Source: Calc AB Lecture Notes, University of Central Florida
Tags: extrema, critical numbers, first derivative test, second derivative test, concavity, inflection points, relative maximum, relative minimum, absolute maximum, absolute minimum, Rolle's theorem, Extreme Value Theorem, AP Calculus AB, analysing functions with derivatives
Difficulty: Intermediate | Prerequisites: Differentiation rules (power rule, trig derivatives), continuity, open vs. closed intervals.
This unit is where differentiation stops being an abstract skill and starts doing something useful: telling you the shape and behaviour of a function. You already know how to find f'(x). Now you use it to locate peaks, valleys, and curves in a graph, and to prove those features exist. Everything here feeds directly into optimisation problems (Unit 3 continued) and is tested heavily on the AP exam. If your differentiation mechanics are shaky, shore those up first.
Derivatives let you find where a function reaches its highest and lowest values (extrema), whether it curves upward or downward (concavity), and where the curvature flips (inflection points). Critical numbers are the x-values where these features live. The first and second derivative tests are your two main tools for classifying them.
Extrema (extreme values)
The maximum and minimum values of a function. "Extrema" is just the plural of "extremum." In simple terms, these are the peaks and valleys on a graph.
Minimum
f(c) is a minimum if f(c) ≤ f(x) for all x on the interval (a, b). Think of it as the lowest point in a given region of the graph.
Maximum
f(c) is a maximum if f(c) ≥ f(x) for all x on the interval (a, b). Think of it as the highest point in a given region of the graph.
Relative (local) extremum
A max or min compared only to nearby points, not the entire domain. The function might go higher or lower somewhere else.
Absolute (global) extremum
The single highest or lowest value on the entire interval. On a closed interval, the Extreme Value Theorem guarantees these exist.
Critical number
A value x = c in the domain of f where f'(c) = 0 or f'(c) is undefined. In simple terms, it is any x-value where the slope is flat or the derivative breaks down. All extrema occur at critical numbers (or endpoints), but not every critical number is an extremum.
Concave up
The graph curves upward, like a bowl. The first derivative is increasing across the interval.
Concave down
The graph curves downward, like an upside-down bowl. The first derivative is decreasing across the interval.
Inflection point
A point where the concavity changes (from up to down, or down to up). At an inflection point, f''(x) = 0 or f''(x) is undefined, and the second derivative must change sign.
Take the derivative f'(x).
Set f'(x) = 0 and solve. Also identify where f'(x) is undefined.
Each solution that lies in the domain of f is a critical number.
Critical numbers are candidates for extrema, but you still need a test to confirm.
If f is continuous on a closed interval [a, b], then f has both an absolute maximum and an absolute minimum on that interval.
This only applies to closed intervals. On an open interval, absolute extrema are not guaranteed.
To find them: evaluate f at every critical number in [a, b] and at the endpoints a and b. The largest value is the absolute max; the smallest is the absolute min.
Let f be continuous on [a, b] and differentiable on (a, b).
If f(a) = f(b), then there is at least one value c in (a, b) where f'(c) = 0.
In plain language: if a function starts and ends at the same height, it must have at least one flat (horizontal tangent) point in between.
Use this to classify critical numbers as relative maxima, relative minima, or neither.
If f'(x) changes from negative to positive at a critical number, that point is a relative minimum. The function is falling, then rising.
If f'(x) changes from positive to negative at a critical number, that point is a relative maximum. The function is rising, then falling.
If f'(x) does not change sign, the critical number is neither a max nor a min.
Worked example: f(x) = (1/2)x – sin x on [0, 2π]
f'(x) = 1/2 – cos x = 0, so cos x = 1/2, giving x = π/3 and x = 5π/3.
Test sign of f' in the intervals (0, π/3), (π/3, 5π/3), and (5π/3, 2π).
f' goes from negative to positive at π/3 (relative min) and from positive to negative at 5π/3 (relative max), then negative again toward 2π.
An alternative way to classify critical numbers. Especially useful when checking sign changes is awkward.
Find the critical numbers (where f'(x) = 0).
Plug each critical number into f''(x).
If f''(c) > 0, the graph is concave up at c, so c is a relative minimum.
If f''(c) < 0, the graph is concave down at c, so c is a relative maximum.
If f''(c) = 0, the test is inconclusive. Fall back to the first derivative test.
Memory aid: positive second derivative means the curve smiles (concave up, minimum). Negative second derivative means the curve frowns (concave down, maximum).
Find f''(x). Set f''(x) = 0 and solve; also note where f''(x) is undefined.
These candidates are potential inflection points, but you must verify that f'' actually changes sign there.
Build a sign chart for f'' across the intervals separated by these candidates.
Where f'' changes from positive to negative (or vice versa), you have an inflection point.
Worked example: f(x) = x⁴ – 4x³
f'(x) = 4x³ – 12x²
f''(x) = 12x² – 24x
Set f''(x) = 0: 12x(x – 2) = 0, so x = 0 and x = 2.
