Difficulty: Intermediate to Advanced | Prerequisites: Differentiation Rules notes, limits and continuity
This is where differentiation becomes useful. Instead of just computing derivatives, you now use them to answer questions: Where is a function increasing? Where are its peaks and valleys? What is the equation of the tangent line? How should a company set its price to maximise profit? These are the most heavily tested topics on the final exam because they combine several skills (algebra, limits, differentiation) into multi-step problems. If you can handle these, you can handle the exam.
The first derivative tells you where a function is rising or falling, and its zeros locate potential peaks and valleys. The second derivative tells you about the curve's shape (concave up or down) and locates inflection points. Tangent line problems combine differentiation with the point-slope formula. Optimisation problems ask you to maximise or minimise a real-world quantity by finding critical numbers.
Tangent line
The straight line that just touches a curve at a single point, matching the curve's slope there. Its equation is found using the derivative (the slope) and the point-slope formula: y − y₁ = m(x − x₁).
Critical number
A value of x where f'(x) = 0 or f'(x) is undefined. These are the only places where local maxima or minima can occur. Think of them as candidates for peaks and valleys.
Increasing / decreasing
A function is increasing on an interval where f'(x) > 0, and decreasing where f'(x) < 0. In simple terms, the function is going up where the derivative is positive, and going down where it is negative.
Local maximum / local minimum
A local max is a point higher than everything nearby; a local min is a point lower than everything nearby. Found at critical numbers where the derivative changes sign.
Absolute maximum / absolute minimum
The highest or lowest value of a function on a closed interval. Found by checking the function's value at all critical numbers within the interval and at both endpoints.
Second derivative test
A shortcut for classifying critical numbers. If f''(c) > 0, the critical number c is a local minimum (concave up). If f''(c) < 0, it is a local maximum (concave down). If f''(c) = 0, the test is inconclusive.
Inflection point
A point where the concavity changes (from concave up to concave down, or vice versa). Found where f''(x) = 0 or is undefined, provided the concavity actually changes.
Concavity
Concave up means the curve opens upward (like a cup); f''(x) > 0. Concave down means it opens downward (like a frown); f''(x) < 0.
Average rate of change
The slope of the secant line between two points: [f(b) − f(a)] / (b − a). In simple terms, the overall change divided by the interval length.
Marginal analysis
Using the derivative to estimate the change from producing one additional unit. Marginal cost = C'(x), marginal revenue = R'(x), marginal profit = P'(x).
The tangent line at x = a has slope m = f'(a) and passes through (a, f(a)). Use point-slope form: y − f(a) = f'(a)(x − a), then rearrange to slope-intercept form y = mx + b.
Example: y = 2x³ − 3x² − 12x + 4 at x = 1. f(1) = 2−3−12+4 = −9. f'(x) = 6x²−6x−12, f'(1) = 6−6−12 = −12. Tangent line: y − (−9) = −12(x−1), which gives y = −12x + 3.
A horizontal tangent line occurs where f'(x) = 0. For the function above, 6x²−6x−12 = 0, so x²−x−2 = 0, (x−2)(x+1) = 0, giving x = −1 and x = 2.
Average rate of change = [f(b) − f(a)] / (b − a).
Example: Gas cost $2.85 in 2007, $3.43 in 2014. Average rate of change = (3.43−2.85)/(2014−2007) = 0.58/7 ≈ $0.08 per year.
Find f'(x) and set it equal to zero to find critical numbers.
Build a sign chart: test f'(x) in each interval between critical numbers.
Where f'(x) > 0 the function is increasing; where f'(x) < 0 it is decreasing.
If f' changes from positive to negative at a critical number, that is a local max. If it changes from negative to positive, that is a local min.
Example: f(x) = x⁴ + 4x³ + 15. f'(x) = 4x³ + 12x² = 4x²(x+3). Critical numbers: x = 0 and x = −3. Sign chart: f' < 0 on (−∞,−3), f' > 0 on (−3,0), f' > 0 on (0,∞). Decreasing on (−∞,−3), increasing on (−3,∞). Local min at x = −3: f(−3) = 81−108+15 = −12, so local min at (−3,−12). No local extremum at x = 0 (the sign does not change).
Given critical numbers where f'(c) = 0, evaluate f''(c):
f''(c) > 0: local minimum (concave up).
f''(c) < 0: local maximum (concave down).
f''(c) = 0: inconclusive, use the first derivative test instead.
