Why organize second-derivative signs?
A curve can bend upward on one interval and downward on another. The second derivative gives a way to track that bending, but its sign may differ across the domain. A second-derivative sign chart organizes those signs interval by interval so that the concavity information is clear.
The chart also helps prevent a common error: treating every zero or undefined value of \(f''\) as an inflection point. Those values are useful partition points, but the chart must be interpreted before deciding what they mean for the graph.
What goes into the chart?
Begin with the domain of \(f\), then find where \(f''\) is zero or undefined. These values divide the domain into open intervals. The second derivative has a constant sign on each such interval as long as it is continuous there: a continuous function cannot change sign without passing through zero.
For each interval, choose a test value and evaluate the sign of \(f''\). You usually need only the sign, not the full value. A positive result means concave up throughout that interval; a negative result means concave down.
A zero of \(f''\), or a point where \(f''\) is undefined, is a candidate breakpoint for the chart. It is not automatically an inflection point. An inflection point is a point on the graph where the function is continuous and its concavity changes across that point. Thus, a candidate that is not in the domain cannot be an inflection point of the function, and a candidate where the sign does not change does not mark a concavity change.
Keep track of points where the function itself is not defined.
Differentiate twice, then find where the result is zero or undefined.
Use the candidate breakpoints and domain breaks to create intervals.
Evaluate the sign of \(f''\) at one point in each interval.
Record concave up or down on each interval; assess a breakpoint separately.
Why the sign determines concavity
The link between \(f''\) and concavity follows from how the first derivative changes. Here is a justification for the positive-sign case. Suppose \(f\) is twice differentiable on an interval and \(f''(x)>0\) throughout it. Take any two points \(u<v\) in the interval. By the Mean Value Theorem applied to \(f'\), there is a point \(\xi\) between them such that
Both factors on the right are positive, so \(f'(v)>f'(u)\). Therefore \(f'\) is increasing. To see how that gives concavity up, take any three points \(x<y<z\) in the interval. Applying the Mean Value Theorem to \(f\) on \([x,y]\) and \([y,z]\) shows that the slope from \(x\) to \(y\) is less than the slope from \(y\) to \(z\): the two slopes equal values of the increasing derivative at points in those respective intervals. Hence
Multiplying by the positive denominators and rearranging gives
The right side is the height of the straight chord between \((x,f(x))\) and \((z,f(z))\) at \(x\)-coordinate \(y\). The inequality says the graph lies below that chord, which is the defining chord property of strict concavity up. When \(f''<0\), the inequalities reverse, giving concavity down. This is why one sign test per interval is enough when \(f''\) is continuous and has no zero there.
Worked Example: A zero where the sign does not change
Worked Example: A zero where the sign does not change
Determine the concavity intervals for \(f(x)=x^4\), and decide what the chart says about \(x=0\).
Differentiate twice:
The second derivative is zero at \(x=0\), and it is defined for all real \(x\). Thus the intervals are \((-\infty,0)\) and \((0,\infty)\). Test \(x=-1\) and \(x=1\):
The chart records concavity up on both intervals:
| Interval or point | Sign or status | Conclusion |
|---|---|---|
| \((-\infty,0)\) | \(f''>0\) | Concave up |
| \(0\) | \(f''(0)=0\) | Candidate breakpoint |
| \((0,\infty)\) | \(f''>0\) | Concave up |
Since the sign is positive on both sides, concavity does not change at zero. The point \((0,f(0))=(0,0)\) is not an inflection point. Check: differentiating \(f'(x)=4x^3\) gives \(12x^2\), and the test values confirm the sign on each interval.
Worked Example: A sign change at a point on the graph
Worked Example: A sign change at a point on the graph
Use a sign chart to describe the concavity of \(f(x)=x^3-3x\).
Differentiate:
The second derivative is zero at \(x=0\) and is defined everywhere. Test \(x=-1\) on the left and \(x=1\) on the right:
Therefore the function is concave down on \((-\infty,0)\) and concave up on \((0,\infty)\). The sign changes at zero, and the function is continuous there. Its graph point is \((0,f(0))=(0,0)\), so this is an inflection point. The chart’s role is to show the change in concavity; the point on the curve must also be identified. Check: differentiating \(3x^2-3\) gives \(6x\), and the test values have the stated signs.
Worked Example: A domain break is not a graph point
Worked Example: A domain break is not a graph point
Analyze the concavity intervals of \(f(x)=\frac{1}{x}\).
The domain is \(x\ne0\). Rewrite the function as \(x^{-1}\) and differentiate twice:
The second derivative is undefined at zero, which is also excluded from the domain. This divides the domain into \((-\infty,0)\) and \((0,\infty)\). Test \(x=-1\) and \(x=1\):
Thus \(f\) is concave down on \((-\infty,0)\) and concave up on \((0,\infty)\). Although the signs differ across zero, \(f(0)\) does not exist, so there is no point of the graph at zero and hence no inflection point there. Check: differentiating \(-x^{-2}\) gives \(2x^{-3}\); the two test values verify the interval signs.
How to read a chart carefully
A sign chart summarizes the behavior on intervals; it does not assign a concavity to a candidate point itself. A point where \(f''=0\) can occur inside an otherwise smooth graph without any change of concavity, as \(x^4\) demonstrated. A point where \(f''\) is undefined may be excluded from the function’s domain, as in \(1/x\), or the function itself may still be defined there. In all cases, inspect the neighboring signs and check whether the function has a graph point at the candidate.
For a chart to support a conclusion on an interval, make sure that the interval lies within the domain and that no zero or undefined point of \(f''\) has been skipped. If \(f''\) is expressed as a product or quotient, factoring it often makes the relevant breakpoints and signs easier to see. Do not use a test point from a different interval: signs can differ on opposite sides of a breakpoint.
Check Your Understanding
For each question, use the domain and interval signs—not just the values at candidate points—to justify your answer.
- For \(f(x)=x^4\), where is \(f''\) zero, and what is the sign of \(f''\) on each side?
- For \(g(x)=x^3\), find \(g''\) and determine the concavity intervals.
- For \(h(x)=1/x\), why does a sign change across zero not make zero an inflection point?
- What information does a test value provide in a second-derivative sign chart?
- State two checks needed before calling a chart breakpoint an inflection point.
