Difficulty: Intermediate | Prerequisites: Basic differentiation rules (power rule, product rule, quotient rule, chain rule), MAT 210 Exam 1 and 2 material.
This unit is where calculus stops being about computation and starts being about interpretation. You already know how to take derivatives; now you use them to find where functions peak, bottom out, or flatten, and what that means in applied settings. This material underpins optimisation problems (maximising profit, minimising cost) and related rates, both of which appear later in the exam. If you are comfortable differentiating polynomials, exponentials, and logarithms, you have the prerequisite toolkit. If you are shaky on the chain rule or quotient rule, revisit those before diving in here.
Critical points are x-values where the derivative is zero or undefined (and the function itself is defined). You classify them as relative maxima, relative minima, or neither using the first or second derivative test. On a closed interval, the absolute max and min are found by comparing function values at critical points and at the endpoints.
Critical point
A value x = c where f(c) is defined and either f'(c) = 0 or f'(c) is undefined. Think of it as: any spot on the curve where the slope goes flat or the derivative breaks down.
Relative maximum (local maximum)
A point where the function value is greater than all nearby values. In simple terms, it is a peak in the immediate neighbourhood, though not necessarily the highest point on the entire graph.
Relative minimum (local minimum)
A point where the function value is less than all nearby values. In simple terms, it is a valley in the immediate neighbourhood.
Absolute maximum (global maximum)
The single largest function value over the entire domain or over a specified interval. On a closed interval, this is guaranteed to exist by the Extreme Value Theorem.
Absolute minimum (global minimum)
The single smallest function value over the entire domain or over a specified interval. Also guaranteed on a closed interval.
First derivative test
A method for classifying critical points by checking whether f'(x) changes sign around the point. If f' goes from positive to negative, the critical point is a relative max. If f' goes from negative to positive, it is a relative min. If there is no sign change, the point is neither.
Closed interval method
The 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 compare. The largest value is the absolute max and the smallest is the absolute min.
Take the derivative f'(x).
Set f'(x) = 0 and solve for x. Also identify any x-values where f'(x) is undefined but f(x) is defined.
Each such x is a critical point.
Worked example: f(x) = x³ − 6x + 1
f'(x) = 3x² − 6
Set 3x² − 6 = 0, so x² = 2, giving x = √2 and x = −√2.
f(√2) = (√2)³ − 6√2 + 1 = 2√2 − 6√2 + 1 = 1 − 4√2 ≈ −4.66
f(−√2) = −2√2 + 6√2 + 1 = 1 + 4√2 ≈ 6.66
Once you have a critical point at x = c:
Pick a test value to the left of c and one to the right.
Plug each into f'(x) and note the sign.
Classify:
f' changes from + to − at c: relative maximum
f' changes from − to + at c: relative minimum
No sign change: neither (the curve just flattens and continues)
Worked example (continued): For f(x) = x³ − 6x + 1 with critical points x = −√2 and x = √2:
At x = −√2: test f'(−2) = 3(4) − 6 = 6 > 0 (left) and f'(0) = −6 < 0 (right). Sign change + to −, so x = −√2 is a relative max with value ≈ 6.66.
At x = √2: test f'(0) = −6 < 0 (left) and f'(2) = 6 > 0 (right). Sign change − to +, so x = √2 is a relative min with value ≈ −4.66.
Some functions have no real critical points. For example, f(x) = x³ + 6x + 1 gives f'(x) = 3x² + 6, which is always positive (the discriminant yields no real roots). The function is always increasing, so there are no extrema.
For f(x) = x³ + 2x² + x, f'(x) = 3x² + 4x + 1 = (3x + 1)(x + 1), giving critical points x = −1 and x = −1/3. Checking the sign of f' on either side of each shows that f' is positive, then zero, then positive again between the two points (the function flattens but does not change direction). Both are neither max nor min.
To find absolute extrema on [a, b]:
Find all critical points 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: f(x) = x⁴ − x³ + 1 on [−1, 1]
f'(x) = 4x³ − 3x² = x²(4x − 3). Critical points: x = 0 and x = 3/4.
f(−1) = 1 + 1 + 1 = 3
f(0) = 1
f(3/4) = (3/4)⁴ − (3/4)³ + 1 = 81/256 − 27/64 + 1 = 81/256 − 108/256 + 256/256 = 229/256 ≈ 0.895
f(1) = 1 − 1 + 1 = 1
Absolute max: 3 at x = −1. Absolute min: 229/256 at x = 3/4.
Critical point condition: f'(c) = 0 or f'(c) undefined, with f(c) defined.
