Difficulty: Intermediate | Prerequisites: Limits, continuity, derivative rules
Once you know how to take derivatives, the next question is: what can you do with them? This section covers the main applications: analysing motion (velocity and acceleration), finding where a function is increasing or decreasing, locating maxima and minima, determining concavity and inflection points, and solving optimisation problems. These are the topics that tie the mechanical derivative rules to real problems.
Velocity, v(t)
The first derivative of the position function s(t). It tells you how fast the object is moving and in which direction (positive = forward/right, negative = backward/left).
Acceleration, a(t)
The second derivative of position, or equivalently the first derivative of velocity. It tells you how the velocity is changing.
Speed
The absolute value of velocity, |v(t)|. Speed is always non-negative and ignores direction.
Critical number
A value of x where f'(x) = 0 or f'(x) is undefined. These are the only candidates for local maxima and minima.
Increasing / decreasing
A function is increasing on an interval where f'(x) > 0 and decreasing where f'(x) < 0.
Local maximum / local minimum
A local max occurs where f' changes from positive to negative. A local min occurs where f' changes from negative to positive.
Concave up / concave down
A function is concave up where f''(x) > 0 (the curve opens upward, like a bowl). It is concave down where f''(x) < 0 (opens downward, like a hill).
Inflection point
A point where the concavity changes, i.e. f'' changes sign. The function switches from curving one way to curving the other.
Absolute (global) extrema
The largest and smallest values of a function on a closed interval. Found by checking critical numbers inside the interval and the two endpoints.
Position s(t), velocity v(t) = s'(t), acceleration a(t) = v'(t) = s''(t). These three are always connected by differentiation.
s(t) = 2t³ − 3t² − 6t − 1.
v(t) = 6t² − 6t − 6 m/s.
a(t) = 12t − 6 m/s².
Set v(t) = 0 and solve. For v(t) = 6t² − 6t − 6 = 0, i.e. t² − t − 1 = 0, the positive solution is t ≈ 1.618 seconds.
v(t) > 0. For this example, v(t) > 0 on (1.618, ∞).
The review includes a velocity vs. time graph for a turtle. Key readings:
Moving forward: v(t) > 0. In the example, the turtle never moves forward (velocity is never positive).
Moving backward: v(t) < 0. The turtle moves backward on (0, 2) and (3, 4).
Speeding up: velocity and acceleration have the same sign. On (3, 4), both v and a are negative, so the turtle is speeding up.
Slowing down: velocity and acceleration have opposite signs. On (0, 2), v < 0 but a > 0 (the slope of the v-graph is positive), so the turtle is slowing down.
Acceleration = 0: the slope of the velocity graph is zero, on the interval (2, 3).
Greatest speed: the largest |v(t)|. Here |v(t)| = 15.
Standing still for more than an instant: where v(t) = 0 on an interval, not just a single point. Here, (2, 3).
Changing direction: where v(t) changes sign. Here, it never does.
s(t) = (10/3)t³ + (1/2)t² − 2t. Then v(t) = 10t² + t − 2 = 0 at t = 0.4 seconds. a(t) = 20t + 1, so a(0.4) = 9 m/s².
Find f'(x).
Set f'(x) = 0 and solve for critical numbers.
Build a sign chart: test a value in each interval between critical numbers to determine whether f' is positive (increasing) or negative (decreasing).
Where f' changes from + to −, there is a local maximum. Where f' changes from − to +, there is a local minimum.
f(x) = 2x⁴ − 3x³ − 2x² − 4.
f'(x) = 8x³ − 9x² − 4x = x(8x² − 9x − 4) = 0.
Critical numbers: x = 0, x ≈ 1.466, x ≈ −0.341.
Sign chart gives:
Increasing on (−0.341, 0) and (1.466, ∞).
Decreasing on (−∞, −0.341) and (0, 1.466).
Local max at (0, −4).
Local min at (−0.341, −4.087) and (1.466, −8.513).
f(x) = (x − 6)x^(1/4). (Problem 56 from the review.)
f'(x) = x^(1/4) + (x − 6) · (1/4)x^(−3/4) = 0.
Solving gives critical numbers x = 6/5 and x = 0.
Sign chart gives:
Increasing on (6/5, ∞).
Decreasing on (0, 6/5).
Local min at (6/5, −5.024).
Local max at (0, 0).
If you are given the graph of f' (not f), remember:
f is increasing where f' > 0 (the f'-graph is above the x-axis).
f is decreasing where f' < 0 (the f'-graph is below the x-axis).
From the review (problem 50): f' > 0 on (−2, 0) and (2, ∞), so f is increasing there. f' < 0 on (−∞, −2) and (0, 2), so f is decreasing there.
Find f''(x) and set it equal to zero.
Test the sign of f'' in each interval.
f''(x) > 0: concave up.
f''(x) < 0: concave down.
An inflection point occurs where f'' changes sign. Just because f'' = 0 at a point does not guarantee an inflection point; the sign must change on either side.
f(x) = 2x⁴ − 3x³ − 2x² − 4.
f''(x) = 24x² − 18x − 4 = 0. Solve: 2(12x² − 9x − 2) = 0, giving x ≈ −0.179 and x ≈ 0.929.
Sign chart for f'':
Concave up on (−∞, −0.179) and (0.929, ∞).
Concave down on (−0.179, 0.929).
Inflection points at (−0.179, −4.045) and (0.929, −6.642).
In motion problems, the second derivative of position gives acceleration. A velocity function v(t) = t³eᵗ yields a(t) = 3t²eᵗ + t³eᵗ. At t = 4: a(4) = 3(16)e⁴ + 64e⁴ = 112e⁴.
