Difficulty: Intermediate | Prerequisites: Basic derivative rules (power, product, quotient, chain rule)
Critical points are where a function's behaviour shifts, from rising to falling or vice versa. They are the foundation for every optimisation problem in this course, from maximising revenue to minimising cost. You need a solid grip on derivative rules before this topic makes sense, because finding critical points starts with setting the derivative equal to zero. If you can take a derivative and solve the resulting equation, you can find critical points.
A critical point of f(x) occurs where f is defined and f'(x) = 0 or f'(x) does not exist. To find absolute extrema on a closed interval, evaluate f at every critical point inside the interval and at both endpoints, then compare. Relative (local) extrema happen at interior critical points where the function changes direction.
Critical point
A point (c, f(c)) where f(c) is defined and either f'(c) = 0 or f'(c) does not exist. Think of it as any spot on the curve where the tangent line is flat or where the function has a sharp corner or cusp.
Critical value
The x-coordinate of a critical point. In simple terms, it is the input value where something interesting might happen to the function's direction.
Relative (local) maximum
A point where f(c) is greater than or equal to all nearby function values. In simple terms, a hilltop on the graph, even if higher hills exist elsewhere.
Relative (local) minimum
A point where f(c) is less than or equal to all nearby function values. Think of it as a valley floor, even if deeper valleys exist elsewhere on the graph.
Absolute (global) maximum
The largest value of f(x) on the entire domain or interval under consideration. This is the single highest point, full stop.
Absolute (global) minimum
The smallest value of f(x) on the entire domain or interval under consideration. The single lowest point.
First Derivative Test
A method for classifying critical points by checking whether f'(x) changes sign around the critical value. If f' changes from positive to negative, you have a local max. If f' changes from negative to positive, you have a local min. If the sign does not change, it is neither.
Closed Interval Method
A procedure for finding absolute extrema on a closed interval [a, b]: find all critical points in (a, b), evaluate f at each critical point and at both endpoints, then pick the largest and smallest values.
Compute f'(x) using the appropriate derivative rules.
Set f'(x) = 0 and solve for x. These are your candidate critical values.
Also identify any x-values in the domain where f'(x) does not exist (corners, cusps, vertical tangents).
Confirm that f(x) is defined at each candidate. If f(c) is not defined, x = c is not a critical point.
Write critical points as ordered pairs (c, f(c)).
Worked example: For f(x) = 3x³ − 18x² + 27x − 10 on [−2, 5]:
f'(x) = 9x² − 36x + 27 = 9(x² − 4x + 3) = 9(x − 1)(x − 3)
Setting f'(x) = 0 gives x = 1 and x = 3
f(1) = 3 − 18 + 27 − 10 = 2, so one critical point is (1, 2)
f(3) = 81 − 162 + 81 − 10 = −10, so the other is (3, −10)
Pick test values on each side of the critical value.
If f'(x) changes from + to −, the critical point is a relative maximum.
If f'(x) changes from − to +, the critical point is a relative minimum.
If f'(x) does not change sign, the critical point is neither a max nor a min (e.g. an inflection point).
When f is continuous on a closed interval [a, b], the absolute max and min must occur either at a critical point inside the interval or at an endpoint. The procedure:
Find all critical points of f in the open interval (a, b).
Evaluate f at each critical point and at both endpoints a and b.
The largest value is the absolute maximum. The smallest is the absolute minimum.
Worked example (continued): For f(x) = 3x³ − 18x² + 27x − 10 on [−2, 5]:
Critical points: x = 1 and x = 3 (both in (−2, 5))
f(−2) = 3(−8) − 18(4) + 27(−2) − 10 = −24 − 72 − 54 − 10 = −160
f(1) = 2
f(3) = −10
f(5) = 3(125) − 18(25) + 27(5) − 10 = 375 − 450 + 135 − 10 = 50
Absolute max: (5, 50). Absolute min: (−2, −160).
Relative max: (1, 2). Relative min: (3, −10).
Critical points appear where the curve has a horizontal tangent (flat spot) or a sharp corner.
On a graph, look for peaks (local max), valleys (local min), and flat spots where the curve levels off before continuing.
The absolute extrema on a closed interval include the endpoints. Compare the y-values at every peak, valley, and endpoint.
f'(x) > 0 on intervals where the graph is rising (going uphill left to right).
f'(x) < 0 on intervals where the graph is falling (going downhill left to right).
f'(x) = 0 at points where the graph transitions between rising and falling (horizontal tangent).
