f' to determine intervals of increase and decrease.f'', apply the Second Derivative Test, and read extrema and concavity directly from the graph of f'.A critical point of f is a value x = c in the domain of f where either f'(c) = 0 or f'(c) does not exist. Both cases matter: f'(c) = 0 gives smooth horizontal-tangent turning points; f'(c) undefined captures corners, cusps, and vertical tangents. If f itself is undefined at c, then c is not a critical point no matter what f' does.
Critical points are the only interior candidates for relative extrema — but a critical point need not be an extremum.
f'f'(x) > 0 on an interval, f is increasing there.f'(x) < 0 on an interval, f is decreasing there.Between consecutive critical points, f' cannot change sign, so a sign chart works: find the critical points, mark them on a number line, test one point in each interval, and record only the sign of f'.
Example.
f(x) = x³ − 3x² − 9x + 5,f'(x) = 3(x − 3)(x + 1). Critical pointsx = −1, 3.x = -1 x = 3 f' + )------( − )------( + increasing decreasing increasing
At a critical point c, look at how the sign of f' behaves as x passes through:
- Changes + to − ⇒ relative maximum.
- Changes − to + ⇒ relative minimum.
- No sign change ⇒ neither.
The no-change case is real. For f(x) = x³, f'(x) = 3x² is 0 at x = 0 but positive on both sides — no extremum, just a momentary flattening.
To find the absolute maximum and minimum of a continuous f on a closed interval [a, b] (the Extreme Value Theorem guarantees both exist):
(a, b).a and b.f at every candidate.The endpoints are full candidates and are the most commonly forgotten.
The second derivative f'' measures how the slope is changing:
- f''(x) > 0 ⇒ graph concave up (opens like a cup ∪); equivalently, f' is increasing.
- f''(x) < 0 ⇒ graph concave down (opens like a cap ∩); equivalently, f' is decreasing.
Inflection point. A point where the concavity changes —
f''changes sign — andfis defined there.
Finding where f''(x) = 0 only gives candidates; you must confirm a sign change. The classic trap is f(x) = x⁴: f''(x) = 12x² is 0 at x = 0, but stays positive on both sides, so x = 0 is not an inflection point.
If f'(c) = 0 and f''(c) exists:
- f''(c) > 0 (concave up) ⇒ relative minimum at c.
- f''(c) < 0 (concave down) ⇒ relative maximum at c.
- f''(c) = 0 ⇒ inconclusive; fall back to the First Derivative Test.
Anchor the direction to the shape: a horizontal tangent at the bottom of a cup (f'' > 0) is a low point.
f'A common item shows the graph of f' (the derivative) and asks about f:
On the graph of f' |
What it tells you about f |
|---|---|
f' above the axis (f' > 0) |
f increasing |
f' below the axis (f' < 0) |
f decreasing |
f' crosses + to − |
f has a relative maximum |
f' crosses − to + |
f has a relative minimum |
f' has a local max or min (turns around) |
f has an inflection point |
The biggest error is reading the picture as the graph of f. On the graph of f', extrema of f occur at the zero crossings, and inflection points of f occur where f' turns around.
[GRAPH: y = f'(x) on [-2, 6]
- f' > 0 on (-2, 1) and (4, 6); f' < 0 on (1, 4)
- f' crosses + to − at x = 1 ⇒ f has a relative maximum
- f' crosses − to + at x = 4 ⇒ f has a relative minimum
- label: "Graph of f ′ (the derivative)"]
f' (for increase/decrease) and f'' (for concavity), never from f itself.f' to be an extremum; f'(c) = 0 alone is not enough.f'' and that f be defined there; f''(c) = 0 alone is not enough.f': zero crossings locate extrema of f; turning points of f' locate inflection points of f.f(x) = x³ − 12x are1. B. f'(x) = 3x² − 12 = 3(x − 2)(x + 2) = 0 ⇒ x = ±2. (A) sets only x = 0; (C) mis-solves x² = 16; (D) reads a coefficient; (E) ignores the roots. Fix: set f'(x) = 0 and solve for every root.
