MAC 2233 Calculus for Business | Sections 4.1, 4.2 | Source: Exam 3 Review
This section covers how to use the first derivative to determine where a function is increasing or decreasing, how to find critical points, and how to classify local maxima and minima using the first derivative test. You will also interpret graphs of f'(x) to deduce the behaviour of f(x), and apply these ideas to cost and production problems.
Difficulty: Intermediate
Prerequisites: All differentiation techniques from Part 1 (chain rule, product rule, quotient rule, log/exponential derivatives). You need to be able to find f'(x) quickly and correctly before you can analyse its sign.
Knowing the derivative of a function tells you its slope at every point. A positive derivative means the function is climbing; a negative derivative means it is falling. The points where the derivative is zero or undefined are the critical points, where the function might switch from rising to falling (a local maximum) or from falling to rising (a local minimum). This section turns the algebraic skill of Part 1 into a tool for understanding the shape and behaviour of functions, which is the core of what calculus is for in business applications.
Critical point (critical number)
A value x = c in the domain of f where f'(c) = 0 or f'(c) does not exist. These are the only candidates for local extrema.
In simple terms, a critical point is where the curve either flattens out or has a sharp corner or vertical tangent.
Increasing function
f(x) is increasing on an interval if f'(x) > 0 for all x in that interval. The graph moves upward from left to right.
Think of it as: positive derivative means the function is climbing.
Decreasing function
f(x) is decreasing on an interval if f'(x) < 0 for all x in that interval. The graph moves downward from left to right.
Think of it as: negative derivative means the function is falling.
Local maximum
A point where f changes from increasing to decreasing. f'(x) changes sign from positive to negative.
In simple terms, the top of a hill.
Local minimum
A point where f changes from decreasing to increasing. f'(x) changes sign from negative to positive.
In simple terms, the bottom of a valley.
First derivative test
A method for classifying critical points: check the sign of f'(x) on either side of the critical point. If the sign changes from + to -, it is a local max. If from - to +, it is a local min. If the sign does not change, it is neither.
Think of it as: watch what the slope is doing on the left and right of the critical point.
Sign chart (sign diagram)
A number line showing where f'(x) is positive, negative, or zero. You build it by testing one value in each interval between critical points.
In simple terms, a quick visual summary of where the function goes up and where it goes down.
Step-by-step method
Find f'(x).
Set f'(x) = 0 and solve. Also identify where f'(x) is undefined (but f(x) is defined).
These x-values are the critical numbers.
Build a sign chart: pick test values in each interval between critical numbers, plug them into f'(x), and record the sign.
Apply the first derivative test at each critical number.
Example: f(x) = 3x² - 12x + 2
f'(x) = 6x - 12 = 6(x - 2). Set equal to zero: x = 2.
Test x = 0: f'(0) = -12 < 0 (decreasing). Test x = 3: f'(3) = 6 > 0 (increasing).
Sign changes from - to + at x = 2, so (2, f(2)) = (2, -10) is a local minimum.
Increasing on (2, ∞). Decreasing on (-∞, 2). No local max.
Example: f(x) = 4x³ - x⁴
f'(x) = 12x² - 4x³ = 4x²(3 - x). Set equal to zero: x = 0, x = 3.
Test intervals: f'(-1) > 0, f'(1) > 0, f'(4) < 0.
At x = 0, sign does not change (+ to +), so x = 0 is not a local extremum.
At x = 3, sign changes from + to -, so (3, 27) is a local maximum.
Increasing on (-∞, 3) (or more precisely (-∞, 0) ∪ (0, 3) since x = 0 is in the domain). Decreasing on (3, ∞).
Example: f(x) = x²/(x - 3)
Use the quotient rule: f'(x) = x(x - 6)/(x - 3)².
Critical numbers: x = 0, x = 6. Also f'(x) is undefined at x = 3 (not in the domain of f).
