MAC 2233 Calculus for Business | Sections 4.2, 4.4 | Source: Exam 3 Review
The second derivative tells you how the curve bends. Positive f''(x) means concave up (like a bowl), negative f''(x) means concave down (like a hill). Inflection points are where the concavity switches. Combined with first derivative analysis, this gives you a complete graphing strategy: domain, intercepts, asymptotes, increase/decrease, concavity, extrema, inflection points, and sketch.
Difficulty: Intermediate to Advanced
Prerequisites: All of Part 1 (differentiation techniques) and Part 2 (first derivative analysis, sign charts, critical points). You need to differentiate twice and manage sign charts for both f' and f''.
Part 2 answered "where is the function going up or down?" This section answers "how is it curving while it does so?" A function can be increasing and concave up (accelerating upward, like early-stage revenue growth) or increasing and concave down (slowing down, like diminishing returns). The second derivative captures that distinction. The graphing strategy problems (38 to 46) put everything together: you produce a complete analysis of a function and sketch its graph. These are the longest problems on the exam but follow a fixed checklist every time.
Second derivative (f''(x))
The derivative of f'(x). It measures the rate of change of the slope, which tells you how the curve bends.
In simple terms, if f'(x) tells you speed, f''(x) tells you acceleration.
Concave up
A curve is concave up on an interval where f''(x) > 0. The curve opens upward, like a bowl or a smile. Tangent lines lie below the curve.
Think of it as: the slope is getting steeper (more positive or less negative) as you move right.
Concave down
A curve is concave down on an interval where f''(x) < 0. The curve opens downward, like a frown or a hill. Tangent lines lie above the curve.
Think of it as: the slope is getting less steep (less positive or more negative) as you move right.
Inflection point
A point where the concavity changes, from up to down or from down to up. At an inflection point, f''(x) = 0 or f''(x) is undefined, and f'' changes sign.
In simple terms, the point where the curve switches from bending one way to bending the other.
Horizontal asymptote (H.A.)
A horizontal line y = L that the graph approaches as x goes to positive or negative infinity. Found by evaluating the limit of f(x) as x approaches ±∞.
In simple terms, the value the function settles toward for very large or very small x.
Vertical asymptote (V.A.)
A vertical line x = a where f(x) goes to positive or negative infinity. Occurs where the denominator is zero (and the numerator is not zero at that same point).
In simple terms, a value of x where the function blows up.
The rule
f''(x) > 0 on an interval means f is concave up there.
f''(x) < 0 on an interval means f is concave down there.
Finding concavity intervals
Find f''(x).
Set f''(x) = 0 and solve. Also note where f''(x) is undefined.
Build a sign chart for f''(x), just as you did for f'(x) in Part 2.
Positive intervals are concave up; negative intervals are concave down.
Example: f(x) = 3x⁴ - 18x² (Problem 29a)
f'(x) = 12x³ - 36x. f''(x) = 36x² - 36 = 36(x² - 1) = 36(x - 1)(x + 1).
f''(x) = 0 at x = -1 and x = 1.
Sign chart: f''(x) > 0 on (-∞, -1) and (1, ∞), f''(x) < 0 on (-1, 1).
Concave up on (-∞, -1) ∪ (1, ∞). Concave down on (-1, 1).
Example: f(x) = 3/(x - 1)² (Problem 29b)
f''(x) turns out to be positive on both (-∞, 1) and (1, ∞).
Concave up everywhere in the domain. No concave down intervals.
Although f''(x) is undefined at x = 1, f itself is also undefined there (vertical asymptote), so no inflection point.
Second derivative of a chain rule expression (Problem 27)
y = (2x - 1)⁴. First derivative: y' = 4(2x - 1)³ · 2 = 8(2x - 1)³. Second derivative: y'' = 8 · 3(2x - 1)² · 2 = 48(2x - 1)².
Definition recap
An inflection point is where f''(x) changes sign (and f is defined there). f''(x) = 0 or f''(x) undefined is necessary but not sufficient; you must confirm the sign actually changes.
Example: f(x) = 3x⁴ - 18x² (Problem 29a)
f''(x) = 36(x - 1)(x + 1). Changes sign at x = -1 and x = 1. f(-1) = 3 - 18 = -15. f(1) = 3 - 18 = -15. Inflection points: (-1, -15) and (1, -15).
Example: f(x) = (2x - 1)/(x + 5) (Problem 29c)
f''(x) is always positive on (-∞, -5) and always negative on (-5, ∞), or vice versa. The sign change happens at x = -5, but f is undefined there (vertical asymptote). So there is no inflection point.
Example: f(x) = 3x⁴ - 8x³ + 6x² + 3 (Problem 32)
f'(x) = 12x³ - 24x² + 12x = 12x(x - 1)². f''(x) = 36x² - 48x + 12 = 12(3x² - 4x + 1) = 12(3x - 1)(x - 1).
f''(x) = 0 at x = 1/3 and x = 1. Sign chart confirms f'' changes sign at both, so inflection points at (1/3, 3.4) and (1, 4).
