f, f', and f'', reading features of one off another.This table is the engine of graph analysis. Read it both directions.
Feature of the graph of f |
What f' does |
What f'' does |
|---|---|---|
f increasing |
f' > 0 (above the axis) |
— |
f decreasing |
f' < 0 (below the axis) |
— |
f has a relative max |
f' = 0, changes + to − |
f'' < 0 |
f has a relative min |
f' = 0, changes − to + |
f'' > 0 |
f concave up |
f' increasing |
f'' > 0 |
f concave down |
f' decreasing |
f'' < 0 |
f has an inflection point |
f' has a local max or min |
f'' changes sign |
Two traps: an x-intercept of f' is a critical point of f, not automatically an extremum (f' must change sign); and a local extremum of f' corresponds to an inflection point of f, because that is where f'' is zero and switching sign.
Matching example. If g(x) = x³ − 3x, then g'(x) = 3x² − 3 — an upward parabola with roots at x = ±1 and vertex (0, −3). The roots of g' are the critical points of g; the vertex of g' (its minimum) marks the inflection point of g.
Work a checklist in order:
1. Domain — exclude denominator zeros and invalid arguments.
2. Intercepts — f(0) for the y-intercept; solve f(x) = 0 for x-intercepts.
3. Asymptotes — vertical at uncancelled denominator zeros; horizontal from lim_{x→±∞} f(x). For a rational function with equal top and bottom degrees, the horizontal asymptote is the ratio of leading coefficients.
4. First-derivative analysis — critical points, increase/decrease, First Derivative Test.
5. Second-derivative analysis — concavity and inflection points.
6. Assemble — plot intercepts, dash the asymptotes, mark extrema and inflections, connect using the sign information.
A concavity sign change at a vertical asymptote is not an inflection point, because f is undefined there.
[GRAPH: y = f'(x) on [-2, 6]
- This is the graph of f ′, the DERIVATIVE of f
- f' < 0 on (-2, 0), f' > 0 on (0, 4), f' < 0 on (4, 6)
- f' has a maximum at x = 2
- reading it: f has a relative minimum at x = 0, a relative maximum at x = 4,
and an inflection point at x = 2 (where f' turns around)]
Optimization asks for the largest or smallest value of a quantity subject to a restriction. The procedure:
Worked (three-sided pen). A farmer has 200 ft of fence for a rectangular pen along a river (no fence on the river side). With sides x (two of them) and y (parallel to the river), the constraint is 2x + y = 200, so y = 200 − 2x and the area is
A(x) = x(200 − 2x) = 200x − 2x², 0 < x < 100
A'(x) = 200 − 4x = 0 ⇒ x = 50
Since A''(x) = −4 < 0, x = 50 is the maximum. Then y = 100 and the maximum area is 5000 ft².
Worked (open box). From a 12 in × 12 in sheet, cut x-inch corner squares and fold: V(x) = x(12 − 2x)². Then V'(x) = (12 − 2x)(12 − 6x) = 0 ⇒ x = 2 (rejecting x = 6). The First Derivative Test confirms a maximum at x = 2, giving V = 128 in³.
A critical point is only a candidate; the justification is what confirms it is a maximum or minimum. Always optimize the objective, never the fixed constraint.
f': sign gives increase/decrease of f, zero crossings give extrema of f, and turning points give inflection points of f.F, the maximum area uses half the fence on the unfenced-parallel side.1. A. With numbers a and 20 − a, P = 20a − a², P' = 20 − 2a = 0 ⇒ a = 10, and P'' = −2 < 0 ⇒ max; product = 10·10 = 100. (B), (C) are non-optimal products; (D) doubles wrongly; (E) squares the half-sum incorrectly. Fix: equal split maximizes a product with fixed sum.
x > 0, the minimum value of f(x) = x + 16/x is2. C. f'(x) = 1 − 16/x² = 0 ⇒ x = 4, f(4) = 4 + 4 = 8, and f''(x) = 32/x³ > 0 ⇒ min. (A) reports the x-value instead of f(x); (B), (D) compute the wrong quantity; (E) is arbitrary. Fix: plug the optimal x back into f to get the value asked for.