Example: f'(x) = 0 at x = 1 and x = 5, and f''(x) = √(2x−1) − 2. f''(1) = √1 − 2 = −1 < 0, so local max at x = 1. f''(5) = √9 − 2 = 1 > 0, so local min at x = 5.
Find f''(x) and set it equal to zero.
Build a sign chart for f''(x).
Inflection points occur where f'' changes sign.
Example: y = 3x⁵ − 10x⁴. y' = 15x⁴ − 40x³. y'' = 60x³ − 120x² = 60x²(x−2). Set y'' = 0: x = 0 and x = 2. Sign chart: y'' changes sign at x = 2 (negative to positive), so (2, −64) is an inflection point. At x = 0, y'' does not change sign (stays negative on both sides through that zero), so (0,0) is not an inflection point by itself. The answer is (2, −64).
Find all critical numbers of f(x) in the interval.
Evaluate f at each critical number and at both endpoints.
The largest value is the absolute max, the smallest is the absolute min.
Example: f(x) = x⁴ − 8x² − 16 on [−1, 4]. f'(x) = 4x³ − 16x = 4x(x²−4) = 4x(x−2)(x+2). Critical numbers in [−1,4]: x = 0 and x = 2 (x = −2 is outside the interval). f(−1) = 1−8−16 = −23. f(0) = −16. f(2) = 16−32−16 = −32. f(4) = 256−128−16 = 112. Absolute max: 112 at x = 4. Absolute min: −32 at x = 2.
General approach:
Define variables and draw a picture if applicable.
Write the quantity to optimise as a function of one variable.
Use a constraint equation to eliminate the second variable.
Differentiate, set equal to zero, solve.
Verify it is a max or min (second derivative test or context).
Example (Q18): Two numbers whose difference is 71, product minimised. Let x − y = 71, so y = x − 71. Product P = x(x−71) = x² − 71x. P' = 2x − 71 = 0, x = 35.5. y = 35.5 − 71 = −35.5. The numbers are 35.5 and −35.5.
Example (Q47): Rectangular fence, area 693 m², three sides at $7/m, fourth side at $15/m. Let the fourth side have length x and the perpendicular sides have length y. Area: xy = 693, so y = 693/x. Cost = 15x + 7x + 2(7y) = 22x + 14(693/x) = 22x + 9702/x. C' = 22 − 9702/x² = 0. x² = 9702/22 = 441, x = 21. y = 693/21 = 33. Dimensions: 21 m by 33 m.
Example (Q48): 300-room hotel at $120/night. Each $2 increase loses 4 rooms. Service cost $10/room. Let n = number of $2 increases. Rooms rented: 300 − 4n. Price per room: 120 + 2n. Revenue = (300−4n)(120+2n). Profit per room after service cost: (110+2n). Gross profit = (300−4n)(110+2n) = 33000 + 600n − 440n − 8n² = 33000 + 160n − 8n². P' = 160 − 16n = 0, n = 10. Price = 120 + 20 = $140. Max profit = 33000 + 1600 − 800 = $33,800.
Average cost: C̄(x) = C(x)/x. Marginal average cost: d/dx[C(x)/x].
Example: C(x) = 4x + 8000. C̄(x) = 4 + 8000/x. C̄'(x) = −8000/x².
Profit function: P(x) = R(x) − C(x).
Revenue function: R(x) = x · p(x) where p(x) is the demand/price function.
Example: p(x) = 300 − 2x, C(x) = 40x + 4000. R(x) = x(300−2x) = 300x − 2x². P(x) = 260x − 2x² − 4000. P'(x) = 260 − 4x. Average profit P̄(x) = (260x − 2x² − 4000)/x = 260 − 2x − 4000/x.
Model: A = A₀ e^(rt) where r is negative for decay.
Example: 700 g decays to 200 g in 50 years. 200 = 700e^(50r). e^(50r) = 2/7. r = ln(2/7)/50 ≈ −0.02506. After 80 years: A = 700e^(80 × −0.02506) ≈ 94 grams.
\text{Tangent line: } y - f(a) = f'(a)(x - a)\text{Average rate of change: } \frac{f(b) - f(a)}{b - a}\text{Profit} = \text{Revenue} - \text{Cost}, \quad R(x) = x \cdot p(x)\bar{C}(x) = \frac{C(x)}{x}, \quad \bar{P}(x) = \frac{P(x)}{x}A = A_0 e^{rt} \quad \text{(exponential growth/decay)}Optimisation problems model real decisions: minimising the cost of materials for a fence, maximising hotel revenue by balancing price against occupancy, finding the production level that maximises profit. Marginal analysis is the core of microeconomic reasoning: "is it worth producing one more unit?"
