Difficulty: Advanced | Prerequisites: All derivative rules (Sections 2.4–3.4)
This chapter brings together everything from the course so far. You already know how to differentiate; now you use the first and second derivatives to fully analyse a function's behaviour: where it rises and falls, where it bends, where it peaks and dips. The graphing strategy is a systematic checklist that produces an accurate sketch without a calculator. Optimisation extends this to real-world problems, where the task is to find the maximum revenue or minimum cost under given constraints.
The first derivative tells you where a function increases or decreases and locates local extrema. The second derivative tells you where the function is concave up or down and locates inflection points. Absolute extrema on a closed interval are found by comparing critical-point values with endpoint values. Optimisation problems require you to build a function from the constraints, then find its absolute maximum or minimum.
Critical point (critical value)
A value of x where f'(x) = 0 or f'(x) is undefined (and x is in the domain of f). These are the only candidates for local extrema.
Increasing / decreasing
f is increasing on an interval where f'(x) > 0 and decreasing where f'(x) < 0.
Local maximum
A point where f changes from increasing to decreasing. f'(x) changes sign from + to –.
Local minimum
A point where f changes from decreasing to increasing. f'(x) changes sign from – to +.
Concave up
The curve bends upward (like a cup). f''(x) > 0 on that interval. The function is "accelerating."
Concave down
The curve bends downward (like a cap). f''(x) < 0 on that interval. The function is "decelerating."
Inflection point
A point where the concavity changes. f''(x) = 0 or is undefined, and the sign of f'' changes on either side.
Second derivative test
At a critical point where f'(c) = 0: if f''(c) > 0, there is a local minimum; if f''(c) < 0, there is a local maximum; if f''(c) = 0, the test is inconclusive.
Absolute (global) extrema
The highest and lowest values of f on a closed interval [a, b]. Found by evaluating f at every critical point in (a, b) and at both endpoints a and b, then comparing.
For any function f(x), work through these steps in order:
Domain. Where is f defined?
Intercepts. x-intercepts: set f(x) = 0. y-intercept: evaluate f(0).
Asymptotes (rational functions). Vertical: where the simplified denominator = 0. Horizontal: limits as x → ±∞.
First derivative analysis. Find f'(x). Set f'(x) = 0 to get critical points. Build a sign chart for f'. Identify intervals of increase (+) and decrease (–). Label local maxima (+ to –) and local minima (– to +).
Second derivative analysis. Find f''(x). Set f''(x) = 0 to get possible inflection points. Build a sign chart for f''. Identify intervals of concave up (+) and concave down (–). Confirm inflection points (sign change in f'').
Plot key points (intercepts, extrema, inflection points, asymptotes) and sketch.
Domain: all real numbers.
x-intercepts: x = 1 (double root), x = –2. y-intercept: f(0) = 1 · 2 = 2.
Expand: f(x) = x³ – 3x + 2. f'(x) = 3x² – 3 = 3(x–1)(x+1). Critical points: x = 1, x = –1.
Sign chart for f': f' > 0 on (–∞, –1), f' < 0 on (–1, 1), f' > 0 on (1, ∞). So f is increasing, then decreasing, then increasing.
Local max at x = –1: f(–1) = (–2)²(1) = 4. Local min at x = 1: f(1) = 0.
f''(x) = 6x. f'' = 0 at x = 0. Sign chart: f'' < 0 on (–∞, 0), f'' > 0 on (0, ∞). Concave down, then concave up.
Inflection point at x = 0: f(0) = 2. The point (0, 2) is where the curve changes from bending down to bending up.
Given f'(x) = –4(x–4)/(x+2)³ and f''(x) = 8(x–7)/(x+2)⁴.
Domain: all x ≠ –2.
x-intercept: x = 1. y-intercept: f(0) = 4(–1)/4 = –1.
Vertical asymptote: x = –2. Horizontal asymptote: degree of numerator < degree of denominator, so y = 0.
f'(x) = 0 at x = 4. Sign chart: f' < 0 on (–∞, –2), f' < 0 on (–2, 4), f' > 0 on (4, ∞). Wait: check signs more carefully. f'(–3) = –4(–7)/(–1)³ = –28/–1 = 28 > 0. f'(0) = –4(–4)/8 = 2 > 0. f'(5) = –4(1)/(7)³ < 0. So: increasing on (–∞, –2), increasing on (–2, 4), decreasing on (4, ∞). Local max at x = 4: f(4) = 4(3)/36 = 1/3.
f''(x) = 0 at x = 7. Inflection point at (7, f(7)) = (7, 4(6)/81) = (7, 8/27).
When given only the graph of f'(x):
Where f'(x) > 0 (above x-axis), f is increasing.
Where f'(x) < 0 (below x-axis), f is decreasing.
Where f'(x) = 0 (crosses x-axis), f has a horizontal tangent. If f' changes from + to –, that is a local max. If – to +, a local min.
Draw a sign chart from the graph, then sketch f(x) accordingly.
Given price-demand p(x) = 2775 – 0.37x² for 0 < x < 85:
Revenue: R(x) = x · p(x) = 2775x – 0.37x³.
R'(x) = 2775 – 1.11x². Set R'(x) = 0: x² = 2500, x = 50 (only positive value in domain).
