Why Build a Concavity Sign Chart?
Concavity describes how a function’s slopes change as \(x\) increases. The second derivative provides a systematic way to analyze that change: where \(f''(x)>0\), the graph is concave up; where \(f''(x)<0\), it is concave down, provided the function is defined on the interval being considered.
A sign chart organizes this information. Instead of repeatedly inspecting a formula, you identify where the sign of \(f''\) could change, divide the domain into intervals, and test one point in each interval. The chart makes the reasoning visible and helps prevent errors involving domain restrictions, signs, and interval notation.
From the Second Derivative to Intervals
To build the chart, start with the domain of \(f\), then determine the domain of \(f''\). Find the \(x\)-values in the domain of \(f\) where \(f''(x)=0\), and the \(x\)-values in the domain of \(f\) where \(f''\) is undefined. These values divide the domain into intervals on which you will check the sign.
A zero or undefined value of \(f''\) is a partition point for the sign analysis. It is not automatically an interval of concavity, and a value outside the domain of \(f\) cannot be treated as a point on the graph. Domain restrictions matter: for example, if the function is undefined at \(x=2\), the intervals to the left and right of 2 must remain separate.
On the intervals used in standard AP Calculus problems, \(f''\) is continuous and nonzero between consecutive partition points. Its sign therefore stays the same throughout each such interval. By the Second Derivative Concavity Theorem, a positive sign means concave up and a negative sign means concave down.
Why One Test Point Can Represent an Interval
The chart relies on a simple sign principle. Suppose \(f''\) is continuous on an interval and has no zero there. If it were positive at one point and negative at another, the Intermediate Value Theorem would force \(f''\) to equal zero somewhere between those points. That contradicts the assumption that the interval contains no zero. Therefore its sign is constant on the interval, and one test point determines the sign throughout.
The reasoning has two parts: first, the partition points separate the places where zeros or undefined values occur; second, on each resulting interval where \(f''\) is continuous, no sign change can occur without a zero. The test point is not a guess about the whole graph. It represents the interval because of this sign argument.
Worked Example: A Polynomial with Two Partition Points
Worked Example: A Polynomial with Two Partition Points
Determine the intervals of concavity for \(f(x)=x^4-4x^2\), and record the signs in a chart.
First differentiate twice:
Set the second derivative equal to zero to find the partition points:
The polynomial and its second derivative are defined for every real \(x\). The two zeros divide the real line into three intervals. Choose \(-2\), \(0\), and \(2\), one from each interval, and evaluate \(f''\):
| Interval | Test value | Sign of \(f''\) | Concavity |
|---|---|---|---|
| \((-\infty,-\sqrt{2/3})\) | \(-2\) | Positive | Concave up |
| \((-\sqrt{2/3},\sqrt{2/3})\) | \(0\) | Negative | Concave down |
| \((\sqrt{2/3},\infty)\) | \(2\) | Positive | Concave up |
Check: The factored expression \(4(3x^2-2)\) is positive when \(x^2>2/3\), which holds on both outer intervals, and negative when \(x^2<2/3\), which holds on the middle interval. This agrees with the test values and chart.
Worked Example: An Undefined Point Separates the Domain
Worked Example: An Undefined Point Separates the Domain
Build a concavity sign chart for \(f(x)=\dfrac{1}{x-2}\).
The function is undefined when \(x-2=0\), so its domain is \((-\infty,2)\cup(2,\infty)\). Differentiate:
The second derivative has no zeros. It is undefined at \(x=2\), which is also excluded from the domain of \(f\). Thus the two intervals to test are \((-\infty,2)\) and \((2,\infty)\).
Choose \(x=1\) on the left and \(x=3\) on the right:
| Interval | Test value | Sign of \(f''\) | Concavity |
|---|---|---|---|
| \((-\infty,2)\) | \(1\) | Negative | Concave down |
| \((2,\infty)\) | \(3\) | Positive | Concave up |
Check: For \(x<2\), \((x-2)^3<0\), so \(2/(x-2)^3<0\). For \(x>2\), \((x-2)^3>0\), so \(2/(x-2)^3>0\). The chart’s signs match the formula on both sides of the domain restriction.
Worked Example: An Undefined Second Derivative on the Domain
Worked Example: An Undefined Second Derivative on the Domain
Determine the concavity intervals for \(f(x)=x^{4/3}\), whose domain is all real numbers.
Differentiate once. The derivative \(f'(x)=\frac{4}{3}x^{1/3}\) is defined for every real \(x\). For \(x\ne 0\), differentiate again:
At \(x=0\), this formula is undefined. In fact, the difference quotient for \(f'\) at zero is \(\frac{f'(h)-f'(0)}{h}=\frac{4}{3}h^{-2/3}\), which has no finite limit as \(h\) approaches zero. Thus \(f''(0)\) is undefined, even though \(f(0)\) exists. This point splits the domain into \((-\infty,0)\) and \((0,\infty)\).
Test \(x=-1\) and \(x=1\). Since \((-1)^{2/3}=(\sqrt[3]{-1})^2=1\), and \(1^{2/3}=1\), the evaluations are:
| Interval | Test value | Sign of \(f''\) | Concavity |
|---|---|---|---|
| \((-\infty,0)\) | \(-1\) | Positive | Concave up |
| \((0,\infty)\) | \(1\) | Positive | Concave up |
Check: For every \(x\ne 0\), \(x^{2/3}=(\sqrt[3]{x})^2>0\). Therefore \(4/(9x^{2/3})>0\) on both intervals, agreeing with the chart.
Common Pitfall: Testing at a Point Outside the Interval
For \(f(x)=\dfrac{1}{x-2}\), a mistaken chart might treat the whole real line as one interval and test at \(x=2\). But \(f''(2)\) is undefined, so this is not a valid test value. It also ignores that the function’s domain is split at 2.
The denominator is zero, so that evaluation does not exist. The correction is to mark \(x=2\) as a boundary and select a test value inside each domain interval, such as \(1\) and \(3\). Then \(f''(1)=-2\) and \(f''(3)=2\), giving concavity down on the left and concavity up on the right.
Using the Chart Carefully
A well-formed chart separates calculation from interpretation. First find the partition points; next determine the intervals and test signs; only then translate those signs into concavity. Keep exact values such as \(\sqrt{2/3}\) when they define interval endpoints, and use a convenient test value rather than a complicated one.
The chart reports concavity on intervals. A partition point is included because it affects the organization of the sign analysis, but it is not itself an interval. In particular, do not infer an additional conclusion about a point merely from the fact that \(f''\) is zero or undefined there. The chart’s direct conclusion is the sign of \(f''\) on the listed intervals and the corresponding concavity there.
Identify where \(f\) is defined so that no chart interval includes values outside its domain.
Locate zeros and undefined values of \(f''\) that lie in the domain of \(f\), along with domain boundaries that separate intervals.
Choose any convenient value inside each interval and determine the sign of \(f''\) there.
Write positive as concave up and negative as concave down, using interval notation.
Check Your Understanding
For each question, show enough work to justify the intervals and signs in your chart.
- For \(f(x)=x^3-3x\), find the partition points and determine the concavity intervals.
- For \(f(x)=\dfrac{1}{x+1}\), state the domain intervals before testing signs of the second derivative.
- Why is a point where \(f''\) is undefined a possible partition point even if \(f\) itself is defined there?
- For \(f''(x)=(x-1)(x+2)\), identify the intervals that should be tested.
- What sign of \(f''\) corresponds to concavity down, and what theorem connects the sign to concavity?
