Difficulty: Intermediate | Prerequisites: Intervals of increase and decrease (see companion notes), power rule, quotient rule, sign charts.
Once you can find where a function increases and decreases, the natural next question is: what happens at the boundaries between those intervals? That is where relative (local) maxima and minima live. This section teaches the First Derivative Test, which uses the sign-change pattern of f'(x) around a critical point to classify it as a relative maximum, a relative minimum, or neither. This is the bridge between "describe the function's behaviour" and "solve optimisation problems," which is where most of the exam weight sits in business calculus.
A critical point where f' changes from positive to negative is a relative maximum (hilltop). A critical point where f' changes from negative to positive is a relative minimum (valley floor). If the sign of f' does not change, the critical point is neither.
Critical point (critical number)
A value c in the domain of f where f'(c) = 0 or f'(c) is undefined. In simple terms, these are the x-values where the function might have a peak, a trough, or a flat spot.
Relative maximum (local maximum)
The point (c, f(c)) where the function value is higher than at all nearby points. Think of it as a local hilltop: the function climbs up to it from the left and drops away from it to the right.
Relative minimum (local minimum)
The point (c, f(c)) where the function value is lower than at all nearby points. This is a local valley floor: the function falls into it from the left and rises away from it to the right.
First Derivative Test
A method for classifying a critical point by examining whether f'(x) changes sign as x passes through it. The direction of the sign change tells you maximum, minimum, or neither.
Neither maximum nor minimum
A critical point where f' has the same sign on both sides (e.g. positive on both sides). The function pauses momentarily but does not reverse direction. An inflection point with a horizontal tangent is a typical example.
Let c be a critical point of f (meaning f'(c) = 0 or f'(c) is undefined, and f(c) exists).
If f'(x) changes from positive to negative at c, then (c, f(c)) is a relative maximum.
The function is rising on the left of c and falling on the right. It peaks at c.
If f'(x) changes from negative to positive at c, then (c, f(c)) is a relative minimum.
The function is falling on the left of c and rising on the right. It bottoms out at c.
If f'(x) does not change sign at c, then (c, f(c)) is neither a relative maximum nor a relative minimum.
Find f'(x).
Find all critical points (where f'(x) = 0 or is undefined).
Build the sign chart for f'(x) across the entire number line.
At each critical point, read the sign change:
to − → relative max
− to + → relative min
to + or − to − → neither
Compute f(c) to get the y-coordinate of the extremum.
The lecture notes show a number line annotated with arrows:
An upward arrow (↑) marks an interval where f' > 0 (increasing).
A downward arrow (↓) marks an interval where f' < 0 (decreasing).
At each critical point, the pattern of arrows on either side tells you the classification:
↑ then ↓ → relative maximum
↓ then ↑ → relative minimum
↑ then ↑ or ↓ then ↓ → neither
Using the rational function f(x) = x² / (x − 2) from the intervals-of-increase notes:
Critical points: x = 0, x = 4. Vertical asymptote at x = 2.
Sign chart of f'(x): positive on (−∞, 0), negative on (0, 2), negative on (2, 4), positive on (4, ∞).
At x = 0: f' changes from + to − → relative maximum at (0, f(0)) = (0, 0).
At x = 4: f' changes from − to + → relative minimum at (4, f(4)) = (4, 8).
At x = 2: this is a vertical asymptote, not a critical point. No extremum here.
First Derivative Test (summary)
f'(x): + → − at c ⟹ relative maximum at (c, f(c))
f'(x): − → + at c ⟹ relative minimum at (c, f(c))
f'(x): no sign change at c ⟹ neither
Finding the y-coordinate
Once you identify a critical point c as a max or min, substitute back into the original function: y = f(c). The answer is the point (c, f(c)), not just the x-value.
In business contexts, relative maxima and minima correspond to peak profit, minimum cost, or maximum revenue. When a company models its profit as a function of units produced, the relative maximum of that function is the production level that maximises profit. The First Derivative Test is the standard method for locating these optima.
Students often assume every critical point is either a max or a min. It can be neither, if the sign of f' does not change. For example, f(x) = x³ has a critical point at x = 0, but f' goes from positive to positive (through zero), so there is no extremum.
A relative maximum does not have to be the highest point on the entire graph. "Relative" means locally: higher than its immediate neighbours. The function may reach higher values elsewhere.
Students sometimes classify the vertical asymptote itself as an extremum. A VA is not in the domain, so it cannot be a max or min.
Forgetting to compute the y-value. The question usually asks for the point (c, f(c)), not just the x-coordinate.
⚠️ The First Derivative Test is a staple exam question. You will almost always be asked to find and classify the critical points of a given function.
⚠️ State the sign change explicitly (e.g. "f' changes from + to − at x = 0, so there is a relative maximum"). Examiners want to see the reasoning, not just the label.
⚠️ Always give the full point (c, f(c)), not just x = c, unless the question specifically asks only for the x-value.
⚠️ If the function has a vertical asymptote, mark it on the sign chart but do not classify it as an extremum.
True or false: A relative maximum occurs where f'(x) changes from negative to positive.
Fill in the blank: If f'(x) changes from positive to negative at x = c, then (c, f(c)) is a relative ______.
True or false: Every critical point is either a relative max or a relative min.
Fill in the blank: To find the y-coordinate of a relative extremum, substitute the critical point back into ______.
True or false: A vertical asymptote can be a relative extremum.
Answers: 1. False (that is a relative minimum). 2. Maximum. 3. False (it can be neither). 4. The original function f(x). 5. False.
Q: For f(x) = 2x³ + 3x² − 12x − 7, find and classify all relative extrema.
A: f'(x) = 6x² + 6x − 12 = 6(x + 2)(x − 1). Critical points: x = −2, x = 1. Sign chart: f' > 0 on (−∞, −2), f' < 0 on (−2, 1), f' > 0 on (1, ∞). At x = −2, f' changes + to − → relative maximum. f(−2) = 2(−8) + 3(4) − 12(−2) − 7 = −16 + 12 + 24 − 7 = 13. Relative max at (−2, 13). At x = 1, f' changes − to + → relative minimum. f(1) = 2 + 3 − 12 − 7 = −14. Relative min at (1, −14).
Q: For g(x) = x⁴ − 4x³, find and classify the critical points.
A: g'(x) = 4x³ − 12x² = 4x²(x − 3). Critical points: x = 0, x = 3. Sign chart: g' < 0 on (−∞, 0), g' < 0 on (0, 3), g' > 0 on (3, ∞). At x = 0, f' is negative on both sides → neither max nor min. At x = 3, f' changes − to + → relative minimum. g(3) = 81 − 108 = −27. Relative min at (3, −27).
Q: A function has f'(x) = (x − 1)(x − 5). Classify the critical points.
A: Critical points at x = 1 and x = 5. For x < 1, f'(0) = (−1)(−5) = 5 > 0. For 1 < x < 5, f'(3) = (2)(−2) = −4 < 0. For x > 5, f'(6) = (5)(1) = 5 > 0. At x = 1: + to − → relative maximum. At x = 5: − to + → relative minimum.
The First Derivative Test builds directly on the intervals-of-increase-and-decrease material (see the companion notes). It also pairs with the Second Derivative Test, which offers a shortcut for classifying critical points when f''(c) is easy to compute. Later in the course, you will use these tools to solve applied optimisation problems: maximising profit, minimising cost, and finding break-even points.
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