Difficulty: Intermediate–Advanced | Prerequisites: Sections 2.1–3.4 (limits, all derivative rules).
Once you can take derivatives, the next question is: what do they tell you about the shape of a graph? The first derivative reveals where a function is climbing or falling, and the second derivative reveals how it curves. Together they let you sketch accurate graphs, find peaks and valleys, and solve real-world problems that ask for the largest or smallest value of something. This material ties together everything from limits through differentiation into a single graphing and optimisation toolkit.
Set f′(x) = 0 to find critical numbers, build a sign chart to determine where f rises and falls, and check for local maxima/minima. Set f″(x) = 0 for inflection points and concavity. On a closed interval, compare function values at critical numbers and endpoints to find absolute extrema. Optimisation problems translate a word problem into a function to maximise or minimise.
Critical number
A value x in the domain of f where f′(x) = 0 or f′(x) is undefined. These are the only places where local extrema can occur.
Partition number
Any value where f′(x) = 0 or f′(x) is undefined. A critical number is a partition number that is also in the domain of f.
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. The first derivative changes from positive to negative.
Local minimum
A point where f changes from decreasing to increasing. The first derivative changes from negative to positive.
Concave up
The graph curves like an upward-opening parabola. f″(x) > 0 on that interval.
Concave down
The graph curves like a downward-opening parabola. f″(x) < 0 on that interval.
Inflection point
A point (c, f(c)) where the concavity changes. f must be defined at c, and f″ must change sign there.
Absolute maximum / minimum
The largest / smallest y-value of f on an interval. On a closed interval [a, b], the extreme value theorem guarantees these exist.
Optimisation
The process of finding the maximum or minimum value of a quantity described in a word problem. Typical signals: "find the greatest," "minimise," "what dimensions will give the largest."
Find f′(x).
Find all partition numbers: values where f′(x) = 0 or f′(x) is undefined.
Build a sign chart: test a value from each interval between partition numbers in f′(x) to determine + or −.
Read the results:
f′ changes from + to − at c: local maximum at (c, f(c)).
f′ changes from − to + at c: local minimum at (c, f(c)).
f′ does not change sign: no extremum (the function just has a flat tangent).
Example: f(x) = −3x² + 12x − 5. f′(x) = −6x + 12. Set equal to zero: x = 2. Test x = 0: f′(0) = 12 > 0 (increasing). Test x = 3: f′(3) = −6 < 0 (decreasing). f′ goes from + to − at x = 2, so there is a local maximum at (2, f(2)) = (2, 7).
A critical number is a partition number that lies in the domain of f. When f has a restricted domain (like a rational function), some partition numbers may not be critical numbers.
Find f″(x), set it to zero, and build a sign chart for f″.
f″(x) > 0: concave up (cup-shaped).
f″(x) < 0: concave down (cap-shaped).
Inflection point: where f″ changes sign and f is defined.
Second Derivative Test for Local Extrema:
Let c be a critical number where f′(c) = 0.
If f″(c) > 0, then f(c) is a local minimum (concave up at c).
If f″(c) < 0, then f(c) is a local maximum (concave down at c).
If f″(c) = 0 or undefined, the test is inconclusive. Fall back to the first derivative test.
Example: f(x) = x³ − 6x² + 12x − 10. f′(x) = 3x² − 12x + 12 = 3(x − 2)². Critical number: x = 2. f″(x) = 6x − 12. f″(2) = 0, so the second derivative test is inconclusive. The first derivative test shows f′ does not change sign (it is always ≥ 0), so x = 2 is neither a max nor a min.
Inflection point: f″(x) = 0 at x = 2, and f″ changes sign there (negative for x < 2, positive for x > 2), so (2, f(2)) = (2, −2) is an inflection point.
Step a: Use f(x) to find domain, x-intercepts, y-intercept, and asymptotes.
Domain: all x where f is defined.
x-intercepts: set f(x) = 0.
y-intercept: evaluate f(0).
Horizontal asymptote: lim x→±∞ f(x).
Vertical asymptote: values that make the denominator zero (after checking for holes).
Step b: Use f′(x) to find intervals of increase/decrease and local extrema.
Find critical numbers, build sign chart for f′, identify local max/min.
Step c: Use f″(x) to find intervals of concavity and inflection points.
Find partition numbers of f″, build sign chart for f″, identify inflection points.
Step d: Sketch the graph.
Plot x-intercepts, y-intercept, local extrema, and inflection points. Draw asymptotes as dashed lines. Use the sign charts to connect the dots with the correct shape (rising/falling, concave up/down). Draw horizontal tangent lines at local extrema.
Your instructor requires all asymptotes to be found using limits. Stating them from memory without showing limit work will receive zero credit.
Example structure for f(x) = (x + 4)/(x − 5):
Domain: all real x, x ≠ 5.
x-int: x = −4 (set numerator = 0). y-int: f(0) = −4/5.
Horizontal asymptote: lim x→±∞ (x + 4)/(x − 5) = 1. So y = 1.
Vertical asymptote: x = 5 (denominator = 0, numerator ≠ 0).
f′(x) = −9/(x − 5)². Always negative, so f is always decreasing. No local extrema.
f″(x) = 18/(x − 5)³. Positive for x > 5 (concave up), negative for x < 5 (concave down). No inflection point (f is not defined at x = 5).