Sign chart across intervals (-∞, 0), (0, 3), (3, 6), (6, ∞) gives +, -, -, +.
Local max at (0, 0). Local min at (6, 12).
The logic
f'(x) > 0 on an interval means f is increasing there.
f'(x) < 0 on an interval means f is decreasing there.
f'(x) = 0 at a point means the tangent is horizontal there.
How to build a sign chart
Find all critical numbers (where f'(x) = 0 or is undefined).
Mark them on a number line.
Choose one test value in each interval.
Evaluate f'(test value). You only need the sign, not the exact number.
Record + or - above each interval.
Reading the chart
Where the sign chart shows +, report that interval as increasing. Where it shows -, report that interval as decreasing. Where f'(x) is undefined because of a discontinuity in f itself, exclude that x-value from the interval.
Example: f(x) = x³ - 3x + 1 (Problem 25)
f'(x) = 3x² - 3 = 3(x - 1)(x + 1). Zeros at x = -1 and x = 1.
Sign chart: (+) on (-∞, -1), (-) on (-1, 1), (+) on (1, ∞).
Increasing on (-∞, -1) and (1, ∞). Decreasing on (-1, 1).
Local max at (-1, 3). Local min at (1, -1).
Problems 20, 23, and 24 give you the graph of f'(x) and ask you to deduce properties of f(x).
Key translations
Where the graph of f'(x) is above the x-axis (positive), f(x) is increasing.
Where the graph of f'(x) is below the x-axis (negative), f(x) is decreasing.
Where the graph of f'(x) crosses the x-axis from above to below, f(x) has a local maximum.
Where the graph of f'(x) crosses the x-axis from below to above, f(x) has a local minimum.
Where the graph of f'(x) touches the x-axis but does not cross it, f(x) has neither a max nor a min at that point.
Example: Problem 20
The graph of f(x) is given and you need to find where f'(x) > 0 (i.e., where f is increasing). Looking at the graph, f is rising on (-∞, -2) and (1, ∞).
Example: Problem 23
The graph of f'(x) is given. f'(x) > 0 on (-∞, -3) and (-1, 2). f'(x) < 0 on (-3, -1) and (2, ∞). So f is increasing on (-∞, -3) and (-1, 2), decreasing on (-3, -1) and (2, ∞). Local max at x = -3 and x = 2. Local min at x = -1.
Example: Problem 24
f'(x) > 0 only on (-∞, -2). f'(x) < 0 on (-2, ∞). So f is increasing on (-∞, -2), decreasing on (-2, ∞). Local max at x = -2. No local min.
Problems 21 and 22 give you a sign chart for f'(x) plus a table of f(x) values, and ask you to sketch the graph.
The method
Plot the given points from the table.
Where f'(x) > 0, draw the curve rising. Where f'(x) < 0, draw it falling.
At points where f'(x) = 0, draw a horizontal tangent. Label it as a local max or local min based on the sign change.
Where f'(x) is "ND" (not defined), the function may have a cusp, corner, or vertical tangent. The function can still be continuous there.
Example: Problem 21
f'(x): + + + 0 + + + 0 - - - with zeros at x = -1 and x = 1. Given points: (-2, -1), (-1, 1), (0, 2), (1, 3), (2, 1). f'(x) is positive on (-∞, -1) and (-1, 1), then zero at x = 1, then negative. So f is always increasing up to x = 1, then decreasing. Local max at (1, 3), no local min (note: x = -1 is a critical point but f' does not change sign there, so no extremum).
Example: Problem 22
f'(x) is negative on (-∞, -1), zero at x = -1, positive on (-1, 0), ND at x = 0, negative on (0, 2), zero at x = 2, negative on (2, ∞). Given points include (-2, 2), (-1, 1), (0, 2), (2, 1), (4, 0). Local min at (-1, 1). Local max at (0, 2). No extremum at x = 2 (sign does not change, stays negative).
