Difficulty: Intermediate | Prerequisites: Chapter 2 (basic differentiation, power rule, product rule)
This chapter sits at the heart of optimisation in a business calculus course. Once you can take a derivative, the next question is: where does the function reach its highest or lowest values, and how do you prove which is which? That matters because business problems (maximising profit, minimising cost) come down to exactly this. If you are comfortable differentiating polynomials and rational functions, you are ready.
Critical points are the x-values where a function's first derivative equals zero (or is undefined), and they mark where relative maxima and minima can occur. The second derivative test gives you a quick way to classify each critical point: if the second derivative is positive there, it is a relative minimum; if negative, a relative maximum. When the second derivative equals zero the test is inconclusive, and you fall back to the first derivative test.
Critical point
A value x = c in the domain of f where f'(c) = 0 or f'(c) is undefined. These are the only candidates for relative extrema.
Think of it as: the spots where the curve levels off or has a sharp corner, so it might be switching from going up to going down (or vice versa).
First derivative (f'(x))
The rate of change of f(x). Setting f'(x) = 0 and solving gives the critical points.
In simple terms, this means: the slope of the tangent line at any point on the curve. Where the slope is zero, the curve is flat.
Second derivative (f''(x))
The derivative of the first derivative. It measures how the slope itself is changing, which tells you about the concavity of the curve.
In simple terms, this means: if f''(x) is positive, the curve bends upward (like a bowl); if negative, it bends downward (like a hill).
Relative maximum (local maximum)
A point where f(c) is greater than or equal to all nearby values of f. The function rises to this point and then falls away.
Think of it as: the top of a hill on the graph.
Relative minimum (local minimum)
A point where f(c) is less than or equal to all nearby values of f. The function falls to this point and then rises.
Think of it as: the bottom of a valley on the graph.
Second derivative test
A method for classifying critical points. If f'(c) = 0 and f''(c) > 0, then x = c is a relative minimum. If f'(c) = 0 and f''(c) < 0, then x = c is a relative maximum. If f''(c) = 0, the test is inconclusive.
In simple terms, this means: check which way the curve bends at a critical point. Bending up = minimum. Bending down = maximum.
Concavity
Describes the direction a curve bends. Concave up means the curve opens upward (f''(x) > 0). Concave down means it opens downward (f''(x) < 0).
Think of it as: concave up holds water like a cup; concave down spills it.
Take the first derivative f'(x) of the given function.
Set f'(x) = 0 and solve for x. Also check where f'(x) is undefined.
The solutions are the critical points (CPs).
Suppose f''(x) exists on an open interval containing x = c, and f'(c) = 0.
If f''(c) > 0, then f has a relative minimum at x = c (the curve is concave up there, so the flat spot is a valley).
If f''(c) < 0, then f has a relative maximum at x = c (the curve is concave down there, so the flat spot is a hilltop).
If f''(c) = 0, the test tells you nothing. Use the first derivative test instead (check sign changes of f' around x = c).
Step 1: Find f'(x)
f'(x) = 6x² + 6x − 12
Factor: f'(x) = 6(x² + x − 2) = 6(x + 2)(x − 1)
Step 2: Set f'(x) = 0
6(x + 2)(x − 1) = 0
Critical points: x = −2 and x = 1
Step 3: Find f''(x)
f''(x) = 12x + 6
Step 4: Evaluate f'' at each critical point
f''(−2) = 12(−2) + 6 = −24 + 6 = −18 < 0, so x = −2 is a relative maximum
f''(1) = 12(1) + 6 = 12 + 6 = 18 > 0, so x = 1 is a relative minimum
f'(c) = 0 \quad \Rightarrow \quad x = c \text{ is a critical point}f''(c) > 0 \quad \Rightarrow \quad \text{relative minimum at } x = cf''(c) < 0 \quad \Rightarrow \quad \text{relative maximum at } x = cf''(c) = 0 \quad \Rightarrow \quad \text{test is inconclusive}In business, critical points are how you find the production level that maximises profit or minimises average cost. If your profit function is a polynomial, setting its derivative to zero gives you the quantity where profit peaks, and the second derivative confirms whether that peak is truly a maximum. The same logic applies to engineering design (minimising material usage) and economics (finding equilibrium prices).
