AP Calculus AB/BC · Unit 5 · Topic 5.6
Determining Concavity of Functions over Their Domains
Use the sign of the second derivative and identify genuine changes in concavity.
1. Topic Focus
Use derivatives to prove existence, classify extrema, analyze monotonicity and concavity, sketch graphs, and solve optimization problems.
This topic: Use the sign of the second derivative and identify genuine changes in concavity.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
An inflection point requires a concavity change, not only f″=0.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
1. Concavity describes changing slopes
Concavity does not describe whether a function is positive or negative. It describes how tangent slopes change as \(x\) increases. Increasing slopes make the graph concave up; decreasing slopes make it concave down.
2. The second-derivative connection
The sign of \(f''\), not the sign of \(f\) or \(f'\), determines concavity.
3. Start with the domain
Concavity intervals must lie inside the domain of \(f\). Holes, vertical asymptotes, and other excluded inputs divide the analysis even when they do not solve \(f''(x)=0\). Never join an interval across an input where the function does not exist.
4. Find possible transition inputs
Possible changes occur where \(f''(x)=0\), where \(f''(x)\) is undefined, or at a boundary between domain pieces. These values partition the domain; they are not automatically inflection inputs.
5. Use a second-derivative sign chart
- Find \(f''\).
- Mark its zeros and undefined points together with domain breaks.
- Choose a test input in every resulting open interval.
- Determine the sign of \(f''\).
- Translate positive to concave up and negative to concave down.
6. Define an inflection point correctly
An inflection point is a point \((c,f(c))\) on the graph where concavity changes across \(c\). Thus \(f(c)\) must exist and the graph must have opposite concavities on the two sides.
7. A zero second derivative is only a candidate
For \(f(x)=x^4\), \(f''(x)=12x^2\) equals zero at \(x=0\), but it is positive on both sides. The graph remains concave up, so \((0,0)\) is not an inflection point.
8. An undefined second derivative can still give inflection
For \(f(x)=x^{1/3}\), the function is continuous at zero while its concavity changes there. Thus \((0,0)\) is an inflection point even though \(f'(0)\) and \(f''(0)\) are undefined.
9. A discontinuity cannot be an inflection point
For \(f(x)=1/x\), concavity differs on the two sides of zero, but \(f(0)\) does not exist. There is no point on the graph at zero, so it is not an inflection input.
10. Read concavity from the graph of the first derivative
When a graph or table of \(f'\) is given, \(f\) is concave up where \(f'\) increases and concave down where \(f'\) decreases. The height of \(f'\) controls the direction of \(f\); the direction of \(f'\) controls the concavity of \(f\).
11. Separate direction from bending
A decreasing graph can be concave up when its negative slopes become less negative.
12. State intervals and points differently
Concavity is reported on open intervals, such as \((1,4)\). Inflection is reported as a point, such as \((2,f(2))\), unless only the input is requested. Keep domain restrictions visible.
13. Common errors
- Using \(f'>0\) to claim concave up.
- Calling every zero of \(f''\) an inflection point.
- Ignoring points where \(f''\) is undefined.
- Calling a discontinuity an inflection point.
- Giving \(c\) without finding \(f(c)\) when a point is requested.
- Joining intervals across a domain break.
6. Detailed Worked Example and Error Check
Example 1: Polynomial sign chart. For \(f(x)=x^4-4x^3\),
The sign pattern is positive, negative, positive across \(0\) and \(2\). Therefore \(f\) is concave up on \((-\infty,0)\) and \((2,\infty)\), and concave down on \((0,2)\). The inflection points are \((0,0)\) and \((2,-16)\).
Example 2: A zero without a change. For \(f(x)=x^4\), \(f''(x)=12x^2\). Although \(f''(0)=0\), it is positive on both sides. The function is concave up over its domain and has no inflection point.
Example 3: A cubic inflection point. Let \(f(x)=x^3-3x^2+2\). Since \(f''(x)=6x-6\), the graph is concave down on \((-\infty,1)\) and concave up on \((1,\infty)\). Because \(f(1)=0\), the inflection point is \((1,0)\).
Example 4: A domain break. For \(f(x)=1/x\), \(f''(x)=2/x^3\). The graph is concave down on \((-\infty,0)\) and concave up on \((0,\infty)\). Zero is not in the domain, so it is not an inflection input.
