AP Calculus AB: Applications of Differential Calculus
This is where derivatives start doing real work: describing how a function behaves, finding its extreme values, and modeling change in the physical world. This free, complete lesson covers slope and critical points, increasing and decreasing functions, concavity and inflection points, relating f, f', and f'' graphically, global extrema (the Closed Interval Test), optimization, motion along a line, local linear approximation, and related rates. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.
1. Slope, Critical Points, and Tangent Lines
The slope of a curve y = f(x) at a point is f'(x) there. A critical point is any c in the domain of f where either f'(c) = 0 or f'(c) is undefined. The equation of the tangent line at (x₁, y₁) is y − y₁ = f'(x₁)(x − x₁).
Worked Example: For f(x) = 2x³ − 3x² − 12x + 5, find the critical points.
f'(x) = 6x² − 6x − 12 = 6(x² − x − 2) = 6(x − 2)(x + 1)
Setting f'(x) = 0: x = 2 and x = −1.
1. For f(x) = x³ − 6x² + 9x, find the critical points.
Explanation: f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), which is 0 at x = 1 and x = 3.
2. Find an equation of the tangent to f(x) = x³ − 2x² at the point (2, 0).
Explanation: f'(x) = 3x² − 4x, so f'(2) = 12 − 8 = 4. Tangent: y − 0 = 4(x − 2), or y = 4x − 8.
3. For g(x) = (x − 3)2/3, is x = 3 a critical point? Why?
Explanation: g'(x) = 23(x − 3)1/3, which is undefined at x = 3 even though g(3) = 0 is defined. A critical point can arise from an undefined derivative, not just a zero one.
4. If the tangent to a curve at x = 5 is horizontal, what must be true?
Explanation: A horizontal tangent line has slope 0, so f'(5) = 0.
2. Increasing and Decreasing Functions
f is increasing on an interval if f(a) < f(b) whenever a < b there; it's decreasing if f(a) > f(b). Where f'(x) ≥ 0, f is increasing; where f'(x) ≤ 0, f is decreasing. The First Derivative Test: if f' changes from + to − at a critical point c, f has a local maximum there; if it changes from − to +, a local minimum; if it doesn't change sign, neither.
Worked Example: For f(x) = x³ − 3x², determine where f is increasing and decreasing.
f'(x) = 3x² − 6x = 3x(x − 2). Critical points: 0, 2.
Testing each interval: f is increasing for x ≤ 0 and x ≥ 2, decreasing for 0 ≤ x ≤ 2.
1. For f(x) = x³ − 12x, on which interval is f decreasing?
Explanation: f'(x) = 3x² − 12 = 3(x − 2)(x + 2). Testing x = 0: f'(0) = −12 < 0, so f decreases on (−2, 2).
2. A function f has f'(x) > 0 for x < 3 and f'(x) < 0 for x > 3, with f'(3) = 0. What does the First Derivative Test conclude at x = 3?
Explanation: A sign change from + to − at a critical point signals a local maximum.
3. For h(x) = 1x − 2, determine the sign of h'(x) for all x ≠ 2.
Explanation: h'(x) = −1(x − 2)². Since (x − 2)² is always positive for x ≠ 2, h'(x) is always negative.
4. If f'(x) ≥ 0 on an interval and f'(x) = 0 only at an isolated point c within it, can c still be included in the interval where f is called increasing?
Explanation: The definition of increasing only requires f(a) < f(b) for a < b in the interval, a momentary flat tangent at one point doesn't violate this.
3. Concavity, Inflection Points, and the Second Derivative Test
A graph is concave upward where f''(x) > 0, and concave downward where f''(x) < 0. A point of inflection is where concavity changes. The Second Derivative Test: at a critical point c where f'(c) = 0, if f''(c) > 0, c yields a local minimum; if f''(c) < 0, a local maximum; if f''(c) = 0, the test fails and the First Derivative Test must be used instead.
Worked Example: For f(x) = x³ − 6x² + 9x + 1, find any local extrema using the Second Derivative Test.
f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). Critical points: 1, 3.
f''(x) = 6x − 12. f''(1) = −6 < 0 (local max). f''(3) = 6 > 0 (local min).
1. For f(x) = x³ − 3x² − 9x + 2, use the Second Derivative Test to classify the critical points.
Explanation: f'(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1); critical points −1, 3. f''(x) = 6x − 6. f''(−1) = −12 < 0 (max); f''(3) = 12 > 0 (min).
2. For f(x) = x⁴ − 4x³, the Second Derivative Test fails at which critical point?
Explanation: f'(x) = 4x³ − 12x² = 4x²(x − 3); critical points 0, 3. f''(x) = 12x² − 24x. f''(0) = 0 (test fails); f''(3) = 36 > 0 (test succeeds, local min).