Students often think every critical point is a maximum or minimum. It is not. If f''(c) = 0, the critical point could be an inflection point (like x = 0 for f(x) = x³), where the curve changes concavity without any extremum.
Students sometimes confuse the sign convention: f''(c) > 0 means concave up, which means minimum (valley), not maximum. A common exam error is writing "positive second derivative = maximum."
Students occasionally forget to check where f'(x) is undefined. Critical points include values where the derivative does not exist, not only where it equals zero.
Students sometimes plug critical points into f(x) when the question only asks to classify them. The second derivative test classifies (max vs. min) without needing the y-value, though the y-value is needed if the question asks for the actual coordinates.
⚠️ "Find and classify all critical points" is one of the most common exam prompts in this chapter. You will almost certainly see a polynomial where you must factor f'(x), find the roots, then apply the second derivative test.
⚠️ Exams often include a function where f''(c) = 0 at one of the critical points, specifically to test whether you know the second derivative test is inconclusive there.
⚠️ Sign errors in computing f''(c) are the most common way students lose marks. Write out each substitution step fully.
⚠️ Some exam questions give you f'(x) already factored and ask you to classify the critical points without finding f(x). Be comfortable working from the derivative alone.
True or false: If f'(c) = 0, then x = c must be a relative maximum or minimum. (False: it could be an inflection point.)
Fill in the blank: If f''(c) > 0, then x = c is a relative ________. (minimum)
True or false: The second derivative test can always determine whether a critical point is a max or min. (False: when f''(c) = 0, the test is inconclusive.)
Fill in the blank: For f(x) = 2x³ + 3x² − 12x − 7, the critical points are x = ________ and x = ________. (x = −2 and x = 1)
True or false: f''(−2) = −18 means x = −2 is a relative minimum for the function above. (False: negative second derivative means relative maximum.)
Q: Find all critical points of f(x) = x³ − 6x² + 9x + 2 and classify each using the second derivative test.
A: f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). Critical points: x = 1 and x = 3. f''(x) = 6x − 12. f''(1) = 6(1) − 12 = −6 < 0, so x = 1 is a relative maximum. f''(3) = 6(3) − 12 = 6 > 0, so x = 3 is a relative minimum.
Q: Suppose f'(2) = 0 and f''(2) = 0. What can you conclude about x = 2?
A: The second derivative test is inconclusive. You cannot determine whether x = 2 is a maximum, minimum, or neither from this information alone. You would need to apply the first derivative test (check the sign of f' on either side of x = 2).
Q: A company's profit function is P(x) = −2x² + 80x − 300, where x is units sold in thousands. Find the production level that maximises profit.
A: P'(x) = −4x + 80. Setting P'(x) = 0 gives x = 20. P''(x) = −4, which is always negative, so x = 20 is a relative maximum. The company maximises profit at 20,000 units.
Q: If f'(x) = 6(x + 2)(x − 1), at which x-values is f increasing?
A: f is increasing where f'(x) > 0. Test intervals: for x < −2, both factors are negative, product is positive (increasing). For −2 < x < 1, (x + 2) is positive and (x − 1) is negative, product is negative (decreasing). For x > 1, both factors are positive, product is positive (increasing). So f is increasing on (−∞, −2) and (1, ∞).
This connects directly to graph sketching (the next topic in Ch. 3), because once you know where the critical points are and whether they are maxima or minima, you have the key landmarks for drawing the curve. It also connects to concavity and inflection points: the second derivative you compute here is the same one you use to determine where the curve changes from bending up to bending down. In later chapters, these ideas extend to optimisation problems, where you build a function from a word problem and then use critical points to find the best answer.
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