Difficulty: Advanced. Prerequisites: Sections 2.1–3.4 (limits, continuity, all differentiation rules).
The first derivative tells you where a function is increasing or decreasing and where local maxima and minima occur. The second derivative tells you about concavity (cupping up or down) and locates inflection points. Together they give a complete graphing strategy. Absolute extrema on a closed interval are found by checking critical points and endpoints. Optimisation problems ask you to build a function from a word problem, then use these tools to find its maximum or minimum.
Critical number
A value x = c in the domain of f where f'(c) = 0 or f'(c) is undefined. These are the only candidates for local extrema. Think of it as: the spots where the slope is flat or the derivative breaks.
Increasing / decreasing intervals
f is increasing on an interval where f'(x) > 0 and decreasing where f'(x) < 0.
First derivative test
A method that uses the sign change of f'(x) around a critical number to classify it as a local max, local min, or neither.
Concave up / concave down
The graph is concave up (cups upward) where f''(x) > 0, and concave down (cups downward) where f''(x) < 0.
Inflection point
A point where the concavity changes. Found where f''(x) = 0 or is undefined, provided the concavity does switch sides.
Second derivative test
If f'(c) = 0 and f''(c) > 0, then f(c) is a local minimum. If f''(c) < 0, it is a local maximum. If f''(c) = 0, the test is inconclusive.
Absolute (global) extrema
The largest (absolute maximum) and smallest (absolute minimum) values of f on an interval. On a closed interval, they must occur at critical points or endpoints.
Extreme Value Theorem
If f is continuous on [a, b], then f has both an absolute max and an absolute min on that interval.
Find f'(x). Set f'(x) = 0 and solve, and note where f'(x) is undefined.
Build a sign chart: pick test values in each interval created by the critical numbers. Plug into f'(x) and record + or −.
f'(x) > 0 means f is increasing on that interval. f'(x) < 0 means f is decreasing.
First derivative test for local extrema:
f' changes from − to +: local minimum at c.
f' changes from + to −: local maximum at c.
f' does not change sign: neither (e.g. an inflection point with a flat tangent).
Find f''(x). Set f''(x) = 0 and solve.
Build a sign chart for f''(x).
f''(x) > 0: concave up. f''(x) < 0: concave down.
Where the sign of f'' changes, the x-value is an inflection point (provided f is defined there).
Step 1, analyse f(x): find the domain, intercepts, and asymptotes.
Step 2, analyse f'(x): find intervals of increase/decrease and local extrema.
Step 3, analyse f''(x): find intervals of concavity and inflection points.
Step 4: draw the asymptotes, plot key points, and sketch the curve using all information.
Example: f(x) = (x + 4)/(x − 5). Domain: all x ≠ 5. Vertical asymptote: x = 5. Horizontal asymptote: y = 1. x-intercept: x = −4. y-intercept: y = −4/5. f'(x) = −9/(x − 5)², which is always negative, so f is always decreasing (on each piece of its domain). No local extrema.
Procedure:
Verify f is continuous on [a, b].
Find critical numbers in (a, b).
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) = x³ − 12x on [−5, 5]. f'(x) = 3x² − 12 = 0 gives x = ±2. Evaluate: f(−5) = −65, f(−2) = 16, f(2) = −16, f(5) = 65. Absolute max: 65 at x = 5. Absolute min: −65 at x = −5.
If f has only one critical number c in an open interval and f'(c) = 0, then:
f''(c) > 0: f(c) is the absolute minimum on that interval.
f''(c) < 0: f(c) is the absolute maximum.
General approach:
Identify the quantity to maximise or minimise (the objective function).
Write it in terms of two variables if needed.
Use a constraint (perimeter, budget, relationship) to eliminate one variable.
Differentiate the single-variable objective function and set the derivative equal to zero.
Verify it is a max or min (second derivative test or context).
Example (bookstore): price = 17 − 2x, quantity = 350 + 25x, cost per book = 5. Profit = (price − cost) × quantity = (12 − 2x)(350 + 25x). Expand, differentiate, set to zero, and solve for x.