3. Continuing the previous problem, is x = 0 a local max, local min, or neither for f(x) = x⁴ − 4x³?
Explanation: f'(x) = 4x²(x − 3). The factor x² is non-negative on both sides of 0, so the sign of f' is controlled entirely by (x − 3), which stays negative on both sides near x = 0. Since f' doesn't change sign, x = 0 yields neither a max nor a min.
4. Find any inflection points of f(x) = x³ − 6x² + 9x + 1.
Explanation: f''(x) = 6x − 12, which is 0 at x = 2 and changes sign (− to +) there, confirming an inflection point.
5. A curve is concave upward on an interval. What does this tell us about f''(x) there?
Explanation: Concave upward is defined by a positive second derivative.
4. Relating f, f', and f'' Graphically
Reading one graph off another is a core AP skill. This table summarizes the key relationships:
| At c or on an interval | What it tells you |
|---|---|
| f'(c) = 0, sign + to − | f has a local maximum at c |
| f'(c) = 0, sign − to + | f has a local minimum at c |
| f'(c) = 0, no sign change | Neither a max nor a min |
| f' is a local min (decreasing to increasing) | f has an inflection point at c |
| f' is a local max (increasing to decreasing) | f has an inflection point at c |
| f' increasing on an interval | f is concave upward there |
| f' decreasing on an interval | f is concave downward there |
Worked Example: If f'(c) = 0 and f' changes from positive to negative as x increases through c, what does this indicate about f at c?
f has a local maximum at c.
1. If f'(c) is a local minimum of f' (that is, f' changes from decreasing to increasing at c), what does this indicate about f at c?
Explanation: When f' itself has a local extremum, the concavity of f is switching, this is precisely a point of inflection of f, not an extremum of f.
2. If f' does not change sign as x increases through a critical point c, what can we conclude about f at c?
Explanation: Without a sign change, f keeps moving in the same direction through c, ruling out both types of local extremum.
3. On an interval where f' is increasing, what must be true of the graph of f?
Explanation: f' increasing means f'' > 0, which is exactly the definition of concave upward.
4. At a point of inflection where f'(c) is a local maximum of f', how does f'' behave as x increases through c?
Explanation: Since f' switches from increasing to decreasing at its own local maximum, f'' (the derivative of f') switches from positive to negative there.
5. Global Extrema: The Closed Interval Test
If f is continuous on [a, b], the Closed Interval Test (Candidates Test) finds global extrema by evaluating f at every critical point in [a, b] and at both endpoints, then comparing the results. The largest value found is the global maximum; the smallest is the global minimum.
Worked Example: Find the global max and global min of f(x) = x³ − 3x on [−2, 3].
f'(x) = 3x² − 3 = 3(x − 1)(x + 1); critical points −1 and 1 (both in range).
f(−2) = −2, f(−1) = 2, f(1) = −2, f(3) = 18
Global max: 18 at x = 3. Global min: −2 (tied at x = −2 and x = 1).
1. Find the global maximum of f(x) = x³ − 12x on [−3, 3].
Explanation: Critical points at ±2. f(−3) = 9, f(−2) = 16, f(2) = −16, f(3) = −9. The largest is 16, at x = −2.
2. Using the same function on the same interval, find the global minimum.
Explanation: From the same four candidate values, −16 (at x = 2) is the smallest.
3. For f(x) = x² − 4x + 1 on [0, 5], the Closed Interval Test requires evaluating f at which x-values?
Explanation: f'(x) = 2x − 4 = 0 gives the critical point x = 2, which falls in [0, 5]. Combined with both endpoints: x = 0, 2, 5.
4. Can a global maximum occur at an endpoint of a closed interval, even if it isn't a critical point?
Explanation: This is exactly why the Closed Interval Test includes endpoints as candidates, a global extremum can occur there even without a horizontal tangent.
6. Optimization
To maximize or minimize a quantity y in terms of x: express y as a function of a single variable, find critical points using y', classify them, and check endpoints if the domain is restricted.
Worked Example: A farmer wants to fence a rectangular field using 200 feet of fencing, with one side of a barn serving as one side of the field (no fencing needed there). If x is the width and y is the length, find the value of x that maximizes the area, given 2x + y = 200.
y = 200 − 2x, so A(x) = x(200 − 2x) = 200x − 2x².
A'(x) = 200 − 4x = 0, so x = 50
1. A rectangular box with a square base has volume 32 cubic feet, with an open top. If x is the side of the square base and h is the height (so x²h = 32), express the surface area S in terms of x alone.
Explanation: h = 32/x². S = x² + 4xh = x² + 4x(32/x²) = x² + 128/x.
2. Using S(x) = x² + 128/x, find the value of x that minimizes the surface area.
Explanation: S'(x) = 2x − 128/x² = 0, so 2x³ = 128, giving x³ = 64, so x = 4.
3. Using the farmer's fencing problem (A(x) = 200x − 2x²) from the worked example, what is the maximum area?
Explanation: A(50) = 200(50) − 2(50)² = 10,000 − 5,000 = 5,000 square feet.
4. In an optimization problem where the domain of the quantity being optimized is a restricted closed interval, what must be checked in addition to critical points?
