Difficulty: Intermediate to Advanced | Prerequisites: Derivative techniques, limits, continuity
Once you can differentiate, the exam shifts to asking what the derivative tells you. This is where the AP exam gets heavier: tangent lines, increasing/decreasing analysis, concavity, inflection points, particle motion, related rates, and the Mean Value Theorem all live here. These topics make up the largest share of both multiple-choice and free-response questions. The underlying logic is always the same: the first derivative encodes slope and velocity; the second derivative encodes concavity and acceleration.
If f' > 0, then f is increasing. If f'' > 0, then f is concave up. A sign change in f' gives a local extremum. A sign change in f'' gives an inflection point. Tangent lines use point-slope form with f'(a) as the slope. Particle motion translates these ideas into position, velocity, and acceleration.
Tangent line
The line that just touches a curve at a point, with slope equal to the derivative at that point. Its equation is y – f(a) = f'(a)(x – a).
Increasing/decreasing
A function is increasing on an interval where f'(x) > 0 and decreasing where f'(x) < 0. Think of it as: positive slope means going uphill.
Concavity
Describes the curvature of a graph. Concave up (f'' > 0) bends upward like a bowl; concave down (f'' < 0) bends downward like a cap.
Inflection point
A point where the concavity changes. This requires f'' to change sign at that point (not merely equal zero).
Critical point
A value of x where f'(x) = 0 or f'(x) is undefined (and x is in the domain of f). Candidates for local maxima and minima.
Mean Value Theorem (MVT)
If f is continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) where f'(c) = [f(b) – f(a)] / (b – a). In simple terms, at some point the instantaneous rate of change equals the average rate of change over the interval.
Particle at rest
A particle is at rest when its velocity is zero, i.e. v(t) = x'(t) = 0.
Speed
The absolute value of velocity, |v(t)|. Speed is always non-negative.
The equation of the tangent line to f at x = a is: y – f(a) = f'(a)(x – a).
Worked example (from Q19): f(x) = x³ – 3x + 5 at x = 2.
f(2) = 8 – 6 + 5 = 7
f'(x) = 3x² – 3, so f'(2) = 12 – 3 = 9
Tangent line: y – 7 = 9(x – 2), which simplifies to y = 9x – 11
The answer is C) y = 9x – 11.
Worked example (from Q3): A line through (2, –2) is tangent to f at (–2, 6). What is f'(–2)?
The slope of the tangent line equals the slope between the two given points:
f'(–2) = (6 – (–2)) / (–2 – 2) = 8 / (–4) = –2
The answer is B) –2.
f(3) = 6, f'(3) = 4, g(x) = f(x)/x. Find the tangent line to g at x = 3.
g(3) = f(3)/3 = 6/3 = 2
g'(x) = [f'(x) · x – f(x)] / x² (quotient rule)
g'(3) = [4 · 3 – 6] / 9 = 6/9 = 2/3
Tangent: y – 2 = (2/3)(x – 3)
The answer is D) y – 2 = (2/3)(x – 3).
Worked example (from Q7): A table of f'(x) values is given, and f' is continuous with exactly two real zeros.
The data shows f'(x) is positive for x < 1, f'(1) = 0, f'(x) is negative for 1 < x < 5, f'(5) = 0, and f'(x) is positive for x > 5.
f is increasing where f' > 0, so f is increasing on x < 1 or x > 5.
The answer is B) x < 1 or x > 5.
Worked example (from Q10): g'(x) = 4/x² – x. When is g increasing?
Set g'(x) > 0: 4/x² – x > 0, so 4/x² > x, which gives 4 > x³ (for x > 0), meaning x < ∛4.
For x < 0, the term 4/x² is positive and –x is also positive, so g'(x) > 0 for all x < 0.
Combining: g is increasing on (–∞, 0) ∪ (0, ∛4), which simplifies to (–∞, ∛4) excluding x = 0 (where g' is undefined). Among the choices, the answer is A) (–∞, ∛4).
Worked example (from Q5): Where is f(x) = xe²ˣ concave up?
f'(x) = e²ˣ + 2xe²ˣ = e²ˣ(1 + 2x)
f''(x) = 2e²ˣ(1 + 2x) + 2e²ˣ = 2e²ˣ(2 + 2x) = 4e²ˣ(1 + x)
Since e²ˣ > 0 always, f'' > 0 when 1 + x > 0, i.e. x > –1
The answer is A) x > –1.
An inflection point occurs where f'' changes sign.
Worked example (from Q13): f''(x) = (x – a)²(x – d).
At x = a, f'' touches zero but does not change sign (even multiplicity). No inflection.
At x = d, f'' crosses zero and changes sign (odd multiplicity). Inflection point.
Points b and c (between a and d on the graph) are not roots of f''.
The answer is E) d only.
Worked example (from Q21): x(t) = t³ – 12t² + 36t – 15. When is the particle at rest?
v(t) = x'(t) = 3t² – 24t + 36 = 3(t² – 8t + 12) = 3(t – 2)(t – 6)
v(t) = 0 when t = 2 or t = 6. The particle is at rest at t = 2 and t = 6.
The answer is B) t = 2 and t = 6.