3. B. 2x + 2y = 100 ⇒ y = 50 − x; A = x(50 − x), A' = 50 − 2x = 0 ⇒ x = 25 = y, a 25×25 square with area 625. (A) uses 50 × 50 (forgets four sides); (C), (E) are non-optimal; (D) confuses perimeter with area. Fix: a fixed perimeter is maximized by the square.
g(x) = x³ − 3x. The graph of g' is4. D. g'(x) = 3x² − 3 is an upward parabola with roots x = ±1 and g'(0) = −3. (A) is linear (wrong degree); (B) has no real roots; (C) opens downward; (E) ignores the x² term. Fix: differentiate, then match both shape and roots.
f', the derivative of f, is positive on (−∞, 2) and negative on (2, ∞). At x = 2, the function f has5. A. f' changes + to − at x = 2, so f has a relative maximum. (B) reverses the test; (C) requires a sign change of f'', not f'; (D) confuses a sign change with an undefined point; (E) ignores the zero of f'. Fix: +-to-− crossing of f' ⇒ maximum of f.
x = c of an objective V(x) gives a maximum volume?6. B. The Second Derivative Test: V'(c) = 0 and V''(c) < 0 ⇒ concave down ⇒ maximum. (A) confirms a minimum; (C), (E) use the value of V, which is irrelevant; (D) contradicts c being a critical point. Fix: negative second derivative at a critical point ⇒ maximum.
x-inch squares from the corners of a 6 in × 6 in sheet and folding up the sides. The volume is maximized when x =7. E. V(x) = x(6 − 2x)², V'(x) = (6 − 2x)(6 − 6x) = 0 ⇒ x = 1 (rejecting x = 3, which gives zero volume). (A) is the rejected root; (B), (C), (D) are miscalculated critical points. Fix: reject roots that fall outside the realistic domain.
f(x) = (2x − 6)/(x + 4) is8. E. Equal degrees give the ratio of leading coefficients, 2/1 = 2, so y = 2. (A) assumes numerator degree is lower; (B) reads the vertical asymptote x = −4; (C) reads a constant term; (D) miscomputes the ratio. Fix: equal-degree rational ⇒ horizontal asymptote = leading-coefficient ratio.
9. B. Three-sided: 2x + y = 80, A = x(80 − 2x), A' = 80 − 4x = 0 ⇒ x = 20, y = 40, A = 800; A'' = −4 < 0 ⇒ max. (A), (C) are non-optimal; (D) forgets a factor; (E) is arbitrary. Fix: for a three-sided pen, half the fence goes on the side parallel to the river.
f''(x) > 0 for every x in (a, b), then on (a, b) the graph of f is10. A. f''(x) > 0 means the graph is concave up, by definition. (B) reverses the sign; (C), (D) confuse the second derivative with the first; (E) would require f'' = 0. Fix: sign of f'' controls concavity, not increase/decrease.
y = 9 − x², symmetric about the y-axis. The value of x that maximizes its area is11. D. Width = 2x, height = 9 − x², so A = 2x(9 − x²) = 18x − 2x³, A' = 18 − 6x² = 0 ⇒ x² = 3 ⇒ x = √3 (A'' = −12x < 0 ⇒ max). (A) is the x-intercept of the curve; (B), (C), (E) are miscalculated roots. Fix: solve A'(x) = 0 on the geometric domain 0 < x < 3.
f, the sign of f' is + on x < 1, + on 1 < x < 4, and − on x > 4; the sign of f'' is + on x < 1, − on 1 < x < 4, and − on x > 4. At x = 1, the function f has12. C. At x = 1, f' stays positive (no extremum) but f'' changes from + to −, so concavity changes — an inflection point. (A), (B) would need a sign change of f'; (D), (E) describe discontinuities the signs do not indicate. Fix: an inflection point is a sign change of f'', even where f' keeps its sign.