Sign chart for f'': f'' is positive on (–∞, 0), negative on (0, 2), positive on (2, ∞).
Both x = 0 and x = 2 are inflection points because f'' changes sign at each.
For the critical numbers from f'(x) = 0: 4x²(x – 3) = 0 gives x = 0 and x = 3. Plug into f'': f''(0) = 0 (inconclusive), f''(3) > 0 (relative minimum at x = 3).
Critical number condition: f'(c) = 0 or f'(c) is undefined (with c in the domain of f)
First Derivative Test (sign-change summary):
f' changes – → + at c: relative min
f' changes + → – at c: relative max
f' does not change sign at c: neither
Second Derivative Test:
f''(c) > 0 at a critical number: relative min
f''(c) < 0 at a critical number: relative max
f''(c) = 0: inconclusive
Inflection point condition: f''(c) = 0 or undefined, AND f'' changes sign at c
Engineers use derivative tests to find the maximum load a beam can carry or the minimum material needed for a container. In economics, finding where a profit function's derivative equals zero tells you the production level that maximises profit. Any time you hear "the most efficient" or "the least cost," a derivative test is doing the work behind the scenes.
Students often assume every critical number is an extremum. It is not. A critical number is only a candidate; you must apply a derivative test to confirm.
Students confuse "f'(c) = 0" with "f has a max or min at c." A flat tangent can also be a saddle point or a point of inflection (if f' does not change sign).
Students forget to check endpoints on closed intervals. The Extreme Value Theorem requires evaluating f at the endpoints too, not just at critical numbers.
Students mix up the second derivative test's sign convention. Positive f'' means concave up (minimum), not maximum.
⚠️ The first derivative test and second derivative test are among the most commonly tested concepts on the AP Calculus AB exam. You need to be able to justify your answer with correct sign-change language.
⚠️ Free-response questions frequently ask you to identify intervals of increase/decrease and concavity from a graph of f' or f''. Make sure you can read derivative graphs, not just compute derivatives.
⚠️ Rolle's Theorem and the Extreme Value Theorem appear as justification steps. The AP exam expects you to state the hypotheses (continuous on [a, b], differentiable on (a, b)) before applying the conclusion.
⚠️ Inflection points require a sign change in f''. Candidates where f'' = 0 but does not change sign are not inflection points.
True or false: Every critical number of f is either a relative max or a relative min.
Fill in the blank: If f'(x) changes from positive to negative at x = c, then c is a __________.
True or false: An inflection point occurs wherever f''(x) = 0.
Fill in the blank: The Extreme Value Theorem requires f to be continuous on a __________ interval.
True or false: If f''(c) > 0 at a critical number c, then c is a relative maximum.
Answers: 1. False. 2. Relative maximum. 3. False (f'' must also change sign). 4. Closed. 5. False (it is a relative minimum).
Q: What is the definition of a critical number, and why are critical numbers important?
A: A critical number of f is a value c in the domain of f where f'(c) = 0 or f'(c) is undefined. Critical numbers are important because all relative extrema of f occur at critical numbers (by Fermat's theorem), so they are the starting point for finding maxima and minima.
Q: Explain the difference between the first derivative test and the second derivative test. When would you prefer one over the other?
A: The first derivative test examines whether f' changes sign at a critical number (positive-to-negative for a max, negative-to-positive for a min). The second derivative test plugs the critical number into f'' (positive means min, negative means max). Use the second derivative test when f'' is easy to compute. Use the first derivative test when f''(c) = 0 (making the second test inconclusive) or when building a full sign chart is needed anyway.
Q: Find the inflection points of f(x) = x⁴ – 4x³.
A: f'(x) = 4x³ – 12x², f''(x) = 12x² – 24x = 12x(x – 2). Setting f''(x) = 0 gives x = 0 and x = 2. Testing signs: f'' > 0 on (–∞, 0), f'' < 0 on (0, 2), f'' > 0 on (2, ∞). Since f'' changes sign at both x = 0 and x = 2, both are inflection points.
Q: State the hypotheses of Rolle's Theorem and explain what it guarantees.
A: Rolle's Theorem requires that f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b). Under these conditions, there exists at least one c in (a, b) such that f'(c) = 0.
Q: On the interval [0, 2π], find the relative extrema of f(x) = (1/2)x – sin x.
A: f'(x) = 1/2 – cos x. Setting f'(x) = 0 gives cos x = 1/2, so x = π/3 and x = 5π/3. By the first derivative test, f' changes from negative to positive at x = π/3 (relative minimum) and from positive to negative at x = 5π/3 (relative maximum).
This material connects directly to optimisation (the next section of Unit 3), where you use derivative tests to solve applied max/min problems with constraints. It also connects to curve sketching, where combining first and second derivative information lets you draw an accurate graph of any function. In later calculus courses, these ideas extend to functions of several variables (partial derivatives and the multivariable second derivative test).
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