f(x) = x³ − 12x decreasing?2. C. f'(x) = 3(x − 2)(x + 2) is negative between its roots, on (−2, 2). (A), (B) are the increasing intervals; (D) ignores the sign change; (E) is a partial interval. Fix: f decreases where the sign chart of f' reads negative.
f'(x) = (x − 1)²(x + 4), then f has3. E. At x = −4, f' changes − to + (the factor (x − 1)² is positive), giving a relative minimum; at x = 1 the squared factor produces no sign change, so no extremum. (A), (C), (D) wrongly treat the repeated root x = 1 as an extremum; (B) compounds the error. Fix: a squared (even-power) factor gives no sign change ⇒ no extremum there.
f'(x) = x²(x − 2). At x = 0, f has4. B. f'(x) = x²(x − 2): near x = 0 the factor x² stays positive and (x − 2) < 0, so f' is negative on both sides — no sign change, hence neither. (A), (C) assume f'(0) = 0 forces an extremum; (D) confuses this with a concavity change; (E) is determinable. Fix: f'(c) = 0 needs a sign change to give an extremum.
f'(c) = 0 and f''(c) > 0. By the Second Derivative Test, f has at x = c5. C. Concave up at a horizontal tangent (f'' > 0) is the bottom of a cup — a relative minimum. (A) reverses the direction; (B), (D) ignore the test's conclusion; (E) over-claims a global result from local data. Fix: positive second derivative ⇒ minimum (cup); negative ⇒ maximum (cap).
f(x) = x³ − 3x² + 4 is6. E. f''(x) = 6x − 6 = 0 ⇒ x = 1, and f'' changes sign there. (A) confuses it with a critical point of f'; (B), (C), (D) misread the second derivative. Fix: inflection candidates come from f'' = 0 with a confirmed sign change.
f on [1, 5] has critical points at x = 2 and x = 4, with f(1) = 3, f(2) = 7, f(4) = −1, and f(5) = 5. The absolute maximum value of f on [1, 5] is7. A. Compare f at all candidates: f(1) = 3, f(2) = 7, f(4) = −1, f(5) = 5; the largest is 7. (B) is a non-maximal endpoint; (C) another endpoint; (D) is the minimum; (E) is an x-value, not an f-value. Fix: the absolute max is the largest f-value among critical points and endpoints.
f', the derivative of f, is positive on (−∞, −1), negative on (−1, 3), and positive on (3, ∞). Then f has8. D. f' changes + to − at x = −1 (relative max) and − to + at x = 3 (relative min). (A) swaps them; (B), (C) misread the sign changes; (E) treats zero crossings as inflection points. Fix: on a graph of f', a +-to-− crossing is a max of f.
f(x) = x⁴, consider the point x = 0, where f''(0) = 0. Which statement is correct?9. D. f''(x) = 12x² ≥ 0, so f is concave up everywhere; f'(x) = 4x³ changes − to + at x = 0, giving a relative minimum, and since f'' does not change sign, x = 0 is not an inflection point. (A) treats f'' = 0 as sufficient; (B), (C) misclassify the extremum; (E) adds a false inflection point. Fix: f''(c) = 0 without a sign change is not an inflection point.
f(x) = x³ − 3x² concave up?10. C. f''(x) = 6x − 6 > 0 ⇒ x > 1, so concave up on (1, ∞). (A) is the concave-down interval; (B), (D) use f' roots by mistake; (E) ignores the sign change at x = 1. Fix: concave up where f'' > 0.
f(x) = x + 2sin(x) in the open interval (0, 2π) is11. B. f'(x) = 1 + 2cos(x) = 0 ⇒ cos x = −1/2 ⇒ x = 2π/3, 4π/3, both in (0, 2π) — two critical points. (A), (C) undercount; (D), (E) overcount the solutions of cos x = −1/2. Fix: count all solutions of f'(x) = 0 inside the interval.
f', the derivative of f, has a relative maximum at x = 4. Then at x = 4 the graph of f has12. A. A relative maximum of f' is where the slope of f is greatest and f'' changes sign — an inflection point of f. (B), (C) read an extremum of f' as an extremum of f; (D), (E) are unrelated. Fix: turning points of f' are inflection points of f, not extrema.
