is no “sell more forever”. - Sketch. Downward parabola, crossing the axis at 63.4 and 236.6, peak (150, 30,000), intercept π(0) = −60,000. Label all four.

D2. (a) [0, +∞); C(0) = 25. That is fixed cost: what the firm pays having produced nothing. (b) AC(q) = 0.4q + 3 + 25/q, domain (0, +∞) — zero output has no average. lim (q→0⁺) AC = +∞ (fixed cost spread over almost nothing). lim (q→∞) AC = +∞ (the 0.4q term takes over). So AC falls, then rises: it has a minimum. (c) AC'(q) = 0.4 − 25/q² = 0 → q² = 62.5 → q ≈ 7.91. AC(7.91) = 3.16 + 3 + 3.16 = **9.32**; C'(7.91) = 0.8(7.91) + 3 = **9.32**. They are equal. Average cost is at its lowest exactly where marginal cost crosses it. Worth stating out loud: it is not a coincidence, it is what “average” means.

D2b. (a) ln(x + 3) needs x + 3 > 0, so the domain is (−3, +∞) — open, because ln 0 is undefined. (b) As x → −3⁺, ln(x+3) → −∞, and x² − 3x → 18, so f(x) → **−∞**. The line x = −3 is a vertical asymptote: the curve plunges down it. (c) As x → +∞ the x² term dominates everything, so f(x) → **+∞**. (d) f'(x) = 2x − 3 + 7/(x + 3) = (2x − 1)(x + 2)/(x + 3), zero at x = −2 and x = ½. So the curve comes up from −∞ along x = −3, reaches a local maximum at (−2, 10), dips to a local minimum at (½, ≈ 7.52), then rises to +∞. Two turns, not one. Draw the asymptote as a dashed vertical line: the marker looks for it.

D3. (a) f'(q) = 3q² − 24q + 36 = 3(q² − 8q + 12) = 3(q − 2)(q − 6) → q = 2, 6. f''(q) = 6q − 24: f''(2) = −12 < 0 → maximum, f(2) = 8 − 48 + 72 = 32. f''(6) = 12 > 0 → minimum, f(6) = 216 − 432 + 216 = 0. (b) f'' = 0 at q = 4, f(4) = 64 − 192 + 144 = 16; the curvature changes sign there, so it is an inflection. (c) Rises to (2, 32), falls to (6, 0), rises again; at q = 10, f = 1000 − 1200 + 360 = 160.

D3b. (a) f'(q) = 3q² − 12q + 9, f''(q) = 6q − 12. f'' = 0 at q = 2. (b) f''(1) = −6 < 0: concave on [0, 2). f''(3) = 6 > 0: convex on (2, ∞). The sign changes, so q = 2 is a genuine inflection and not just a zero. (c) f(2) = 8 − 24 + 18 + 4 = 6: inflection at (2, 6). Both coordinates, or the point is not located. (d) f'(q) = 3(q − 1)(q − 3) = 0 → q = 1, 3. f''(1) = −6 < 0 → maximum, f(1) = 8. f''(3) = 6 > 0 → minimum, f(3) = 4. Max at 1, inflection at 2, min at 3: the inflection sits between the turning points, where the fall is steepest. (e) “The curve is concave up to q = 2 and convex beyond it, with an inflection at (2, 6): the decline that begins at the maximum is steepest there, and slows afterwards until the minimum at q = 3.”

D4. (a) π(q) = 30 ln(q + 1) − 0.4q² − 3q − 25. (b) π'(q) = 30/(q + 1) − 0.8q − 3. (c) 30/(q + 1) = 0.8q + 3. Multiply by (q + 1): 30 = (0.8q + 3)(q + 1) = 0.8q² + 3.8q + 3 → 0.8q² + 3.8q − 27 = 0. Multiply by 5: 4q² + 19q − 135 = 0. Δ = 361 + 2160 = 2521, √Δ ≈ 50.21. q = (−19 ± 50.21)/8 → q ≈ 3.90 (the negative root is discarded: a quantity). (d) π''(q) = −30/(q + 1)² − 0.8 < 0 everywhere → maximum. (e) π(3.90) = 30 ln(4.90) − 0.4(15.21) − 3(3.90) − 25 = 47.68 − 6.08 − 11.70 − 25 = **€4.89k**. Profitable, but only just — the window is narrow, which is the point.