If dy/dx = (3x + y)/(x − 3y) and you need d²y/dx², differentiate dy/dx with respect to x using the quotient rule, substituting dy/dx wherever it appears. From the review (problem 53): at (2, 0), dy/dx = 3, and d²y/dx² = 15.
Find f'(x) and set it to zero to get critical numbers.
Evaluate f at each critical number that lies inside the interval.
Evaluate f at both endpoints.
The largest value is the absolute maximum; the smallest is the absolute minimum.
y = −3x³ + 2x − 1 on [−1, 3].
y' = −9x² + 2 = 0, so x² = 2/9, giving x = ±√(2/9) ≈ ±0.471.
Evaluate:
f(−1) = 0.
f(−√(2/9)) ≈ −1.629.
f(√(2/9)) ≈ −0.371.
f(3) = −76.
Absolute maximum: (−1, 0). Absolute minimum: (3, −76).
The key insight is that local extrema found by setting f' = 0 are not necessarily absolute extrema. On a closed interval, you must also check the endpoints. The absolute max or min can occur at an endpoint where f' is not zero.
Identify the quantity to maximise or minimise. Write it as a function of one variable.
If there are two variables, use a constraint equation to eliminate one.
Differentiate and set equal to zero.
Verify the result is a max or min (using a sign chart or the second derivative test).
The difference of two numbers is 30. Maximise the product of one number and the square of the other.
Let the two numbers be x and y with x − y = 30, so x = 30 + y.
Product: P = xy² = (30 + y)y².
P' = 60y + 3y² = 3y(20 + y) = 0.
y = 0 or y = −20. Since we want a maximum, test both.
At y = −20: x = 10, P = 10 · 400 = 4000.
A sign chart confirms y = −20 gives a maximum. The maximum product is 4000.
f(x) = 3x/(x + 4). Where does the tangent have slope 1/3?
f'(x) = 12/(x + 4)². Set 12/(x + 4)² = 1/3, so (x + 4)² = 36, giving x + 4 = ±6, hence x = 2 or x = −10.
The points are (2, 1) and (−10, 5).
Students confuse "speeding up" with "positive velocity." Speeding up means |v(t)| is increasing, which happens when velocity and acceleration have the same sign, even if both are negative.
Students forget to check endpoints when finding absolute extrema on a closed interval. The absolute max or min can easily be at an endpoint, not at a critical number.
Students assume f''(x) = 0 guarantees an inflection point. It does not. You must verify that f'' changes sign at that point.
In optimisation, students sometimes try to maximise or minimise a function of two variables without using the constraint to eliminate one. You need a single-variable function before you can differentiate.
⚠️ Motion problems are almost always on the exam. Know how to go from position to velocity to acceleration, and be able to interpret sign charts for when the object is moving forward, backward, speeding up, or slowing down.
⚠️ Expect a question asking you to find intervals of increase/decrease and local extrema using the first derivative test.
⚠️ Concavity and inflection points from f'' are commonly tested alongside the increasing/decreasing analysis.
⚠️ The closed interval method for absolute extrema is a standard exam question. The method is straightforward, so this is a free-points question if you remember to check the endpoints.
⚠️ Horizontal tangents (where f'(x) = 0) appear frequently. For y = 5x³ − 2x² − 3x, set y' = 15x² − 4x − 3 = 0 and solve (possibly using the quadratic formula or factoring).
True or false: If v(t) < 0 and a(t) < 0, the particle is slowing down. (False. Same sign means speeding up.)
Fill in the blank: A critical number is where f'(x) = ___ or f'(x) is ___. (0; undefined)
True or false: To find absolute extrema on [a, b], you only need to check where f'(x) = 0. (False. You must also check the endpoints.)
Fill in the blank: f is concave up where f''(x) ___ 0. (>)
True or false: An inflection point always occurs where f''(x) = 0. (False. f'' = 0 is necessary but not sufficient; f'' must change sign.)
Q: A particle's position is s(t) = 2t³ − 3t² − 6t − 1. Find velocity and acceleration at any time t.
A: v(t) = s'(t) = 6t² − 6t − 6. a(t) = v'(t) = 12t − 6.
Q: Find the absolute extrema of y = −3x³ + 2x − 1 on [−1, 3].
A: y' = −9x² + 2 = 0 gives x = ±√(2/9). Evaluate f at x = −1, −√(2/9), √(2/9), and 3. Absolute max at (−1, 0). Absolute min at (3, −76).
Q: The difference of two numbers is 30. What is the maximum product of one number times the square of the other?
A: Let x − y = 30. P = (30 + y)y². P' = 3y(20 + y) = 0 gives y = −20, x = 10. Maximum product is 4000.
Q: For f(x) = 2x⁴ − 3x³ − 2x² − 4, find the intervals of concavity and any inflection points.
A: f''(x) = 24x² − 18x − 4 = 0 at x ≈ −0.179 and x ≈ 0.929. Concave up on (−∞, −0.179) and (0.929, ∞). Concave down on (−0.179, 0.929). Inflection points at approximately (−0.179, −4.045) and (0.929, −6.642).
Q: At what x-values does the curve y = 5x³ − 2x² − 3x have a horizontal tangent?
A: y' = 15x² − 4x − 3 = 0. Factor: (5x − 3)(3x + 1) = 0. So x = 3/5 or x = −1/3.
Motion problems connect directly to integration later in the course: if differentiation takes you from position to velocity, integration reverses the process and takes you from velocity back to position. The first and second derivative tests for extrema and concavity are the foundation of full curve sketching, which combines everything (domain, intercepts, asymptotes, increasing/decreasing, concavity) into one complete picture of the function. Optimisation problems appear throughout applied mathematics, physics, and engineering whenever you need the best outcome subject to a constraint.
Related Terms / Search Tags
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