Difficulty: Intermediate to Advanced. Prerequisites: Parts 1–3 of these notes (Limits, Derivatives Fundamentals, Advanced Derivative Rules).
The first derivative tells you where a function is increasing or decreasing and locates local peaks and valleys. The second derivative tells you about the curve's shape (concave up or down) and finds inflection points. Together, they give you a complete graphing strategy. Optimisation applies these tools to find the maximum profit, minimum cost, or best dimensions in a word problem. Integration reverses differentiation: the antiderivative recovers a total quantity from a rate, and the Fundamental Theorem of Calculus evaluates definite integrals to compute areas, totals, and averages.
Critical number
A value x in the domain of f where f'(x) = 0 or f'(x) is undefined. These are the only candidates for local extrema.
Think of it as: "Where the function might change direction."
Local maximum / local minimum (local extremum)
A function value f(c) that is higher (max) or lower (min) than all nearby values. Found using the first or second derivative test.
Increasing / decreasing
f is increasing on an interval where f'(x) > 0 and decreasing where f'(x) < 0.
Concave up / concave down
The graph is concave up where f''(x) > 0 (the curve opens upward, like a cup). It is concave down where f''(x) < 0 (the curve opens downward, like a frown).
Think of it as: "Concave up holds water. Concave down spills it."
Inflection point
A point where the concavity changes (from up to down or down to up). Found where f''(x) = 0 or is undefined, provided the concavity actually switches.
Absolute maximum / absolute minimum (absolute extremum)
The single highest or lowest value of f on a given domain. On a closed interval [a, b], check critical numbers and endpoints.
Second derivative test
At a critical number c where f'(c) = 0: if f''(c) > 0, then f(c) is a local minimum. If f''(c) < 0, then f(c) is a local maximum.
Antiderivative
A function F whose derivative is f: F'(x) = f(x). The general antiderivative includes a constant of integration C.
Indefinite integral
The notation ∫f(x) dx = F(x) + C, representing the family of all antiderivatives of f.
Definite integral
The integral ∫ from a to b of f(x) dx = F(b) − F(a). Gives a specific numerical value, often representing a total accumulation or area.
Fundamental Theorem of Calculus
If f is continuous on [a, b] and F is any antiderivative of f, then ∫ from a to b of f(x) dx = F(b) − F(a). Connects differentiation and integration.
Average value of a function
Over [a, b]: (1/(b−a)) ∫ from a to b of f(x) dx. This is the "height" that would produce the same total if f were constant across the interval.
Find f'(x). Set f'(x) = 0 and solve for critical numbers.
Build a sign chart: pick test values in each interval between critical numbers, plug into f'(x), and record the sign.
Where f'(x) > 0, f is increasing. Where f'(x) < 0, f is decreasing.
At a critical number: if f' changes from negative to positive, there is a local minimum. If f' changes from positive to negative, there is a local maximum. If f' does not change sign, there is no extremum.
Example: f(x) = x² − 6x + 10. f'(x) = 2x − 6 = 0 at x = 3. Sign chart: f'(x) < 0 on (−∞, 3), f'(x) > 0 on (3, ∞). Local minimum at x = 3, f(3) = 1.
Find f''(x). Set f''(x) = 0 and solve.
Build a sign chart for f''(x).
Where f''(x) > 0, the graph is concave up. Where f''(x) < 0, it is concave down.
Where f''(x) changes sign, there is an inflection point (provided f is defined there).
Example: f(x) = 2x³ − 3x² − 36x − 7. f''(x) = 12x − 6 = 0 at x = 1/2. For x < 1/2, f''(x) < 0 (concave down). For x > 1/2, f''(x) > 0 (concave up). Inflection point at x = 1/2.
For polynomials:
Find domain, x-intercepts (set f(x) = 0), y-intercept (evaluate f(0)).
Find intervals of increase/decrease and local extrema (first derivative).
Find concavity and inflection points (second derivative).
Sketch the graph.
For rational functions, add:
Horizontal asymptotes (limit at infinity).
Vertical asymptotes or holes (set denominator = 0, test for 0/0 vs n/0).
Note: the function is not defined at vertical asymptotes, so these break the domain and create separate intervals in your sign charts.
At a critical number c where f'(c) = 0:
If f''(c) > 0, then f(c) is a local minimum (graph is concave up at c).
If f''(c) < 0, then f(c) is a local maximum (graph is concave down at c).
Special note: if only one critical number exists on the interval, the local extremum is also the absolute extremum.