Key point: if f'' = 0 at a point but does not change sign there, it is not an inflection point. Always verify with the sign chart.
Problems 38 to 43 use the same checklist. Memorise it.
The checklist
Domain: where is f defined? Exclude values that make a denominator zero.
x-intercept(s): set f(x) = 0 and solve.
y-intercept: evaluate f(0).
Horizontal asymptote(s): evaluate the limit of f(x) as x approaches ±∞. Compare the degrees of numerator and denominator.
Vertical asymptote(s): values of x where the denominator is zero (and the numerator is not zero there). If both are zero, check for a hole instead.
f'(x) and sign chart: find intervals of increase/decrease and local extrema.
f''(x) and sign chart: find intervals of concavity and inflection points.
Sketch: plot intercepts, asymptotes, extrema, and inflection points, then draw a smooth curve respecting all the information.
Example: f(x) = (x + 3)/(x - 3) (Problem 38)
Domain: all x except x = 3.
x-int: x = -3. y-int: y = -1.
H.A.: y = 1 (degrees are equal, ratio of leading coefficients = 1/1).
V.A.: x = 3.
f'(x) = -6/(x - 3)². Always negative, so f is always decreasing on its domain.
No local extrema.
f''(x) = 12/(x - 3)³. Positive when x > 3 (concave up), negative when x < 3 (concave down).
No inflection points (concavity changes at x = 3, but f is undefined there).
Example: f(x) = 1/(x² + 1) (Problem 40)
Domain: all real numbers (denominator is always positive).
No x-intercept. y-int: y = 1.
H.A.: y = 0. No V.A.
f'(x) = -2x/(x² + 1)². Positive when x < 0, negative when x > 0. Local max at x = 0.
Inflection points at x = ±√3/3 (approximately ±0.6).
Example: f(x) = (x - 1)/(x² - 1) (Problem 41)
Simplify: f(x) = 1/(x + 1) with a hole at (1, 1/2).
Domain: all x except x = -1 and x = 1.
V.A. at x = -1. Hole at x = 1.
H.A.: y = 0.
Always decreasing on each piece of the domain. No local extrema.
These problems give you sign charts for both f'(x) and f''(x), a table of points, and asymptote information, then ask you to sketch.
How to combine the information
f' positive + f'' positive = increasing and concave up (curving upward, accelerating).
f' positive + f'' negative = increasing and concave down (curving over a hill top, decelerating).
f' negative + f'' positive = decreasing and concave up (approaching a valley bottom).
f' negative + f'' negative = decreasing and concave down (falling and curving away).
Example: Problem 31
Given: f(-2) = -4, f(1) = 3, f(3) = 5. f'(-2) = 0 and f'(3) = 0. f' > 0 on (-2, 3). f' < 0 on (-∞, -2) ∪ (3, ∞). f''(1) = 0. f'' > 0 on (-∞, 1). f'' < 0 on (1, ∞).
Local min at (-2, -4). Local max at (3, 5). Inflection point at (1, 3). The curve falls to (-2, -4), then rises and is concave up until x = 1 where it switches to concave down, continues rising to (3, 5), then falls.
Example: Problem 37
Domain: all real x except x = -2. H.A.: y = 1. V.A.: x = -2. f' is positive on (-∞, -2) and (-2, 1), zero at x = 1, negative after. f'' changes sign at x = -1, x = 0, and x = 3. The sketch must show the vertical asymptote, the horizontal asymptote, a local max at (1, 3), and three inflection points.
Revenue and profit (Problems 33, 34)
Problem 33: demand function p = 2600 - 0.26x². Revenue R(x) = x · p(x) = 2600x - 0.26x³.
R'(x) = 2600 - 0.78x². Set to zero: x ≈ 58. Maximum revenue of about $100,071 at 58 laptops.
R''(x) = -1.56x. Since R''(x) < 0 for all x > 0, the revenue function is concave down on the entire domain (0, 100). No concave up interval.
Problem 34: profit P(x) = R(x) - C(x) = 2610x - 0.2x³ - 970 - 450x = 2160x - 0.2x³ - 970.
P'(x) = 2160 - 0.6x². Set to zero: x = 60. Maximum profit of $85,430 at 60 units.
P''(x) = -1.2x < 0 on (0, 110), so concave down throughout. No concave up interval.
Bacteria growth (Problem 35)
N(t) = 1000 + 30t² - t³. The question asks when the rate of growth N'(t) is increasing or decreasing, not when N(t) itself is.
N'(t) = 60t - 3t². N''(t) = 60 - 6t. Set N''(t) = 0: t = 10.
N''(t) > 0 on (0, 10), so N'(t) is increasing (rate of growth accelerates) on (0, 10).