1. A. With numbers a and 20 − a, P = 20a − a², P' = 20 − 2a = 0 ⇒ a = 10, and P'' = −2 < 0 ⇒ max; product = 10·10 = 100. (B), (C) are non-optimal products; (D) doubles wrongly; (E) squares the half-sum incorrectly. Fix: equal split maximizes a product with fixed sum.
2. C. f'(x) = 1 − 16/x² = 0 ⇒ x = 4, f(4) = 4 + 4 = 8, and f''(x) = 32/x³ > 0 ⇒ min. (A) reports the x-value instead of f(x); (B), (D) compute the wrong quantity; (E) is arbitrary. Fix: plug the optimal x back into f to get the value asked for.
3. B. 2x + 2y = 100 ⇒ y = 50 − x; A = x(50 − x), A' = 50 − 2x = 0 ⇒ x = 25 = y, a 25×25 square with area 625. (A) uses 50 × 50 (forgets four sides); (C), (E) are non-optimal; (D) confuses perimeter with area. Fix: a fixed perimeter is maximized by the square.
4. D. g'(x) = 3x² − 3 is an upward parabola with roots x = ±1 and g'(0) = −3. (A) is linear (wrong degree); (B) has no real roots; (C) opens downward; (E) ignores the x² term. Fix: differentiate, then match both shape and roots.
5. A. f' changes + to − at x = 2, so f has a relative maximum. (B) reverses the test; (C) requires a sign change of f'', not f'; (D) confuses a sign change with an undefined point; (E) ignores the zero of f'. Fix: +-to-− crossing of f' ⇒ maximum of f.
6. B. The Second Derivative Test: V'(c) = 0 and V''(c) < 0 ⇒ concave down ⇒ maximum. (A) confirms a minimum; (C), (E) use the value of V, which is irrelevant; (D) contradicts c being a critical point. Fix: negative second derivative at a critical point ⇒ maximum.
7. E. V(x) = x(6 − 2x)², V'(x) = (6 − 2x)(6 − 6x) = 0 ⇒ x = 1 (rejecting x = 3, which gives zero volume). (A) is the rejected root; (B), (C), (D) are miscalculated critical points. Fix: reject roots that fall outside the realistic domain.
8. E. Equal degrees give the ratio of leading coefficients, 2/1 = 2, so y = 2. (A) assumes numerator degree is lower; (B) reads the vertical asymptote x = −4; (C) reads a constant term; (D) miscomputes the ratio. Fix: equal-degree rational ⇒ horizontal asymptote = leading-coefficient ratio.
9. B. Three-sided: 2x + y = 80, A = x(80 − 2x), A' = 80 − 4x = 0 ⇒ x = 20, y = 40, A = 800; A'' = −4 < 0 ⇒ max. (A), (C) are non-optimal; (D) forgets a factor; (E) is arbitrary. Fix: for a three-sided pen, half the fence goes on the side parallel to the river.
10. A. f''(x) > 0 means the graph is concave up, by definition. (B) reverses the sign; (C), (D) confuse the second derivative with the first; (E) would require f'' = 0. Fix: sign of f'' controls concavity, not increase/decrease.
11. D. Width = 2x, height = 9 − x², so A = 2x(9 − x²) = 18x − 2x³, A' = 18 − 6x² = 0 ⇒ x² = 3 ⇒ x = √3 (A'' = −12x < 0 ⇒ max). (A) is the x-intercept of the curve; (B), (C), (E) are miscalculated roots. Fix: solve A'(x) = 0 on the geometric domain 0 < x < 3.
12. C. At x = 1, f' stays positive (no extremum) but f'' changes from + to −, so concavity changes — an inflection point. (A), (B) would need a sign change of f'; (D), (E) describe discontinuities the signs do not indicate. Fix: an inflection point is a sign change of f'', even where f' keeps its sign.