Find all critical numbers of f on [a, b].
Evaluate f at each critical number and at both endpoints a and b.
The largest value is the absolute maximum. The smallest is the absolute minimum.
Example: f(x) = x³ − 6x² on [−7, 2]. f'(x) = 3x² − 12x = 3x(x−4) = 0 at x = 0 and x = 4. Only x = 0 is in [−7, 2]. Compare f(−7), f(0), f(2): f(−7) = −637, f(0) = 0, f(2) = −16. Absolute max = 0 at x = 0. Absolute min = −637 at x = −7.
Identify what quantity to maximise or minimise. Write it as a function of one variable.
If there is a constraint (e.g., fixed area), use it to eliminate one variable.
Differentiate, set equal to zero, solve, and verify it is a max or min (using the second derivative test or endpoint analysis).
Example: A company sells x headphones per week. Price-demand: p = 400 − 0.5x. Cost: C(x) = 2000 + 200x. Revenue: R(x) = 400x − 0.5x². R'(x) = 400 − x = 0 at x = 400. Max revenue at 400 units, price = $200, revenue = $80,000. For profit: P(x) = R(x) − C(x) = −0.5x² + 200x − 2000. P'(x) = −x + 200 = 0 at x = 200. Max weekly profit at 200 units.
A function F is an antiderivative of f if F'(x) = f(x).
The indefinite integral ∫f(x) dx = F(x) + C represents the entire family of antiderivatives (they differ by a constant).
The integral sign ∫ and the dx always appear together.
Power rule: ∫x^n dx = x^(n+1)/(n+1) + C, provided n ≠ −1.
Constant rule: ∫k dx = kx + C.
Exponential rule: ∫e^x dx = e^x + C.
Reciprocal rule: ∫(1/x) dx = ln|x| + C, for x ≠ 0.
Constant multiple: ∫k·f(x) dx = k∫f(x) dx.
Sum/difference: ∫[f(x) ± g(x)] dx = ∫f(x) dx ± ∫g(x) dx.
As with differentiation, rewrite roots and fractions as powers before integrating.
Example: ∫(5/x − 4e^x) dx = 5 ln|x| − 4e^x + C.
Example: ∫(x² − 2)(x + 3) dx. Expand first: ∫(x³ + 3x² − 2x − 6) dx = x⁴/4 + x³ − x² − 6x + C.
Integrate to get the general solution F(x) + C.
Use the initial condition (e.g., f(1) = 5) to solve for C.
Example: f'(x) = 4/x² − 3/x + 2, f(1) = 5. Integrate: f(x) = −4/x − 3 ln|x| + 2x + C. Substitute x = 1: 5 = −4 − 0 + 2 + C, so C = 7. Then f(x) = −4/x − 3 ln|x| + 2x + 7.
If f is continuous on [a, b] and F is any antiderivative of f, then ∫ from a to b of f(x) dx = F(b) − F(a).
There is no +C in a definite integral because the constants cancel.
If the upper and lower limits are equal, the integral is 0.
Example: ∫ from −1 to 2 of 2x dx = [x²] from −1 to 2 = 4 − 1 = 3.
Example: ∫ from 0 to 2 of 3e^x dx = [3e^x] from 0 to 2 = 3e² − 3.
Average value of f on [a, b] = (1/(b−a)) ∫ from a to b of f(x) dx.
This is distinct from average cost (C(x)/x). The average value formula integrates the function over the interval, then divides by the interval length.
Example: Average value of f(x) = 6x² − 2x on [−2, 3]. Compute (1/5)∫ from −2 to 3 of (6x² − 2x) dx = (1/5)[2x³ − x²] from −2 to 3 = (1/5)[(54 − 9) − (−16 − 4)] = (1/5)[45 + 20] = 13.
If you know the marginal cost C'(x) and marginal revenue R'(x), integrate to recover the profit function, using an initial condition.
Rate-of-change problems: if a price is increasing at a rate p'(t), integrate to find the price function, using the current price as the initial condition.
A first-order differential equation dy/dx = f(x) is solved by integrating both sides.
General solution: y = ∫f(x) dx = F(x) + C.
Particular solution: use an initial condition y(a) = b to find C.
\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1\int e^x\,dx = e^x + C \qquad \int \frac{1}{x}\,dx = \ln|x| + C\int_a^b f(x)\,dx = F(b) - F(a)\text{Average value} = \frac{1}{b-a}\int_a^b f(x)\,dxStudents often forget the +C on indefinite integrals. Every indefinite integral must include the constant of integration.