Average cost minimisation (Problem 26)
Given C(x) = 0.08x² + 30x + 450, the average cost per jacket is C̄(x) = C(x)/x = 0.08x + 30 + 450/x.
To minimise average cost, find C̄'(x) = 0.08 - 450/x². Set it to zero: 0.08 = 450/x², so x² = 5625, giving x = 75.
Sign chart: C̄'(x) < 0 on (0, 75) and C̄'(x) > 0 on (75, 200). So average cost is decreasing until 75 jackets and increasing after. Local minimum at x = 75 with C̄(75) = $42 per jacket.
Why this matters in business: a manufacturer wants to know the production level that minimises cost per unit. The first derivative of the average cost function gives exactly that answer.
Students often assume that f'(c) = 0 automatically means there is a local extremum at x = c. It does not. You must check whether f' actually changes sign. For example, f(x) = x³ has f'(0) = 0 but no local extremum at x = 0, because f' does not change sign.
Students report intervals of increase/decrease using the critical point itself (e.g., writing "increasing on [2, ∞)" with a closed bracket). Use open intervals: increasing on (2, ∞).
When f'(x) is undefined at a point where f(x) is also undefined (like x = 3 in a rational function with x - 3 in the denominator), that point is not a critical number because it is not in the domain. Only points in the domain of f can be critical.
Students sometimes test f(x) values instead of f'(x) values in the sign chart. You must plug test values into f'(x), not f(x).
⚠️ Problems 17 to 19 are the bread-and-butter exam question: find f'(x), build a sign chart, state intervals of increase/decrease, and classify extrema. Expect at least two of these.
⚠️ Problems 20, 23, 24 test graph reading. You may be given the graph of f'(x) and asked about f(x). Know the translations cold.
⚠️ Problems 21 and 22 test graph sketching from sign chart data. Practice drawing smooth curves that respect the given sign information and pass through the given points.
⚠️ Problem 26 is an applied problem. On the exam, an applied first-derivative problem will likely involve cost, revenue, or production. The method is the same: find the derivative, set it to zero, build a sign chart.
True or False: If f'(5) = 0, then f has a local extremum at x = 5. False. f' must also change sign at x = 5.
Fill in the blank: f(x) is increasing on intervals where f'(x) is _____. Answer: positive (f'(x) > 0).
True or False: A critical number must be in the domain of f. True.
Fill in the blank: If f' changes from negative to positive at x = c, then f has a local _____ at x = c. Answer: minimum.
True or False: The function f(x) = 4x³ - x⁴ has a local minimum at x = 0. False. f' does not change sign at x = 0.
Q: Find the critical points, intervals of increase/decrease, and local extrema for f(x) = x³ - 3x + 1.
A: f'(x) = 3x² - 3 = 3(x + 1)(x - 1). Critical points at x = -1 and x = 1. Increasing on (-∞, -1) and (1, ∞). Decreasing on (-1, 1). Local max at (-1, 3). Local min at (1, -1).
Q: Given f(x) = 3x² - 12x + 2, where is f increasing?
A: f'(x) = 6(x - 2). f'(x) > 0 when x > 2, so f is increasing on (2, ∞).
Q: If the graph of f'(x) is above the x-axis on (-∞, -3) and (-1, 2), what does this tell you about f?
A: f is increasing on (-∞, -3) and (-1, 2). f is decreasing wherever f'(x) is below the x-axis.
Q: A company's average cost is C̄(x) = 0.08x + 30 + 450/x. At what production level is average cost minimised?
A: Set C̄'(x) = 0.08 - 450/x² = 0, giving x = 75 jackets. The minimum average cost is $42 per jacket.
The first derivative analysis you learn here is half of the full graphing strategy. Part 3 adds the second derivative (concavity and inflection points), which completes the picture. Together, first and second derivative analysis let you sketch accurate graphs of any function without a graphing calculator. The applied problems (cost, revenue, profit) connect directly to business optimisation, which is the reason this course exists in a business calculus sequence.
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