N''(t) < 0 on (10, 20), so N'(t) is decreasing (rate of growth decelerates) on (10, 20).
Inflection point at (10, 3000). This is where the population growth switches from speeding up to slowing down.
Drug concentration (Problem 46)
C(t) = 0.5t/(t² + 4). C'(t) = 0.5(4 - t²)/(t² + 4)². Set to zero: t = 2.
Increasing on (0, 2), decreasing on (2, 8). Max concentration at t = 2 hours: C(2) = 1/8 = 0.125 ppm.
C(6) = 3/40 = 0.075 ppm.
Dive watch production (Problem 44)
N(t) = 100t/(t + 9). Always increasing on (0, ∞), always concave down. H.A.: y = 100 (the employee approaches but never reaches 100 watches per day). A classic diminishing-returns model.
Students confuse concavity with increase/decrease. A function can be increasing and concave down at the same time (it is rising but the rate of rise is slowing). These are independent properties.
Students assume f''(x) = 0 always produces an inflection point. You must verify that f'' changes sign. For example, f(x) = x⁴ has f''(0) = 0 but no inflection point, because f'' is positive on both sides of x = 0.
When reading a graph of f(x) to determine f''(x), students look for where the curve crosses the x-axis. That tells you about f(x) = 0, not f''(x). To read f''(x) from a graph of f(x), look at the curvature: bowl-shaped regions have f'' > 0, hill-shaped regions have f'' < 0.
In the graphing strategy, students often forget to check for holes. When both numerator and denominator are zero at the same x-value, you get a hole, not a vertical asymptote. Problem 41 tests exactly this: (x - 1)/(x² - 1) simplifies to 1/(x + 1) with a hole at x = 1.
⚠️ The graphing strategy (Problems 38 to 43) is the most time-consuming problem type on the exam. Practice the full checklist until it is automatic. The answer key shows exactly what is expected for each function.
⚠️ Problem 28 tests your ability to read f''(x) from the graph of f(x). Know the visual cues: concave up looks like a bowl, concave down looks like a hill.
⚠️ Applied problems (33 to 35, 44 to 46) wrap the same calculus in a word problem. The exam will likely include one or two. Read carefully to determine whether the question asks about f, f', or f'': Problem 35 asks about N'(t), not N(t).
⚠️ Problem 32 is a full analysis problem (increase/decrease, concavity, extrema, inflection points) for a polynomial. This is a high-probability exam question.
True or False: If f''(x) > 0, then f(x) is increasing. False. f''(x) > 0 means f is concave up, not necessarily increasing.
Fill in the blank: An inflection point occurs where f''(x) changes _____. Answer: sign.
True or False: Every rational function has a vertical asymptote. False. For example, f(x) = 1/(x² + 1) has no vertical asymptote because x² + 1 is never zero.
Fill in the blank: To find the horizontal asymptote of a rational function, evaluate the limit as x approaches _____. Answer: ±∞ (positive or negative infinity).
True or False: If a revenue function is concave down on its entire domain, there is no maximum revenue. False. A concave-down function with a critical point has a local (and often absolute) maximum there.
Q: Determine the intervals of concavity and inflection points for f(x) = 3x⁴ - 18x².
A: f''(x) = 36(x - 1)(x + 1). Concave up on (-∞, -1) ∪ (1, ∞). Concave down on (-1, 1). Inflection points at (-1, -15) and (1, -15).
Q: For f(x) = (x + 3)/(x - 3), state the domain, asymptotes, and intervals of increase/decrease.
A: Domain: (-∞, 3) ∪ (3, ∞). H.A.: y = 1. V.A.: x = 3. Decreasing on (-∞, 3) and (3, ∞). No local extrema.
Q: A drug concentration is C(t) = 0.5t/(t² + 4). When is the concentration at its maximum, and what is that maximum?
A: C'(t) = 0 at t = 2. Max concentration at 2 hours: C(2) = 0.125 ppm.
Q: For N(t) = 1000 + 30t² - t³, when is the rate of growth N'(t) increasing?
A: N''(t) = 60 - 6t > 0 when t < 10. So N'(t) is increasing on (0, 10).
Q: Find the second derivative of y = (2x - 1)⁴.
A: y'' = 48(2x - 1)². Apply the chain rule twice: first derivative is 8(2x - 1)³, second derivative is 48(2x - 1)².
Concavity and the second derivative connect directly to optimisation. In business calculus, the second derivative test is a quick way to confirm whether a critical point is a max or min: if f''(c) < 0, it is a local max; if f''(c) > 0, it is a local min. The graphing strategy ties together every skill from the entire course: algebra (factoring, simplifying), limits (asymptotes), first derivatives (increase/decrease), and second derivatives (concavity). If you can complete a full graphing strategy problem, you have demonstrated mastery of the course material up to this point.
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