Worked example (from Q22): Given the graph of f', determine which statement is true.
f' > 0 from x = –3 to x = –1 (f' is positive), so f is increasing on that interval.
f' changes from positive to negative at x = –1, giving f a local maximum there, not a minimum.
f' < 0 from x = –1 to x = 1, so f is decreasing (not increasing) there.
The answer is D) f is increasing from x = –3 to x = –1.
Q27: f'(x) > 0 (always increasing), f''(x) < 0 (concave down). f(2) = 5, f(3) = 9.
Since f is concave down and increasing, the rate of increase is slowing. The secant slope from 2 to 3 is 4, so the slope from 3 to 4 must be less than 4. Thus f(4) < 9 + 4 = 13. Also, f is increasing so f(4) > 9. The possible value is C) 12.
Q28: f(x) > 0, f'(x) > 0, f''(x) > 0 everywhere. The graph must be positive, increasing, and concave up. The answer is D), the graph that rises and bends upward while staying above the x-axis.
A circle's radius increases at 0.3 m/s. Find dA/dt when the circumference is 6π metres.
Circumference = 2πr = 6π, so r = 3 m
A = πr², so dA/dt = 2πr(dr/dt) = 2π(3)(0.3) = 1.8π m²/s
The answer is C) 1.8π m²/s.
f is continuous on [–1, 3], differentiable on (–1, 3), f(–1) = 1, f(3) = 3.
The MVT guarantees some c in (–1, 3) with f'(c) = (3 – 1) / (3 – (–1)) = 2/4 = 1/2.
The answer is E) There exists c in (–1, 3) such that f'(c) = 1/2.
Tangent line: y – f(a) = f'(a)(x – a)
Slope from a differential equation (Q9): if f'(x) = 4x – 4, then f(x) = 2x² – 4x + C. Use the given point to find C.
Related rates: differentiate the geometric formula with respect to t, substitute known values, solve for the unknown rate.
MVT: f'(c) = [f(b) – f(a)] / (b – a)
Tangent lines approximate function values locally (linearisation). Increasing/decreasing analysis tells engineers where a system output is rising or falling. Concavity indicates whether growth is accelerating or decelerating, which matters in everything from population models to investment returns. Related rates let you connect the rates of change of two linked quantities: how fast a shadow lengthens, how fast a water level rises given an inflow rate.
Students assume f'(x) = 0 always means a local extremum. It does not; f' must change sign. A point where f' = 0 but does not change sign (e.g. f(x) = x³ at the origin) is not an extremum.
An inflection point requires f'' to change sign, not merely equal zero. If f'' = 0 but does not change sign (e.g. f''(x) = (x – a)²), there is no inflection at x = a.
Speed and velocity are different. Speed = |velocity|. A particle with negative velocity that becomes less negative has increasing velocity but decreasing speed.
The MVT guarantees at least one c, not that you can find it. On the exam, the question usually asks what must be true, not what the value of c is.
⚠️ Tangent line equations appear in both MC and FRQ. Always check: did you find f(a) for the y-value and f'(a) for the slope?
⚠️ The distinction between speed and velocity is tested almost every year. If velocity is negative and becoming less negative, speed is decreasing.
⚠️ Inflection point questions require a sign change in f''. Zero alone is not enough.
⚠️ The MVT shows up as "which of the following must be true." The guaranteed statement is about the existence of c, not a specific value.
True or false: If f'(3) = 0 and f''(3) < 0, then f has a local maximum at x = 3.
Fill in the blank: A particle is at rest when ______ = 0.
True or false: If f'' changes from negative to positive at x = 2, there is an inflection point at x = 2.
Fill in the blank: The MVT guarantees f'(c) = ______ for some c in (a, b).
True or false: Speed can be negative.
Answers: 1. True (second derivative test). 2. v(t) or x'(t). 3. True. 4. [f(b) – f(a)] / (b – a). 5. False; speed = |velocity| ≥ 0.
Q: f(x) = x³ – 3x + 5. What is the equation of the tangent line at x = 2?
A: y = 9x – 11. The point is (2, 7), the slope is f'(2) = 9.
Q: x(t) = t³ – 12t² + 36t – 15. At what times is the particle at rest?
A: t = 2 and t = 6. Set v(t) = 3(t – 2)(t – 6) = 0.
Q: Where is f(x) = xe²ˣ concave up?
A: x > –1. The second derivative is 4e²ˣ(1 + x), which is positive when x > –1.
Q: A circle's radius increases at 0.3 m/s. What is dA/dt when the circumference is 6π?
A: 1.8π m²/s. Since r = 3, dA/dt = 2πr(dr/dt) = 2π(3)(0.3).
Q: f'(x) > 0 and f''(x) < 0 everywhere. f(2) = 5, f(3) = 9. What is a possible value for f(4)?
A: 12. The function is increasing but concave down, so the rate of increase slows. f(4) must be between 9 and 13.
Applications of derivatives connect backwards to the derivative rules (you need to compute f' and f'' accurately) and forwards to applications of integrals (recovering position from velocity is integration). Related rates problems are fundamentally implicit differentiation with respect to time. The MVT also serves as the theoretical backbone for the FTC and for error bounds in numerical approximations.
tangent line, increasing decreasing intervals, concavity, inflection point, critical point, particle motion, velocity, acceleration, speed, related rates, Mean Value Theorem, MVT, second derivative test, AP Calculus AB applications