Increasing, decreasing and concavity practice problems
Sign charts for f′ and f″: where a graph rises, falls, and bends. 70 problems with worked solutions; in 70 of them every
equation is proved by a computer algebra system.
20 worked problems
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
For \( \displaystyle f(x) = 3 x^{3} + 18 x^{2} + 27 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of -2 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,-3) \cup (-1,\infty);\ \downarrow (-3,-1);\ \text{concave down } (-\infty,-2),\ \text{up } (-2,\infty);\ \text{inflection at } x=-2 \)
Top: the graph of f = 3x³ + 18x² + 27x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 9x² + 36x + 27. The tangent is flat at x = -3, -1.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = 2 x^{3} - 18 x^{2} + 48 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of 3 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,2) \cup (4,\infty);\ \downarrow (2,4);\ \text{concave down } (-\infty,3),\ \text{up } (3,\infty);\ \text{inflection at } x=3 \)
Top: the graph of f = 2x³ - 18x² + 48x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 6x² - 36x + 48. The tangent is flat at x = 2, 4.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = 3 x^{3} - \frac{27 x^{2}}{2} \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' = 0 at x = \frac{3}{2}, where it changes sign.✓ Proved
f'' < 0 to the left of \frac{3}{2} (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,0) \cup (3,\infty);\ \downarrow (0,3);\ \text{concave down } (-\infty,\frac{3}{2}),\ \text{up } (\frac{3}{2},\infty);\ \text{inflection at } x=\frac{3}{2} \)
Top: the graph of f = 3x³ - 27x²/2. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 9x² - 27x. The tangent is flat at x = 0, 3.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = x^{3} - \frac{3 x^{2}}{2} - 18 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' = 0 at x = \frac{1}{2}, where it changes sign.✓ Proved
f'' < 0 to the left of \frac{1}{2} (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,-2) \cup (3,\infty);\ \downarrow (-2,3);\ \text{concave down } (-\infty,\frac{1}{2}),\ \text{up } (\frac{1}{2},\infty);\ \text{inflection at } x=\frac{1}{2} \)
Top: the graph of f = x³ - 3x²/2 - 18x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 3x² - 3x - 18. The tangent is flat at x = -2, 3.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = 3 x^{3} - 36 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of 0 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,-2) \cup (2,\infty);\ \downarrow (-2,2);\ \text{concave down } (-\infty,0),\ \text{up } (0,\infty);\ \text{inflection at } x=0 \)
Top: the graph of f = 3x³ - 36x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 9x² - 36. The tangent is flat at x = -2, 2.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = 3 x^{3} - 9 x^{2} - 27 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of 1 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,-1) \cup (3,\infty);\ \downarrow (-1,3);\ \text{concave down } (-\infty,1),\ \text{up } (1,\infty);\ \text{inflection at } x=1 \)
Top: the graph of f = 3x³ - 9x² - 27x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 9x² - 18x - 27. The tangent is flat at x = -1, 3.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = x^{3} + 3 x^{2} - 9 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of -1 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,-3) \cup (1,\infty);\ \downarrow (-3,1);\ \text{concave down } (-\infty,-1),\ \text{up } (-1,\infty);\ \text{inflection at } x=-1 \)
Top: the graph of f = x³ + 3x² - 9x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 3x² + 6x - 9. The tangent is flat at x = -3, 1.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = 3 x^{3} + \frac{27 x^{2}}{2} - 36 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
Top: the graph of f = 3x³ + 27x²/2 - 36x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 9x² + 27x - 36. The tangent is flat at x = -4, 1.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = x^{3} - \frac{15 x^{2}}{2} + 12 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' = 0 at x = \frac{5}{2}, where it changes sign.✓ Proved
f'' < 0 to the left of \frac{5}{2} (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,1) \cup (4,\infty);\ \downarrow (1,4);\ \text{concave down } (-\infty,\frac{5}{2}),\ \text{up } (\frac{5}{2},\infty);\ \text{inflection at } x=\frac{5}{2} \)
Top: the graph of f = x³ - 15x²/2 + 12x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 3x² - 15x + 12. The tangent is flat at x = 1, 4.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = x^{3} + 3 x^{2} \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of -1 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,-2) \cup (0,\infty);\ \downarrow (-2,0);\ \text{concave down } (-\infty,-1),\ \text{up } (-1,\infty);\ \text{inflection at } x=-1 \)