First derivative test sign chart:
f'(x) left of c | f'(x) right of c | Classification |
|---|---|---|
+ | − | Relative maximum |
− | + | Relative minimum |
+ | + | Neither |
− | − | Neither |
Closed interval method: Evaluate f at all critical points in (a, b) and at endpoints a and b. Compare values.
Derivative rules you will need repeatedly:
Power rule: d/dx [xʳ] = r·xʳ⁻¹
Product rule: d/dx [f·g] = f'g + fg'
Quotient rule: d/dx [f/g] = (f'g − fg') / g²
Chain rule: d/dx [f(g(x))] = f'(g(x))·g'(x)
d/dx [eˣ] = eˣ
d/dx [ln x] = 1/x
Finding extrema is how businesses determine the production level that maximises profit or minimises cost. The same mathematics applies to engineering (minimising material usage in a design), medicine (finding peak drug concentration in the bloodstream), and economics (identifying optimal pricing). On the exam, expect word problems that frame these ideas as cost, revenue, or population models.
Students often assume every critical point is either a max or a min. It can be neither, as with inflection points where f' touches zero but does not change sign.
Students frequently forget to check endpoints when finding absolute extrema on a closed interval. The absolute max or min often occurs at an endpoint, not at a critical point.
A relative maximum is not necessarily the highest point on the graph. "Relative" means locally, within a neighbourhood. The absolute max may be elsewhere, or at an endpoint.
Setting f(x) = 0 instead of f'(x) = 0 is a common procedural error. Critical points come from the derivative, not from the original function.
⚠️ Problems 1 through 6 on the study guide all test the same skill: differentiate, set equal to zero, classify. Be fluent in this sequence.
⚠️ Problems 8 through 13 test the closed interval method. You must evaluate at endpoints every time.
⚠️ Problem 7 (finding the value of a so that f(x) = xe^(ax) has a critical point at x = 3) requires setting up f'(3) = 0 and solving for the parameter. This is a common exam twist.
⚠️ Watch for functions involving square roots (problem 4) and exponentials (problem 6), where the domain matters and the product rule is needed.
True or False: If f'(c) = 0, then x = c is always a relative extremum.
False. It could be neither (e.g. an inflection point).
Fill in the blank: To find absolute extrema on [a, b], evaluate f at all ______ in (a, b) and at both ______.
Critical points; endpoints.
True or False: A relative maximum must be the highest point on the entire graph.
False. It is the highest point only in its immediate neighbourhood.
Fill in the blank: If f' changes from negative to positive at x = c, then c is a relative ______.
Minimum.
True or False: If f'(x) is always positive, the function has no relative extrema.
True. The function is strictly increasing.
Q: Find the critical points of f(x) = 3x⁵ − 5x³ and classify each.
A: f'(x) = 15x⁴ − 15x² = 15x²(x² − 1) = 15x²(x − 1)(x + 1). Critical points at x = 0, x = 1, x = −1. By the first derivative test: x = −1 is a relative max, x = 1 is a relative min, and x = 0 is neither (f' does not change sign there).
Q: Find the critical points of f(x) = x²eˣ and classify each.
A: f'(x) = 2xeˣ + x²eˣ = xeˣ(2 + x). Since eˣ is never zero, set x(2 + x) = 0, giving x = 0 and x = −2. By the first derivative test (or second derivative test): x = −2 is a relative max with f(−2) = 4e⁻² ≈ 0.541, and x = 0 is a relative min with f(0) = 0.
Q: Find the absolute maximum and minimum of f(x) = x − ln x on [0.5, 3].
A: f'(x) = 1 − 1/x. Set equal to zero: 1 − 1/x = 0, so x = 1. Evaluate: f(0.5) = 0.5 − ln(0.5) ≈ 0.5 + 0.693 = 1.193, f(1) = 1 − 0 = 1, f(3) = 3 − ln 3 ≈ 1.901. Absolute min is 1 at x = 1. Absolute max is approximately 1.901 at x = 3.
Q: Find the value of a so that f(x) = xe^(ax) has a critical point at x = 3.
A: f'(x) = e^(ax) + axe^(ax) = e^(ax)(1 + ax). Set f'(3) = 0: e^(3a)(1 + 3a) = 0. Since e^(3a) is never zero, 1 + 3a = 0, so a = −1/3.
This material feeds directly into optimisation, where you find critical points of an objective function to maximise or minimise a real-world quantity. It also connects to concavity and the second derivative test, which gives a faster classification method for many problems. Related rates problems (later in this exam) rely on the same differentiation skills, applied to equations involving time.
Critical points, relative extrema, local max, local min, absolute extrema, global max, global min, first derivative test, sign chart, closed interval method, extreme value theorem, stationary points, turning points, endpoints, MAT 210, brief calculus, ASU, exam 3