D5. (a) π' = R' − C' = 0 means R'(q) = C'(q): produce up to the point where the revenue from one more unit equals what that unit costs. (b) R'(2) = 30/3 = **10**; C'(2) = 0.8(2) + 3 = **4.6**. Marginal revenue is larger, so each extra unit still adds 10 − 4.6 = 5.4 — produce more, up to q ≈ 3.90.

D6. h = 32/x², so S = x² + 4x(32/x²) = x² + 128/x. S' = 2x − 128/x² = 0 → x³ = 64 → x = 4, h = 2. S'' = 2 + 256/x³ > 0 → minimum. S = 16 + 32 = **48 m²**.

D7. (a) Split first: (q⁵ − 3q³ + 2)/q² = q³ − 3q + 2q⁻². ∫ = q⁴/4 − 3q²/2 − 2/q + c. (b) q⁴ − 3q² + 5q + c. (c) ∫ 2/q dq = 2 ln|q|; ∫ 3q^{1/2} dq = 3 · (2/3) q^{3/2} = 2q^{3/2}. Total: 2 ln|q| + 2q^{3/2} + c. The exam’s integral is always this shape: divide term by term first, then each term is a power or a log.

D8. f(q) = 4q³ − 3q² + q + c. f(2) = 32 − 12 + 2 + c = 20 → c = −2. f(q) = 4q³ − 3q² + q − 2.

D8b. f(x) = 4x³ − 2x² + c. Now substitute carefully: f(−3) = 4(−27) − 2(9) + c = −108 − 18 + c = −126 + c = 17 → c = 143. f(x) = 4x³ − 2x² + 143. The trap is (−3)³ = −27 and (−3)² = +9. Two signs, two marks.

D8c. Ask each time: is the top the derivative of the bottom? (a) bottom differentiates to 1 → ln|q − 4| + c. (b) bottom differentiates to 2q, top is q, so a factor ½ is missing → ½ ln(q² + 5) + c. (No modulus needed: q² + 5 > 0 always.) (c) bottom differentiates to 2q + 3, which is exactly the top → ln|q² + 3q − 1| + c. (d) bottom differentiates to 2, top is 5, so 5/2 comes out → (5/2) ln|2q + 1| + c.

D9. (a) [0.4q² + 3q]₀²⁰ = 160 + 60 = **220**. Total variable cost of the first 20 bikes: adding up marginal cost gives total cost. (b) [6 ln q]₁⁴ = 6 ln 4 − 6 ln 1 = **6 ln 4 ≈ 8.32**.

D9b. Each term is a log. ∫ 1/(x−1) dx = ln|x−1|; ∫ x/(x²−1) dx = ½ ln|x²−1|. So the antiderivative is ln|x−1| + ½ ln|x²−1|, and

at ½ : ln(½) + ½ ln(¾) = −0.6931 + ½(−0.2877) = −0.8370 at 0 : ln 1 + ½ ln 1 = 0

= −0.8370, or exactly ln(½) + ½ ln(¾). Note the modulus: between 0 and ½ both x − 1 and x² − 1 are negative, and without the absolute value the logs are undefined. The exam’s model answer expects it.

D10. (a) ∫₁₀₀¹⁵⁰ (1200 − 8q) dq = [1200q − 4q²]₁₀₀¹⁵⁰ = (180,000 − 90,000) − (120,000 − 40,000) = 90,000 − 80,000 = **€10,000**. (b) π(150) = 30,000; π(100) = −40,000 + 120,000 − 60,000 = 20,000. Difference = **€10,000**. ✔ The integral of the marginal is the change in the total. That sentence is the whole of integration.