\text{Critical point condition: } f'(c) = 0 \text{ or } f'(c) \text{ does not exist, with } f(c) \text{ defined}\text{First Derivative Test sign chart: test values of } f'(x) \text{ on each side of } c\text{Closed interval: evaluate } f(a),\; f(c_1),\; f(c_2),\; \ldots,\; f(b) \text{ and compare}Critical points are the mathematical backbone of any decision about "what value of x gives the best result." Engineers use them to find the dimensions that minimise material cost for a given volume. Economists use them to find the price that maximises revenue or profit.
Students often think that f'(c) = 0 is the only condition for a critical point. It is not. If f'(c) does not exist but f(c) is defined, x = c is still a critical point. Sharp corners and cusps count.
Students often assume every critical point is a local max or min. That is false. A critical point can be an inflection point (e.g. f(x) = x³ at x = 0). The first derivative test is how you tell the difference.
Students sometimes forget the endpoints when finding absolute extrema on a closed interval. The absolute max or min can occur at an endpoint, and often does.
Students sometimes confuse "critical point" with "critical value." The critical value is the x-coordinate alone. The critical point is the ordered pair (x, f(x)).
⚠️ The definition of critical point (f(c) defined AND f'(c) = 0 or DNE) is tested directly on multiple-choice questions. Know it precisely.
⚠️ The closed interval method is a guaranteed free-response topic. Practise it end to end: derivative, critical values, endpoint evaluation, comparison.
⚠️ Graph-reading questions test whether you can identify relative max, relative min, absolute extrema, and intervals of increase/decrease just by looking at the curve. Practise reading peaks, valleys, and slopes.
True or False: If f'(c) does not exist but f(c) is defined, then x = c is a critical point. Answer: True.
True or False: Every critical point is either a relative maximum or a relative minimum. Answer: False. A critical point can be neither (e.g. an inflection point).
Fill in the blank: To find absolute extrema on [a, b], evaluate f at all critical points in ______ and at both ______. Answer: the open interval (a, b); endpoints.
True or False: If f'(x) changes from negative to positive at x = c, then f has a relative maximum at x = c. Answer: False. That pattern gives a relative minimum.
Fill in the blank: The x-coordinate of a critical point is called the ______. Answer: critical value.
Q: Which condition must be true for x = c to be a critical point of g(x)?
A: g(c) must be defined, and either g'(c) = 0 or g'(c) does not exist.
Q: If h(2) exists but h'(2) does not exist, what can we conclude about x = 2?
A: x = 2 is a critical point. The function is defined there and the derivative does not exist, which satisfies the definition.
Q: Let f(x) = 3x³ − 18x² + 27x − 10 on [−2, 5]. Find all critical points.
A: f'(x) = 9x² − 36x + 27 = 9(x − 1)(x − 3). Setting f'(x) = 0 gives x = 1 and x = 3. The critical points are (1, 2) and (3, −10).
Q: Using the function above, find the absolute maximum and absolute minimum on [−2, 5].
A: Evaluate f at the critical points and endpoints: f(−2) = −160, f(1) = 2, f(3) = −10, f(5) = 50. Absolute max is (5, 50). Absolute min is (−2, −160).
Q: A function f(x) is graphed on [−2, 2]. The graph rises from (−2, −1) to a peak near x = −0.5, then falls to (2, −1). Where does f have a relative maximum, and where does it have an absolute minimum?
A: Relative maximum near x = −0.5 (the interior peak). Absolute minimum at x = −2 and x = 2 (both endpoints share the lowest value of −1).
Q: Consider f(x) = xe^(2x). Find the critical point(s).
A: Using the product rule: f'(x) = e^(2x) + x(2e^(2x)) = e^(2x)(1 + 2x). Since e^(2x) is never zero, set 1 + 2x = 0, giving x = −1/2.
Critical points rely entirely on derivative computation, so everything from the power rule through the chain rule feeds directly into this topic. Optimisation problems (maximising revenue, minimising cost) are applied versions of the closed interval method or the first derivative test. Second derivatives, covered later, give another way to classify critical points (the second derivative test).
critical point, critical value, critical number, relative maximum, relative minimum, local max, local min, absolute maximum, absolute minimum, global extrema, first derivative test, sign chart, closed interval method, extreme value theorem, increasing and decreasing intervals, optimisation, optimization, turning point, stationary point, MAT 210, brief calculus, ASU