Students often confuse "average cost" (C(x)/x) with "average value of the cost function" ((1/(b−a))∫ C(x) dx). These are different calculations with different meanings.
Students often try to use the power rule when n = −1. The integral of 1/x is ln|x| + C, not x⁰/0.
Students often forget to expand a product before integrating. There is no "product rule" for integration in this course. Expand or simplify first, then integrate term by term.
Students often apply the graphing strategy incompletely. A common exam error is finding critical numbers but forgetting to test the sign of f'(x) in each interval, or finding inflection point candidates without verifying a concavity change.
⚠️ The graphing strategy (domain, intercepts, asymptotes, increase/decrease, concavity, sketch) is a frequent exam question. Practise it on both polynomials and rational functions.
⚠️ Optimisation: expect a word problem requiring you to build the function, differentiate, set equal to zero, and interpret the answer in context.
⚠️ Absolute extrema on closed intervals: evaluate f at critical numbers and endpoints. Compare all values.
⚠️ Definite integrals: the Fundamental Theorem makes these straightforward, but watch the substitution and sign work when limits include negative numbers.
⚠️ Average value: know the formula and how it differs from average cost.
True or false: If f'(c) = 0, then f has a local extremum at c.
Fill in the blank: The graph is concave up where f''(x) __________ 0.
True or false: ∫x^(−1) dx = x⁰/0 + C.
Fill in the blank: ∫ from a to a of f(x) dx = __________.
True or false: To find absolute extrema on [a, b], you must check endpoints.
Answers: 1. False (f' could change sign or not; it might be a saddle point). 2. > (is greater than). 3. False (it is ln|x| + C). 4. 0. 5. True.
Q: Find the intervals of increase/decrease and local extrema of f(x) = x³ + 3x² − 240x − 10.
A: f'(x) = 3x² + 6x − 240 = 3(x² + 2x − 80) = 3(x+10)(x−8). Critical numbers: x = −10 and x = 8. f'(x) > 0 on (−∞, −10) and (8, ∞): increasing. f'(x) < 0 on (−10, 8): decreasing. Local max at x = −10, local min at x = 8.
Q: Find the absolute max and min of f(x) = x³ − 6x² on [−7, 2].
A: f'(x) = 3x² − 12x = 0 at x = 0 and x = 4. Only x = 0 is in [−7, 2]. f(−7) = −637, f(0) = 0, f(2) = −16. Absolute max = 0 at x = 0. Absolute min = −637 at x = −7.
Q: Evaluate ∫ from −2 to 3 of (4x³ − 6x² + 10x) dx.
A: Antiderivative: x⁴ − 2x³ + 5x². At x = 3: 81 − 54 + 45 = 72. At x = −2: 16 + 16 + 20 = 52. Result: 72 − 52 = 20.
Q: Find the average value of f(x) = 0.06t² − 0.2t + 6 on [0, 6].
A: (1/6)∫ from 0 to 6 of (0.06t² − 0.2t + 6) dt = (1/6)[0.02t³ − 0.1t² + 6t] from 0 to 6 = (1/6)[4.32 − 3.6 + 36] = (1/6)(36.72) = $6.12 per pound.
Q: A manufacturer has price-demand p = 400 − 0.5x and cost C(x) = 2000 + 200x. Find the production level that maximises profit.
A: R(x) = 400x − 0.5x². P(x) = R(x) − C(x) = −0.5x² + 200x − 2000. P'(x) = −x + 200 = 0 at x = 200. P''(x) = −1 < 0, confirming a maximum. Max profit at 200 units.
Graph analysis ties directly to optimisation: locating extrema on a cost or revenue curve is how you find the profit-maximising output. Integration reverses differentiation and recovers total quantities from marginal rates, closing the loop between Parts 2–3 and Part 4. The Fundamental Theorem of Calculus is the bridge between the derivative (local rate) and the integral (accumulated total). Differential equations extend this further in later courses.
Tags: critical number, local maximum, local minimum, increasing, decreasing, first derivative test, second derivative test, concavity, concave up, concave down, inflection point, graphing strategy, absolute maximum, absolute minimum, closed interval, optimisation, maximize revenue, maximize profit, minimize cost, antiderivative, indefinite integral, definite integral, Fundamental Theorem of Calculus, average value, differential equation, integration rules, power rule for integrals, MAC 2233, business calculus, FIU