Example (fencing): maximise area A = (20 + x) × y subject to the fencing constraint x + 2y + x = 160 (or 2x + 2y = 160, depending on the wall configuration). Solve for y, substitute, differentiate A(x), set to zero.
\text{First derivative test: sign change of } f'(x) \text{ around } cf''(c) > 0 \implies \text{local min}; \quad f''(c) < 0 \implies \text{local max}\text{Absolute extrema on } [a,b]: \text{ compare } f(a),\, f(c_1),\, f(c_2),\, \ldots,\, f(b)Students often think every critical number is a local extremum. A critical number where f' does not change sign (e.g. x = 0 for f(x) = x³) is neither a max nor a min.
Students sometimes check only the critical points and forget the endpoints when finding absolute extrema on a closed interval.
Students confuse concave up with increasing. A function can be increasing and concave down at the same time (it rises but bends over).
Students apply the second derivative test when f''(c) = 0 and declare "no extremum." When f''(c) = 0 the test is inconclusive; you must fall back to the first derivative test.
⚠️ Full graphing-strategy problems (domain, intercepts, asymptotes, first and second derivative analysis, sketch) are major exam questions, often worth many points.
⚠️ Absolute extrema on a closed interval is a standard 5-minute exam problem. Make the procedure automatic.
⚠️ Optimisation word problems (maximise profit, minimise cost, maximise area) appear on the final. The hardest part is setting up the equation; the calculus is usually straightforward.
⚠️ Know the second derivative test and when it fails (f''(c) = 0 or undefined).
Fill in the blank: f is decreasing where f'(x) is ________. Negative.
True or False: An inflection point occurs wherever f''(x) = 0. False. The concavity must change at that point.
True or False: On a closed interval, the absolute max must occur at a critical point. False. It can also occur at an endpoint.
Fill in the blank: In the second derivative test, f''(c) > 0 means f(c) is a local ________. Minimum.
True or False: Setting f'(x) = 0 finds horizontal tangent lines and potential local extrema. True.
Q: Find the intervals of increase and decrease for f(x) = −3x² + 12x − 5.
A: f'(x) = −6x + 12. Setting f'(x) = 0: x = 2. Sign chart: f'(x) > 0 for x < 2 (increasing), f'(x) < 0 for x > 2 (decreasing). Local max at x = 2, f(2) = 7.
Q: Find the absolute max and min of f(x) = x³ − 12x on [−5, 5].
A: f'(x) = 3x² − 12 = 0 at x = ±2. Evaluate: f(−5) = −65, f(−2) = 16, f(2) = −16, f(5) = 65. Absolute max = 65 at x = 5. Absolute min = −65 at x = −5.
Q: Find the local extrema of f(x) = x³ − 9x² + 24x − 10 using the second derivative test.
A: f'(x) = 3x² − 18x + 24 = 3(x − 2)(x − 4) = 0 at x = 2 and x = 4. f''(x) = 6x − 18. f''(2) = −6 < 0: local max at x = 2, f(2) = 10. f''(4) = 6 > 0: local min at x = 4, f(4) = 6.
Q: Determine concavity and inflection points for f(x) = x³ − 2x² − 12x + 3.
A: f''(x) = 6x − 4. Set f''(x) = 0: x = 2/3. Sign chart: f''(x) < 0 for x < 2/3 (concave down), f''(x) > 0 for x > 2/3 (concave up). Inflection point at x = 2/3.
The graphing strategy pulls together limits (for asymptotes), continuity, and both derivative tests into one comprehensive procedure. Optimisation extends marginal analysis (Section 2.7) to more complex real-world problems. The integral calculus section (5.1, 5.5) reverses differentiation, asking "given f', find f."
Critical number, critical point, increasing, decreasing, first derivative test, local maximum, local minimum, concave up, concave down, inflection point, second derivative test, graphing strategy, absolute maximum, absolute minimum, extreme value theorem, closed interval method, optimisation, maximise profit, minimise cost, area maximisation, MAC 2233, business calculus, FIU