Critical points are where a function's derivative equals zero or is undefined, and they are the candidates for local and absolute extrema. To find absolute extrema on a closed interval, evaluate the function at every critical point inside the interval and at both endpoints, then compare. The second derivative test is a shortcut for classifying critical points as local maxima, local minima, or neither.
Difficulty: Intermediate
Prerequisites: Basic differentiation rules (power rule, product rule, chain rule), familiarity with evaluating functions, and comfort with interval notation.
Critical points sit at the heart of what calculus does for business students: they tell you where a function changes direction, which is exactly the information you need for maximising profit, minimising cost, or finding when revenue peaks. This topic bridges differentiation (which you already know) and optimisation (which comes next). If you can find and classify critical points confidently, the applied problems in economics and production become straightforward. If you missed the last few weeks of derivatives, revisit the power rule and product rule before diving in here.
Critical point
A value x = c in the domain of f where f'(c) = 0 or f'(c) is undefined. In simple terms, it is any spot where the slope of the curve is flat or the derivative does not exist.
Absolute (global) maximum
The largest value a function takes on an entire interval, including the endpoints. Think of it as the highest point on the curve within the region you are examining.
Absolute (global) minimum
The smallest value a function takes on an entire interval, including the endpoints. Think of it as the lowest point on the curve within the region you are examining.
Local (relative) maximum
A point where f(c) is greater than or equal to f(x) for all x near c. In simple terms, it is a hilltop, though there may be a taller hill elsewhere.
Local (relative) minimum
A point where f(c) is less than or equal to f(x) for all x near c. Think of it as a valley floor, even if a deeper valley exists somewhere else on the curve.
Second derivative test
A method that uses f''(c) to classify a critical point. If f''(c) > 0, the curve is concave up at c and the point is a local minimum. If f''(c) < 0, the curve is concave down at c and the point is a local maximum. If f''(c) = 0, the test tells you nothing and you need another method.
Closed interval
An interval [a, b] that includes both of its endpoints. Absolute extrema are guaranteed to exist on a closed interval for any continuous function (this is the Extreme Value Theorem).
Concavity
The direction a curve bends. Concave up means the curve opens upward (like a bowl); concave down means it opens downward (like an arch). Concavity is determined by the sign of the second derivative.
The procedure is mechanical and always the same:
Take the derivative f'(x).
Set f'(x) = 0 and solve for x. Also note any x where f'(x) is undefined.
Discard any critical points that fall outside [a, b].
Evaluate f(x) at each remaining critical point and at both endpoints a and b.
The largest output is the absolute maximum; the smallest is the absolute minimum.
Worked example: f(x) = x^4 - 6x^2 on [1, 3].
f'(x) = 4x^3 - 12x = 4x(x^2 - 3). Setting this to zero gives x = 0 and x = ±√3.
Only x = √3 ≈ 1.73 lies inside [1, 3], so discard 0 and -√3.
Evaluate: f(1) = 1 - 6 = -5, f(√3) = 9 - 18 = -9, f(3) = 81 - 54 = 27.
Absolute minimum is -9 at x = √3. Absolute maximum is 27 at x = 3.
Other functions to practise with (same method applies to each):
f(x) = 5 ln(x) - 4x on [1, 3]
f(x) = x^3 - 27x + 1 on [-4, 0]
f(x) = 30/x^2 on [1, 4]
f(x) = 5xe^x on [0, 2]
Once you have a critical point c (where f'(c) = 0), compute f''(c):
f''(c) > 0: the curve is concave up, so c is a local minimum.
f''(c) < 0: the curve is concave down, so c is a local maximum.
f''(c) = 0: the test is inconclusive. Fall back to the first derivative test (check the sign of f' on either side of c).
This test is faster than the first derivative test when the second derivative is easy to compute, which is often the case with polynomial and exponential functions in this course.
When the interval is open or extends to infinity, there are no endpoints to evaluate. You rely entirely on critical points and the behaviour of f as x approaches the boundaries.
Example: f(x) = 20 - 4x - 16/x^3 on (0, ∞).
Find f'(x), set it to zero, and classify with the second derivative test.
Check the limits as x → 0+ and x → ∞ to confirm the critical point gives an absolute extremum rather than just a local one.
Example: f(x) = 8x^3 - 4x^5 on (0, ∞).
f'(x) = 24x^2 - 20x^4 = 4x^2(6 - 5x^2). Setting f'(x) = 0 gives x = 0 and x = √(6/5).
Since the interval is (0, ∞), examine x = √(6/5) with the second derivative test.
