Short on time? This page is the recap. For the reasoning, the diagram and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- Relative extrema can only happen at critical points: x in the domain of f with f′(x) = 0 or f′(x) undefined.
- Not every critical point is an extremum. The First Derivative Test sorts them.
- The test uses the sign chart from Topic 5.3: increasing then decreasing gives a peak; decreasing then increasing gives a valley.
- Location is the x-value c. Value is f(c).
Key relationships
| Sign of f′ left of c | Sign of f′ right of c | f at c (f continuous at c) |
|---|---|---|
| + | − | Relative maximum |
| − | + | Relative minimum |
| + | + | Neither |
| − | − | Neither |
| On a graph of f′ at x = c | Meaning for f |
|---|---|
| Crosses the axis downwards | Relative maximum |
| Crosses the axis upwards | Relative minimum |
| Touches the axis and turns back | Neither |
| Peak or valley away from the axis | Not an extremum of f |
Assumptions behind the test
- c must be in the domain of f, and f must be continuous at c.
- f′ must keep one sign on an interval just left of c and one sign just right of c.
- The test is for interior points. Endpoints are compared in Topic 5.5.
Mistakes to avoid
- Justifying with “f′(c) = 0”. Say how the sign of f′ changes.
- Missing critical points where f′ is undefined (cusps such as 3x^(2/3) − 2x at x = 0).
- Giving the x-value when the question asks for the value, or the reverse.
- Substituting into f′ instead of f to get the extremum value.
- Treating a peak of the f′ graph as a maximum of f.
- Applying the test at a jump or a hole, where f is not continuous.
Quick self-check
- m(x) = x³ − 3x. Classify the critical points. (m′(x) = 3(x − 1)(x + 1). m′ changes + to − at x = −1: relative maximum, value m(−1) = 2. m′ changes − to + at x = 1: relative minimum, value m(1) = −2.)
- f′(x) = (x − 2)³. Does f have a relative extremum at x = 2? (Yes, a relative minimum: f′ < 0 for x < 2 and f′ > 0 for x > 2.)
- The graph of g′ has a peak at x = 5 where g′(5) = 4. Does g have a maximum at x = 5? (No. g′(5) ≠ 0, so x = 5 is not a critical point; g is increasing there.)
Next: practice questions.