1. B. f'(x) = 3x² − 12 = 3(x − 2)(x + 2) = 0 ⇒ x = ±2. (A) sets only x = 0; (C) mis-solves x² = 16; (D) reads a coefficient; (E) ignores the roots. Fix: set f'(x) = 0 and solve for every root.
2. C. f'(x) = 3(x − 2)(x + 2) is negative between its roots, on (−2, 2). (A), (B) are the increasing intervals; (D) ignores the sign change; (E) is a partial interval. Fix: f decreases where the sign chart of f' reads negative.
3. E. At x = −4, f' changes − to + (the factor (x − 1)² is positive), giving a relative minimum; at x = 1 the squared factor produces no sign change, so no extremum. (A), (C), (D) wrongly treat the repeated root x = 1 as an extremum; (B) compounds the error. Fix: a squared (even-power) factor gives no sign change ⇒ no extremum there.
4. B. f'(x) = x²(x − 2): near x = 0 the factor x² stays positive and (x − 2) < 0, so f' is negative on both sides — no sign change, hence neither. (A), (C) assume f'(0) = 0 forces an extremum; (D) confuses this with a concavity change; (E) is determinable. Fix: f'(c) = 0 needs a sign change to give an extremum.
5. C. Concave up at a horizontal tangent (f'' > 0) is the bottom of a cup — a relative minimum. (A) reverses the direction; (B), (D) ignore the test's conclusion; (E) over-claims a global result from local data. Fix: positive second derivative ⇒ minimum (cup); negative ⇒ maximum (cap).
6. E. f''(x) = 6x − 6 = 0 ⇒ x = 1, and f'' changes sign there. (A) confuses it with a critical point of f'; (B), (C), (D) misread the second derivative. Fix: inflection candidates come from f'' = 0 with a confirmed sign change.
7. A. Compare f at all candidates: f(1) = 3, f(2) = 7, f(4) = −1, f(5) = 5; the largest is 7. (B) is a non-maximal endpoint; (C) another endpoint; (D) is the minimum; (E) is an x-value, not an f-value. Fix: the absolute max is the largest f-value among critical points and endpoints.
8. D. f' changes + to − at x = −1 (relative max) and − to + at x = 3 (relative min). (A) swaps them; (B), (C) misread the sign changes; (E) treats zero crossings as inflection points. Fix: on a graph of f', a +-to-− crossing is a max of f.
9. D. f''(x) = 12x² ≥ 0, so f is concave up everywhere; f'(x) = 4x³ changes − to + at x = 0, giving a relative minimum, and since f'' does not change sign, x = 0 is not an inflection point. (A) treats f'' = 0 as sufficient; (B), (C) misclassify the extremum; (E) adds a false inflection point. Fix: f''(c) = 0 without a sign change is not an inflection point.
10. C. f''(x) = 6x − 6 > 0 ⇒ x > 1, so concave up on (1, ∞). (A) is the concave-down interval; (B), (D) use f' roots by mistake; (E) ignores the sign change at x = 1. Fix: concave up where f'' > 0.
11. B. f'(x) = 1 + 2cos(x) = 0 ⇒ cos x = −1/2 ⇒ x = 2π/3, 4π/3, both in (0, 2π) — two critical points. (A), (C) undercount; (D), (E) overcount the solutions of cos x = −1/2. Fix: count all solutions of f'(x) = 0 inside the interval.
12. A. A relative maximum of f' is where the slope of f is greatest and f'' changes sign — an inflection point of f. (B), (C) read an extremum of f' as an extremum of f; (D), (E) are unrelated. Fix: turning points of f' are inflection points of f, not extrema.