At a critical point c where f'(c) = 0, you compute f''(c) = 0. What is the correct conclusion?
Ac is an inflection point, not a local extremum
Bc is neither a maximum nor a minimum
CThe second derivative test is inconclusive; you must use the first derivative test
Dc is a local minimum because the concavity is neutral
When f''(c) = 0, the second derivative test gives no information — it is inconclusive. All three behaviors are possible: f(x) = x⁴ has a local minimum at 0, f(x) = −x⁴ has a local maximum at 0, and f(x) = x³ has neither (an inflection point). In every case f''(0) = 0. The correct fallback is the first derivative test: check the sign of f' on each side of c.
Question 2 Multiple Choice
A function has f'(2) = 0 and f''(2) = −7. Which statement best explains why x = 2 is a local maximum?
Af' is negative at x = 2, so the function is falling there
Bf'' is negative, meaning the function is concave down at x = 2 — like the top of a hill with zero slope
Cf'' < 0 means f' is decreasing, so the function must be at a minimum
DThe negative second derivative shows the function crosses zero at that point
The second derivative test classifies via concavity: f''(2) < 0 means the graph is concave down at x = 2 — curving like a dome. Combined with f'(2) = 0 (a flat tangent), this gives a hilltop — a local maximum. Option C contains a true fact (f'' < 0 does mean f' is decreasing) but draws the wrong conclusion; a decreasing f' at a zero-slope point means the function went from rising to falling, confirming a maximum.
Question 3 True / False
The second derivative test can classify a critical point without examining the sign of f' on both sides of the critical point.
TTrue
FFalse
Answer: True
True. The second derivative test only requires evaluating f''(c) at the critical point itself. If f''(c) > 0, it's a local min; if f''(c) < 0, it's a local max — no need to check f' values nearby. This is the test's main advantage over the first derivative test. The exception is when f''(c) = 0, in which case you must fall back to examining f' on both sides.
Question 4 True / False
If f''(c) = 0, then c is an inflection point of f.
TTrue
FFalse
Answer: False
False. f''(c) = 0 is a necessary but not sufficient condition for an inflection point. It also means the second derivative test is inconclusive about whether c is an extremum. For example, f(x) = x⁴ has f''(0) = 0, but x = 0 is a local minimum, not an inflection point. An inflection point requires f'' to change sign at c — not merely to equal zero.
Question 5 Short Answer
Explain why the second derivative test works geometrically: what do f'(c) = 0 and f''(c) > 0 together tell you about the shape of the graph near c?
Think about your answer, then reveal below.
Model answer: f'(c) = 0 means the tangent line at c is horizontal — the slope is zero. f''(c) > 0 means the function is concave up at c — the graph curves like the bottom of a bowl. A horizontal tangent at the bottom of a bowl is a local minimum: the function falls approaching from either side and the point c is the lowest nearby value.
The geometric picture is the key insight. Concavity tells you the direction the curve is bending: concave up (f'' > 0) means the curve bends upward, like a valley floor. Zero slope at a valley floor means you're at the bottom — a local minimum. Concave down (f'' < 0) means the curve bends downward, like a hill peak. Zero slope at a hilltop means you're at the top — a local maximum. This is why the second derivative test works without checking both sides of the critical point.