For f(x) = x³ − 3x, we find f'(x) = 3x² − 3, which equals zero at x = ±1. The sign of f' changes from positive (for x < −1) to negative (for −1 < x < 1) at x = −1. What does the first derivative test conclude about x = −1?
Ax = −1 is an inflection point because the derivative equals zero there
Bx = −1 is a local minimum because f'(−1) = 0 and the function is decreasing afterward
Cx = −1 is a local maximum because f' changes from positive to negative
DWe cannot classify x = −1 without also computing f''(−1)
The first derivative test classifies critical points by sign change, not by the value of f'' or anything else. When f' changes from positive (function rising) to negative (function falling), the function peaked — it is a local maximum. f'(−1) = 0 only tells us x = −1 is a critical point; the sign change is what determines it is a local maximum. The second derivative test is an alternative, not a requirement — the first derivative test is self-contained.
Question 2 Multiple Choice
For the function g(x) where g'(x) = (x − 2)² for all x, the point x = 2 is a critical point since g'(2) = 0. What does the first derivative test say about x = 2?
AIt is a local minimum, since (x − 2)² ≥ 0 means g' is never negative near x = 2
BIt is a local maximum, since g' = 0 at x = 2 and positive everywhere else
CIt is neither a local maximum nor a local minimum, since g' does not change sign at x = 2
DThe first derivative test is inconclusive here; we must use the second derivative test
Since (x − 2)² ≥ 0 for all x, g'(x) ≥ 0 on both sides of x = 2 — positive before, zero at, positive after. There is no sign change. The function is non-decreasing on both sides; it simply pauses its ascent momentarily without ever falling. This is the classic pattern of a 'saddle-like' critical point (like x³ at 0 in one dimension): f'(c) = 0 but no extremum. The first derivative test is decisive here — 'no sign change' is a valid conclusion, not an inconclusive one.
Question 3 True / False
Nearly every point where f'(c) = 0 is either a local maximum or a local minimum of f.
TTrue
FFalse
Answer: False
f'(c) = 0 is a necessary condition for a local extremum at an interior point, but not sufficient. The classic counterexample is f(x) = x³ at c = 0: f'(0) = 0, yet x = 0 is neither a maximum nor a minimum — the function is increasing on both sides. The first derivative test shows this: f'(x) = 3x² ≥ 0 on both sides of 0, so there is no sign change and no extremum. f'(c) = 0 means the function is momentarily flat; whether it turns around depends on what happens to the sign of f' nearby.
Question 4 True / False
The first derivative test requires examining the sign of f' on both sides of a critical point, not just confirming that f'(c) = 0.
TTrue
FFalse
Answer: True
This is the entire content of the first derivative test. f'(c) = 0 (or undefined) merely identifies c as a critical point — a candidate for classification. The classification itself comes from the sign pattern: positive-to-negative means local max, negative-to-positive means local min, same sign on both sides means neither. A student who only verifies f'(c) = 0 has done half the work and drawn no valid conclusion about extrema.
Question 5 Short Answer
Explain why f'(c) = 0 is a necessary but not sufficient condition for a local extremum. Give a concrete example where the condition fails and explain what additional information the first derivative test provides.
Think about your answer, then reveal below.
Model answer: f'(c) = 0 is necessary because at a smooth local extremum the tangent must be horizontal — the function cannot be rising or falling at a peak or valley. But it is not sufficient because the function could be momentarily flat without reversing direction. Example: f(x) = x³ at c = 0. f'(x) = 3x², so f'(0) = 0, yet x = 0 is not an extremum — the function increases on both sides. The first derivative test provides the missing information: examine the sign of f' just left and just right of c. Only if the sign changes (positive-to-negative or negative-to-positive) is there an extremum. If the sign is the same on both sides, f' = 0 is a pause, not a reversal.
The intuition is that a local extremum requires the function to change from rising to falling (max) or falling to rising (min). f'(c) = 0 says the instantaneous slope is zero, but says nothing about whether the slope was positive before and is negative after. The sign chart captures exactly this reversal.