Example 5: An undefined derivative at inflection. For \(f(x)=x^{1/3}\),
The second derivative is positive for \(x<0\) and negative for \(x>0\). Since \(f\) is continuous at zero, \((0,0)\) is an inflection point.
Example 6: Two concavity changes. For \(f(x)=x^4-6x^2\), \(f''(x)=12(x^2-1)\). The graph is concave up on \((-\infty,-1)\) and \((1,\infty)\), and concave down on \((-1,1)\). Since \(f(\pm1)=-5\), the inflection points are \((-1,-5)\) and \((1,-5)\).
Example 7: Concavity in context. A height model is \(h(t)=t^3-6t^2+9t+4\) for \(0\le t\le5\). Since \(h''(t)=6t-12\), the graph is concave down on \((0,2)\) and concave up on \((2,5)\). Its velocity decreases before \(t=2\) and increases after it. Because \(h(2)=6\), the model has an inflection point at \((2,6)\).
7. AP Reasoning Routine
State theorem hypotheses, make sign charts on domain intervals, include endpoints when required, and connect derivative signs to function behavior.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Determine all concavity intervals and inflection points, or answer the stated conceptual question.
(a) \(f(x)=x^3-6x^2+9x\).
(b) \(f(x)=x^4-4x^2\).
(c) \(f(x)=x^5-5x^3\).
(d) Explain why \(f(x)=x^4\) has no inflection point at zero.
(e) Analyze \(f(x)=1/x\), including the role of the domain.
(f) Analyze \(f(x)=x^{1/3}\), including the derivative failure at zero.
(g) Suppose \(f'(x)=(x-1)^2(x+2)\). Determine the concavity intervals of \(f\).
(h) A continuous function has \(f''>0\) on \((-\infty,-2)\), \(f''<0\) on \((-2,3)\), and \(f''>0\) on \((3,\infty)\). State the concavity intervals and possible inflection inputs.
(i) Explain why opposite signs of \(f''\) around \(x=1\) do not make \(x=1\) an inflection input for \(f(x)=1/(x-1)\).
(j) If \(f'(a)<0\) and \(f''(a)>0\), describe the local direction and concavity of \(f\) near \(a\).
Check the solution
In part (a), \(f''(x)=6x-12\). The graph is concave down on \((-\infty,2)\), concave up on \((2,\infty)\), and has inflection point \((2,2)\). In part (b), \(f''(x)=12x^2-8\), which vanishes at \(x=\pm\sqrt{2/3}\). The graph is concave up outside those inputs and concave down between them. Since \(f(\pm\sqrt{2/3})=-20/9\), the inflection points are \((\pm\sqrt{2/3},-20/9)\). In part (c), \(f''(x)=10x(2x^2-3)\). Let \(r=\sqrt{3/2}\). The graph is concave down on \((-\infty,-r)\), concave up on \((-r,0)\), concave down on \((0,r)\), and concave up on \((r,\infty)\). The inflection points are \((-r,21r/4)\), \((0,0)\), and \((r,-21r/4)\). In part (d), \(f''(x)=12x^2\) is positive on both sides of zero, so concavity does not change. In part (e), \(f''(x)=2/x^3\), giving concave down on \((-\infty,0)\) and concave up on \((0,\infty)\), but no inflection point because \(f(0)\) is undefined. In part (f), \(f''(x)=-2/(9x^{5/3})\) is positive to the left and negative to the right. Continuity at zero makes \((0,0)\) an inflection point despite the derivative failure. In part (g), \(f'(x)=x^3-3x+2\), so \(f''(x)=3(x-1)(x+1)\). Thus \(f\) is concave up on \((-\infty,-1)\) and \((1,\infty)\), and concave down on \((-1,1)\). Without values of \(f\), only the inflection inputs \(-1\) and \(1\) can be reported. In part (h), the function is concave up on the first and third intervals and concave down on the middle interval. If the stated continuous function is defined at those inputs, \(-2\) and \(3\) are inflection inputs. In part (i), \(f(1)\) does not exist, so there is no graph point where inflection can occur. In part (j), \(f\) is decreasing because its slope is negative and concave up because its slopes are increasing.