Extreme Value Theorem: A function continuous on a closed interval [a, b] is guaranteed to have both an absolute maximum and an absolute minimum on that interval.
Closed interval method: To find absolute extrema of f on [a, b]:
Find f′(x) and all critical numbers in (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 − 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.
Single critical number on an open interval: If there is only one critical number on an open interval and it is a local extremum, then it is automatically the absolute extremum.
Example: f(x) = 2x⁴ − 8x³ on (0, ∞). f′(x) = 8x³ − 24x² = 8x²(x − 3). In (0, ∞), x = 3 is the only critical number. Test: f′ changes from negative to positive at x = 3, so it is a local min. By the single critical number theorem, f(3) = 2(81) − 8(27) = 162 − 216 = −54 is the absolute minimum.
Optimisation problems describe a situation in words and ask you to find the maximum or minimum of some quantity. The general approach:
Identify the quantity to maximise or minimise. Write its formula.
If the formula has two variables, use a constraint (a second equation from the problem) to eliminate one variable.
Differentiate, set the derivative equal to zero, and solve.
Verify your critical number gives a max or min (using the first or second derivative test).
Answer the question the problem asked (not just the x-value).
Example (pricing): Hats sell at $8 each, 6000 sold monthly. For each $2 price increase, 500 fewer sell. Production cost is $4/hat. Let x = the number of $2 increases.
Price = 8 + 2x.
Quantity = 6000 − 500x.
Revenue = (8 + 2x)(6000 − 500x).
Cost = 4(6000 − 500x).
Profit P(x) = Revenue − Cost.
Expand, differentiate, set P′(x) = 0, solve for x, then compute the optimal price.
Example (fencing): A 20m stone wall forms one side of a rectangle. 160m of fencing is available for the other three sides (two widths plus the portion of the length beyond the wall). Let x = the extra length beyond the wall on each side, y = the width.
Fencing constraint: 2x + y + (20 + 2x − 20) rearranges based on the diagram. Solve for y in terms of x.
Area = length × width, expressed in one variable.
Differentiate, set A′(x) = 0, solve, and report the dimensions.
Always test your critical numbers. If you skip the verification step, you cannot confirm whether you found a maximum or a minimum.
Students assume every critical number is a local extremum. It is only an extremum if f′ changes sign. If f′ does not change sign (e.g., f(x) = x³ at x = 0), no extremum exists there.
Students confuse partition numbers with critical numbers. A partition number must also be in the domain of f to be a critical number.
Students forget to include endpoints when finding absolute extrema on a closed interval.
Students skip the verification step in optimisation. You must confirm (via first or second derivative test) that a critical number is a max or min.
⚠️ Sign charts for f′ and f″ must include all partition numbers (both where the derivative equals zero and where it is undefined).
⚠️ The graphing strategy (domain, intercepts, asymptotes via limits, sign charts, sketch) is the format expected on exams.
⚠️ Optimisation problems are guaranteed on the final. Make sure you can translate word problems into functions and verify your answer.
⚠️ Draw horizontal tangent lines at every local extremum on your graph.
True or false: If f′(c) = 0, then f has a local extremum at c.
Fill in the blank: f is concave up where f″ is ________.
True or false: On a closed interval, absolute extrema can only occur at critical numbers or endpoints.
Fill in the blank: If there is only one critical number on an open interval and it is a local min, it is also the ________.
True or false: An inflection point is where f′(x) = 0.
Answers: 1. False. 2. Positive. 3. True. 4. Absolute minimum. 5. False (it is where f″ changes sign).
Q: Find the intervals of increase and decrease for f(x) = x³ − 6x² + 9x − 7.
A: f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). Critical numbers: x = 1, x = 3. Sign chart: f′ > 0 on (−∞, 1) and (3, ∞), f′ < 0 on (1, 3). Increasing on (−∞, 1) and (3, ∞). Decreasing on (1, 3). Local max at x = 1, local min at x = 3.
Q: Find the absolute max and min of f(x) = 3x/(x² + 9) on [−1, 4].
A: f′(x) = (3(x² + 9) − 3x(2x))/(x² + 9)² = (27 − 3x²)/(x² + 9)². Set numerator = 0: x² = 9, x = ±3. Only x = 3 is in [−1, 4]. Evaluate: f(−1) = −3/10, f(3) = 9/18 = 1/2, f(4) = 12/25. Absolute max = 1/2 at x = 3. Absolute min = −3/10 at x = −1.
Curve sketching combines everything from limits (asymptotes) through derivatives (shape). Optimisation is the business payoff of all the derivative rules. Integration (Sections 5.1, 5.5) reverses this process: given the rate of change, it recovers the total quantity.
first derivative test, second derivative test, critical number, partition number, sign chart, increasing, decreasing, local maximum, local minimum, concave up, concave down, inflection point, absolute extrema, extreme value theorem, closed interval method, optimisation, optimization, graphing strategy, curve sketching, horizontal tangent, MAC 2233, business calculus, FIU, Sections 4.1, 4.2, 4.4, 4.5, 4.6