AP Calculus AB · Lesson 3 of 5
Applications of derivatives
Use derivatives to justify increasing behavior, concavity, and extreme values on an interval.
Extrema
A critical point is an interior point where the derivative is zero or undefined. Candidates for an absolute maximum or minimum on a closed interval also include the endpoints. Evaluating only the critical points can miss an extreme value that occurs at an endpoint.
If f′ changes from positive to negative at c, f has a local maximum at c. If f′ changes from negative to positive, f has a local minimum. A zero derivative by itself does not prove a maximum or a minimum, because the derivative might not change sign.
Related rates and meaning
In a related-rates problem, name each quantity and what its derivative represents. Differentiate the relationship with respect to time, then substitute the values that hold at the instant in question. Do not substitute a constant length before differentiating if that length is actually changing.
The second derivative describes concavity: f″ > 0 on an interval means the graph is concave up there. A justification should name the sign and the interval, not only say “the graph looks like a cup.”
Worked example
f(x) = x³ − 3x on [−2, 2]. Where do the absolute maximum and minimum occur?
- f′(x) = 3x² − 3 = 3(x − 1)(x + 1), so the critical points are x = −1 and x = 1.
- Evaluate f(−2) = −2, f(−1) = 2, f(1) = −2, and f(2) = 2.
- The absolute maximum value is 2, at x = −1 and x = 2. The absolute minimum value is −2, at x = −2 and x = 1.
Why this works. On a closed interval, compare critical values with both endpoints.
Check your understanding
- List critical points and endpoints before choosing an absolute extremum.
- Explain a local extremum by a sign change of the first derivative.
- Differentiate a related-rates equation before substituting an instant’s values.