Students set f'(x) = 0 and immediately call the result a maximum or minimum without checking. You must verify using the first or second derivative test, or check the sign change.
When finding absolute extrema on a closed interval, students forget to check the endpoints. The absolute max or min can occur at an endpoint, not just at a critical number.
Students confuse inflection points with extrema. An inflection point is where the concavity changes, not where the function reaches a peak or valley. A function can be increasing through an inflection point.
In optimisation, students sometimes try to optimise a function of two variables without using the constraint to eliminate one. You must reduce to a single-variable function before differentiating.
⚠️ Tangent line problems appear multiple times. Always give the answer in slope-intercept form (y = mx + b) unless told otherwise.
⚠️ Optimisation word problems (fence, hotel) carry heavy marks. Show all steps: define variables, write the objective function, use the constraint, differentiate, solve, verify.
⚠️ "Interpret your results using words" means stating what P(30) and P'(30) mean in context, not just giving numbers.
⚠️ Curve sketching questions combine increasing/decreasing, concavity, extrema and inflection points in one problem. Practise the full procedure.
True or false: A critical number always produces a local extremum. (False. The derivative must also change sign.)
Fill in the blank: If f''(c) > 0 at a critical number c, then c is a local ______. (Minimum.)
True or false: To find absolute extrema on [a,b], you only need to check critical numbers. (False. You must also check the endpoints.)
Fill in the blank: A horizontal tangent line occurs where f'(x) = ______. (0.)
True or false: An inflection point is where f'(x) = 0. (False. It is where f''(x) = 0 or is undefined and the concavity changes.)
Q: Find the equation of the tangent line to y = 3x − 6√(x³) + e^(4x) at x = 0.
A: f(0) = 0 − 0 + 1 = 1. f'(x) = 3 − 9√x + 4e^(4x). f'(0) = 3 − 0 + 4 = 7. Tangent line: y = 7x + 1.
Q: The profit function is P(x) = 5x² − 100x − 3000. Find and interpret P(30) and P'(30).
A: P(30) = 5(900) − 3000 − 3000 = −1500. P'(x) = 10x − 100, P'(30) = 200. At 30 units, the company loses $1,500, but profit is increasing by $200 per additional unit.
Q: R(x) = −36x² + 7200x. Find the actual revenue from the 61st unit and the marginal estimate.
A: Actual: R(61) − R(60) = [−36(3721) + 7200(61)] − [−36(3600) + 7200(60)] = 305,244 − 302,400 = $2,844. Marginal estimate: R'(x) = −72x + 7200. R'(60) = −4320 + 7200 = $2,880.
Q: Find the absolute extrema of f(x) = x⁴ − 8x² − 16 on [−1, 4].
A: f'(x) = 4x³ − 16x = 4x(x−2)(x+2). Critical numbers in [−1,4]: x = 0, x = 2. f(−1) = −23, f(0) = −16, f(2) = −32, f(4) = 112. Absolute max: 112 at x = 4. Absolute min: −32 at x = 2.
Q: Use the second derivative test to find local extrema of f(x) = 2x − 6 + 98/x.
A: f'(x) = 2 − 98/x². Set f' = 0: x² = 49, x = ±7. f''(x) = 196/x³. f''(7) = 196/343 > 0, local min at x = 7: f(7) = 14 − 6 + 14 = 22. f''(−7) = 196/(−343) < 0, local max at x = −7: f(−7) = −14 − 6 − 14 = −34.
Q: Two numbers differ by 71. Find the pair whose product is minimised.
A: Let x − y = 71. Product P = x(x−71) = x² − 71x. P' = 2x − 71 = 0, x = 35.5. y = −35.5. The numbers are 35.5 and −35.5.
Everything in these notes depends on the differentiation rules from the previous set of notes. The first derivative test uses the sign of f'(x), which requires you to solve equations and factor. Optimisation problems combine derivative techniques with algebra and word-problem translation. Curve sketching ties together limits (for asymptotes), first derivatives (for increasing/decreasing) and second derivatives (for concavity) into a single comprehensive problem. Integration, the next topic, will reverse much of what is done here.
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