As x → ∞, f(x) → -∞, so the critical point is a local (and in this case absolute) maximum on (0, ∞).
f'(c) = 0 \quad \text{or} \quad f'(c) \text{ undefined} \implies c \text{ is a critical point}f''(c) > 0 \implies \text{local minimum at } cf''(c) < 0 \implies \text{local maximum at } cPower rule reminder:
\frac{d}{dx}[x^n] = nx^{n-1}For functions involving ln(x):
\frac{d}{dx}[\ln(x)] = \frac{1}{x}For functions involving e^x:
\frac{d}{dx}[e^x] = e^xBusinesses use critical points every time they set a price or decide how many units to produce. The revenue function R(x) typically has one critical point that marks the production level where revenue is highest, and the second derivative test confirms it is a maximum. Similarly, a cost function's critical point reveals the output level where average cost per unit is minimised, which directly affects pricing strategy.
Students often assume that every critical point is an extremum. It is not. A critical point can be an inflection point (e.g., f(x) = x^3 at x = 0, where f'(0) = 0 but there is no max or min).
Students frequently forget to check the endpoints when finding absolute extrema on a closed interval. The absolute max or min is often at an endpoint, not at a critical point.
Confusing the second derivative test result: f''(c) > 0 means concave up, which gives a local minimum (a bowl), not a maximum. Think "positive second derivative = smile = minimum."
Applying the second derivative test on open or infinite intervals and concluding you have an absolute extremum without checking the function's behaviour at the boundaries.
⚠️ "Find the absolute extrema of f(x) on [a, b]" is one of the most common exam question formats. The procedure is always the same: derivative → critical points → evaluate at critical points and endpoints → compare.
⚠️ You will almost certainly see a question requiring the second derivative test. Know the three outcomes cold: positive = min, negative = max, zero = inconclusive.
⚠️ Questions on infinite intervals (0, ∞) appear in business calculus to model scenarios where production can scale without a hard upper bound. Be ready to check limits at the boundary.
True or false: A critical point always produces a local maximum or minimum. (False)
Fill in the blank: If f''(c) > 0, then x = c is a local ______. (minimum)
True or false: To find absolute extrema on [a, b], you only need to check the critical points. (False, you must also check endpoints)
Fill in the blank: If f''(c) = 0, the second derivative test is ______. (inconclusive)
True or false: A continuous function on a closed interval is guaranteed to have both an absolute maximum and an absolute minimum. (True)
Q: Find the absolute maximum and minimum of f(x) = 5 ln(x) - 4x on [1, 3].
A: Take f'(x) = 5/x - 4. Set to zero: 5/x = 4, so x = 5/4. Evaluate f(1) = 0 - 4 = -4, f(5/4) = 5 ln(1.25) - 5 ≈ -3.88, f(3) = 5 ln(3) - 12 ≈ -6.51. Absolute max is f(5/4) ≈ -3.88. Absolute min is f(3) ≈ -6.51.
Q: Use the second derivative test to classify the critical points of f(x) = x^4 - 6x^2.
A: f'(x) = 4x^3 - 12x = 4x(x^2 - 3). Critical points: x = 0, x = √3, x = -√3. f''(x) = 12x^2 - 12. f''(0) = -12 < 0, so x = 0 is a local max. f''(√3) = 36 - 12 = 24 > 0, so x = √3 is a local min. f''(-√3) = 24 > 0, so x = -√3 is a local min.
Q: Find the absolute extrema of f(x) = x^3 - 27x + 1 on [-4, 0].
A: f'(x) = 3x^2 - 27 = 3(x^2 - 9). Critical points: x = 3 and x = -3. Only x = -3 is in [-4, 0]. Evaluate f(-4) = -64 + 108 + 1 = 45, f(-3) = -27 + 81 + 1 = 55, f(0) = 1. Absolute max is 55 at x = -3. Absolute min is 1 at x = 0.
Q: A function is defined on (0, ∞). After finding a single critical point, how do you confirm it gives an absolute extremum rather than just a local one?
A: Check the limits of f(x) as x → 0+ and x → ∞. If the function value at the critical point is larger (or smaller) than both limiting values, it is the absolute maximum (or minimum) on the interval.
This connects directly to optimisation problems (maximising revenue, minimising cost), which apply the same critical-point procedure to economic functions. It also connects to integration: once you understand where a function increases and decreases, you can reason about the area under the curve and the net change over an interval. The second derivative test reappears when classifying equilibrium points in differential equations.
Critical points, extrema, absolute maximum, absolute minimum, local max, local min, relative extrema, second derivative test, concavity, closed interval method, Extreme Value Theorem, inflection point, f'(x) = 0, open interval extrema, business calculus critical points, MATH 201, calculus for business, FIU calculus