Explanation: Just as with the Closed Interval Test, a restricted domain means the true optimum could occur at a boundary rather than at an interior critical point.
7. Motion Along a Line
If a particle's position is s(t), its velocity is v(t) = s'(t) and its acceleration is a(t) = v'(t) = s''(t). The particle's speed is |v(t)|. Speed is increasing when v and a share the same sign, and decreasing when they have opposite signs.
Worked Example: A particle moves with position s(t) = t³ − 6t² + 9t, t ≥ 0. Find the velocity and determine when the particle moves right.
v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3). v = 0 at t = 1, 3.
v > 0 (moving right) for t < 1 and t > 3; v < 0 (moving left) for 1 < t < 3
1. A particle's position is s(t) = t³ − 9t² + 24t. Find the velocity function v(t).
Explanation: v(t) = s'(t) = 3t² − 18t + 24.
2. Using v(t) = 3t² − 18t + 24 = 3(t − 2)(t − 4), for what t ≥ 0 is the particle moving to the right?
Explanation: v > 0 outside the roots 2 and 4, so the particle moves right for 0 ≤ t < 2 and for t > 4.
3. A particle has position s(t) = t² − 6t + 5, so v(t) = 2t − 6 and a(t) = 2 (constant). For what t is the particle's speed increasing?
Explanation: Speed increases when v and a share the same sign. Since a is always positive, speed increases wherever v > 0, that is, t > 3.
4. A particle moves with position s(t) = t³ − 3t. What is its acceleration at t = 2?
Explanation: v(t) = 3t² − 3, so a(t) = 6t. a(2) = 12.
5. If a particle's velocity and acceleration have opposite signs at some instant, what is happening to its speed?
Explanation: Opposite signs of v and a mean the particle is decelerating, its speed is decreasing.
8. Local Linear Approximation
For values of x close to a, the tangent-line (local linear) approximation is:
f(x) ≈ f(a) + f'(a)(x − a)
Worked Example: Find the tangent-line approximation for f(x) = √x at a = 9, and use it to estimate √9.3.
f(9) = 3. f'(x) = 1/(2√x), so f'(9) = 1/6.
f(x) ≈ 3 + 16(x − 9). At x = 9.3: 3 + 16(0.3) = 3.05
1. Find the tangent-line approximation for f(x) = x³ at a = 2.
Explanation: f(2) = 8. f'(x) = 3x², so f'(2) = 12. Approximation: 8 + 12(x − 2).
2. Using f(x) ≈ 8 + 12(x − 2) for f(x) = x³, estimate the value of (2.1)³.
Explanation: 8 + 12(0.1) = 8 + 1.2 = 9.2. (The true value, 9.261, is close, confirming the approximation is reasonable for x near a.)
3. Using the local linearization (1 + x)k ≈ 1 + kx for x near 0, estimate (1.02)⁵.
Explanation: With k = 5 and x = 0.02: 1 + 5(0.02) = 1.1.
4. If a function f is concave up on an interval containing a, is its tangent-line approximation at a an overestimate or underestimate of f(x) for x near a?
Explanation: A concave-up curve always lies above its tangent lines, so the tangent-line approximation underestimates the true value.
Common Mistakes to Avoid
- Forgetting that a critical point can come from an undefined derivative, not just f'(x) = 0. Always check for x-values where f' fails to exist, not only where it equals zero.
- Confusing a local extremum with a global extremum. A local max or min only needs to beat its immediate neighbors; a global extremum must beat every value in the entire interval, which is why the Closed Interval Test also checks the endpoints.
- Mixing up which sign change of f' signals a max versus a min. Plus-to-minus is a local maximum; minus-to-plus is a local minimum, it's easy to flip these under time pressure.
- Treating "speed is increasing" and "velocity is increasing" as the same thing. Speed increases only when velocity and acceleration share the same sign; velocity alone increasing whenever acceleration is positive says nothing about speed if velocity is negative.
- Substituting known numerical values before differentiating in a related rates problem. Every variable that changes with time must stay a variable until after you differentiate, only then do you plug in the specific values for that instant.
Frequently Asked Questions
When should I use the First Derivative Test instead of the Second Derivative Test?
Use the Second Derivative Test as a quick first check, it's usually faster. But whenever f''(c) = 0 at a critical point, the test is inconclusive, and you must fall back on the First Derivative Test's sign analysis instead.
How is an inflection point different from a local extremum?
A local extremum is where f itself changes from increasing to decreasing (or vice versa), found from the sign of f'. An inflection point is where the concavity of f changes, found from the sign of f''. They describe different features of the same curve.
What's the general strategy for setting up an optimization or related rates word problem?
Draw a diagram if possible, identify which quantity is changing and which equation relates the variables involved, then differentiate (with respect to the input variable for optimization, or with respect to time for related rates) before substituting any given numerical values.
Where can I practice more problems like these?
The Applications of Differential Calculus quizzes in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.