Top: the graph of f = x³ + 3x². A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 3x² + 6x. The tangent is flat at x = -2, 0.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = 3 x^{3} + 9 x^{2} \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of -1 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,-2) \cup (0,\infty);\ \downarrow (-2,0);\ \text{concave down } (-\infty,-1),\ \text{up } (-1,\infty);\ \text{inflection at } x=-1 \)
Top: the graph of f = 3x³ + 9x². A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 9x² + 18x. The tangent is flat at x = -2, 0.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = 3 x^{3} - \frac{27 x^{2}}{2} + 18 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' = 0 at x = \frac{3}{2}, where it changes sign.✓ Proved
f'' < 0 to the left of \frac{3}{2} (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,1) \cup (2,\infty);\ \downarrow (1,2);\ \text{concave down } (-\infty,\frac{3}{2}),\ \text{up } (\frac{3}{2},\infty);\ \text{inflection at } x=\frac{3}{2} \)
Top: the graph of f = 3x³ - 27x²/2 + 18x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 9x² - 27x + 18. The tangent is flat at x = 1, 2.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = x^{3} + \frac{21 x^{2}}{2} + 36 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
Top: the graph of f = x³ + 21x²/2 + 36x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 3x² + 21x + 36. The tangent is flat at x = -4, -3.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = x^{3} - 6 x^{2} + 9 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of 2 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,1) \cup (3,\infty);\ \downarrow (1,3);\ \text{concave down } (-\infty,2),\ \text{up } (2,\infty);\ \text{inflection at } x=2 \)
Top: the graph of f = x³ - 6x² + 9x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 3x² - 12x + 9. The tangent is flat at x = 1, 3.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = x^{3} - 6 x^{2} \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of 2 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,0) \cup (4,\infty);\ \downarrow (0,4);\ \text{concave down } (-\infty,2),\ \text{up } (2,\infty);\ \text{inflection at } x=2 \)
Top: the graph of f = x³ - 6x². A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 3x² - 12x. The tangent is flat at x = 0, 4.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = x^{3} - 3 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of 0 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,-1) \cup (1,\infty);\ \downarrow (-1,1);\ \text{concave down } (-\infty,0),\ \text{up } (0,\infty);\ \text{inflection at } x=0 \)
Top: the graph of f = x³ - 3x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 3x² - 3. The tangent is flat at x = -1, 1.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = 2 x^{3} + 3 x^{2} - 12 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
Top: the graph of f = 2x³ + 3x² - 12x. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 6x² + 6x - 12. The tangent is flat at x = -2, 1.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = x^{3} - \frac{9 x^{2}}{2} \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' = 0 at x = \frac{3}{2}, where it changes sign.✓ Proved
f'' < 0 to the left of \frac{3}{2} (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,0) \cup (3,\infty);\ \downarrow (0,3);\ \text{concave down } (-\infty,\frac{3}{2}),\ \text{up } (\frac{3}{2},\infty);\ \text{inflection at } x=\frac{3}{2} \)
Top: the graph of f = x³ - 9x²/2. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 3x² - 9x. The tangent is flat at x = 0, 3.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = 2 x^{3} - 6 x^{2} \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
f'' < 0 to the left of 1 (concave down) and f'' > 0 to the right (concave up).
Reviewed
Answer \( \uparrow (-\infty,0) \cup (2,\infty);\ \downarrow (0,2);\ \text{concave down } (-\infty,1),\ \text{up } (1,\infty);\ \text{inflection at } x=1 \)
Top: the graph of f = 2x³ - 6x². A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 6x² - 12x. The tangent is flat at x = 0, 2.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
For \( \displaystyle f(x) = 3 x^{3} + \frac{27 x^{2}}{2} \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
Top: the graph of f = 3x³ + 27x²/2. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = 9x² + 27x. The tangent is flat at x = -3, 0.
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
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