R''(x) = –2.22x. R''(50) = –111 < 0, confirming a local (and absolute) maximum.
Maximum revenue: R(50) = 2775(50) – 0.37(125,000) = 138,750 – 46,250 = $92,500.
Revenue increases on (0, 50) and decreases on (50, 85).
For f(x) = x(x+3)³ with f'(x) = (x+3)²(4x+3):
Critical points: f'(x) = 0 at x = –3 and x = –3/4.
Find f''(x) and evaluate at each critical point.
f''(–3): need to compute. If f''(–3) = 0, the test is inconclusive at that point; use the first derivative test instead.
f''(–3/4): if positive, local min; if negative, local max.
For f(x) = 3x – 2 + 75/x on [2, 6]:
f'(x) = 3 – 75/x². Set f'(x) = 0: 75/x² = 3, x² = 25, x = 5 (in [2, 6]).
Evaluate f at critical point and endpoints: f(2) = 6 – 2 + 37.5 = 41.5. f(5) = 15 – 2 + 15 = 28. f(6) = 18 – 2 + 12.5 = 28.5.
Absolute max: f(2) = 41.5. Absolute min: f(5) = 28.
Area = 1800 m². Two sides are partly bordered by walls (Wall 1 = 10 m along the top, Wall 2 = 8 m along the left). The side along Wall 1 uses $10/m material; the other three sides use $7/m. Walls themselves need no fencing.
Let x = height (side parallel to Wall 2), y = width (side parallel to Wall 1). Area: xy = 1800, so y = 1800/x.
Fencing needed: top side = y – 10 (at $10/m), bottom side = y (at $7/m), right side = x (at $7/m), left side = x – 8 (at $7/m).
Cost: C(x) = 10(y–10) + 7y + 7x + 7(x–8) = 10(1800/x – 10) + 7(1800/x) + 7x + 7(x–8) = 18,000/x – 100 + 12,600/x + 14x – 56 = 30,600/x + 14x – 156.
C'(x) = –30,600/x² + 14. Set C'(x) = 0: x² = 30,600/14 ≈ 2185.7. x ≈ 46.8 m.
y = 1800/46.8 ≈ 38.5 m.
First derivative test: f' changes + to – = local max; – to + = local min.
Second derivative test: f'(c) = 0 and f''(c) > 0 = local min; f''(c) < 0 = local max; f''(c) = 0 = inconclusive.
Absolute extrema on [a, b]: compare f(a), f(b), and f(c) for every critical point c in (a, b).
Revenue: R(x) = x · p(x)
Students assume that every critical point is an extremum. A critical point is only a candidate. You must check the sign change of f' (first derivative test) or the sign of f'' (second derivative test) to confirm.
Students forget to check endpoints when finding absolute extrema on a closed interval. The absolute max or min often occurs at an endpoint.
Students confuse concavity with increasing/decreasing. A function can be increasing and concave down at the same time (it is rising but bending downward, slowing its ascent).
In optimisation problems, students often skip writing the constraint equation. You need both an objective function (what to maximise or minimise) and a constraint (the fixed relationship, such as area = 1800).
⚠️ All work must be shown on graphing problems. Asymptotes must be found using limits.
⚠️ When the problem says "second derivative test," the first derivative test will not receive credit.
⚠️ Revenue optimisation problems typically ask three things: the revenue function, the intervals of increase/decrease, and the number of units that maximises revenue plus the maximum revenue itself.
⚠️ Draw horizontal tangent lines on your sketch wherever f'(x) = 0.
True or False: If f'(c) = 0 and f''(c) > 0, then f has a local minimum at x = c. (True.)
Fill in the blank: An inflection point occurs where __________ changes sign. (f''(x))
True or False: To find absolute extrema on [a, b], you only need to check where f'(x) = 0. (False. You also check f(a) and f(b).)
Fill in the blank: f is concave up where f''(x) __________ 0. (>)
True or False: If f'(x) does not change sign at a critical point, there is no local extremum there. (True.)
Q: For f(x) = (x–1)²(x+2), find the intervals of increase/decrease and all local extrema.
A: f'(x) = 3(x–1)(x+1). Critical points: x = –1, x = 1. Increasing on (–∞, –1) and (1, ∞). Decreasing on (–1, 1). Local max at (–1, 4). Local min at (1, 0).
Q: Find the absolute extrema of f(x) = 3x – 2 + 75/x on [2, 6].
A: f'(x) = 3 – 75/x² = 0 at x = 5. f(2) = 41.5, f(5) = 28, f(6) = 28.5. Absolute max = 41.5 at x = 2. Absolute min = 28 at x = 5.
Q: For p(x) = 2775 – 0.37x², find the number of units that maximises revenue and the maximum revenue.
A: R(x) = 2775x – 0.37x³. R'(x) = 2775 – 1.11x² = 0 at x = 50. R(50) = $92,500.
The graphing strategy combines limits (Ch. 2), derivatives (Ch. 2–3), and now concavity analysis into one unified toolkit. Optimisation is the business-calculus payoff: it answers "how many units should we produce?" and "what dimensions minimise cost?" Integration (Ch. 5) reverses the derivative process and builds on these ideas.
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