Why a Concavity Sign Chart Needs Breakpoints
A second-derivative sign chart records where a function is concave up and where it is concave down. The key step is to divide the domain into intervals on which the sign of \(f''\) can be tested reliably. Testing only one value on an interval is useful because the sign cannot switch there without an appropriate break in the analysis.
The possible breakpoints include zeros of \(f''\) and points where \(f''\) is undefined. Domain restrictions on \(f\) must also be respected: if \(f\) is undefined at a point, that point separates the domain into different pieces even if an algebraic expression for \(f''\) might otherwise appear to have a value there.
Here, “concave up” and “concave down” describe the usual second-derivative criterion. A zero of \(f''\) or a point where \(f''\) is undefined is a place to split the chart; identifying such a point alone does not classify the concavity on either side.
How to Build the Chart
First find the domain of \(f\). Then compute \(f''\) and identify the points in the domain where \(f''=0\), along with points where \(f''\) is undefined. Sort these points from left to right. Together with any domain restrictions, they determine the intervals to test.
Record points where the function itself is undefined. These points cannot be included in a concavity interval for the function.
Locate its zeros and the points where its formula or derivative is undefined. Keep track of which points belong to the domain of \(f\).
Use the ordered breakpoints to create intervals. Do not combine intervals across a zero or an undefined point without further analysis.
Choose a convenient point in each interval, evaluate \(f''\), and use its sign to classify concavity there.
For the common functions used in introductory calculus, \(f''\) is continuous on each interval where it is defined. If it is continuous and nonzero throughout an interval, its sign cannot change there: a change from positive to negative would force a zero by the Intermediate Value Theorem. Thus a single test value determines the sign on that whole interval. If \(f''\) is not continuous on an interval, do not assume one test value is enough; additional analysis may be needed.
Worked Example: A Rational Function with a Zero and a Domain Break
Worked Example: Classifying \(x^2+\frac{1}{x-1}\)
Let \(f(x)=x^2+\frac{1}{x-1}\). The denominator is zero at \(x=1\), so the domain is \(\mathbb{R}\setminus\{1\}\). Differentiate twice:
To find the zeros of \(f''\), combine the terms:
The quadratic factor is always positive because
Therefore the only zero of \(f''\) is \(x=0\), and \(f''\) is undefined at \(x=1\). The intervals to test are \((-\infty,0)\), \((0,1)\), and \((1,\infty)\).
Choose \(x=-1\), \(x=\frac12\), and \(x=2\), respectively:
Thus \(f\) is concave up on \((-\infty,0)\), concave down on \((0,1)\), and concave up on \((1,\infty)\). Check: the test values lie in their stated intervals, and substituting them into the derivative formula gives the displayed signs. The point \(x=1\) is excluded from the function’s domain, so it cannot be part of any of these intervals.
Why the Partition Supports the Sign Test
The partition is not just a convenient way to organize calculations. On an interval with no zeros or undefined points of a continuous \(f''\), the sign must remain fixed. If \(f''\) were positive at one point and negative at another, continuity and the Intermediate Value Theorem would imply that \(f''=0\) somewhere between them. That would be a breakpoint missing from the interval.
Once the sign is established, the second-derivative criterion converts it into concavity. In particular, if \(f''>0\) throughout an interval, then \(f\) is concave up there; if \(f''<0\) throughout, then \(f\) is concave down there. A test value gives the sign for the whole interval only after the breakpoints and the relevant conditions have been checked.
Worked Example: A Function Whose Domain Has Two Components
Worked Example: Concavity of \(\ln(x^2-1)\)
Let \(g(x)=\ln(x^2-1)\). The logarithm requires \(x^2-1>0\), so \(x<-1\) or \(x>1\). The domain has two components: \((-\infty,-1)\) and \((1,\infty)\). Differentiate on these intervals:
The numerator is strictly negative because \(x^2+1>0\), and the denominator is positive wherever the function is defined. Hence \(g''<0\) on both components. For explicit checks, take \(x=-2\) and \(x=2\):
Therefore \(g\) is concave down on \((-\infty,-1)\) and on \((1,\infty)\). Check: both test values lie in the domain, and direct substitution gives negative second derivatives. The points \(-1\) and \(1\) are excluded because \(x^2-1=0\), so there is no domain interval between them to classify.
Worked Example: A Point Where Only the Second Derivative Breaks
Worked Example: Concavity of \(x^{4/3}\)
Let \(h(x)=x^{4/3}\), which is defined for every real \(x\). Its first derivative exists for all real \(x\), including \(0\), and is \(h'(x)=\frac43x^{1/3}\). For \(x\ne0\), differentiate again:
Since \(x^{2/3}>0\) for \(x\ne0\), \(h''(x)>0\) on both intervals \((-\infty,0)\) and \((0,\infty)\). At \(x=0\), the second derivative does not exist: the difference quotient for \(h'\) at zero is
which has no finite limit as \(t\to0\). Thus the sign chart has a breakpoint at \(0\), even though \(h\) itself is defined there. The function is concave up on \((-\infty,0)\) and \((0,\infty)\). Check: substituting \(x=-1\) and \(x=1\) into \(h''\) gives \(h''(-1)=\frac49>0\) and \(h''(1)=\frac49>0\), in agreement with the classification on the two intervals.
A Common Sign-Chart Mistake
A frequent error is to mark only the zeros of \(f''\) and ignore points where \(f''\) is undefined. For the first example, a flawed chart might use only the zero \(x=0\), test \(x=2\), find \(f''(2)=4>0\), and conclude that the function is concave up everywhere to the right of zero.
That conclusion is wrong. The point \(x=1\) is a domain break, and testing \(x=\frac12\) gives \(f''(\frac12)=-14<0\). The correct partition includes both \(0\) and \(1\), producing separate intervals \((0,1)\) and \((1,\infty)\) with different signs. A test value chosen on one side of an omitted breakpoint cannot classify the other side.
Check Your Understanding
For each question, identify the breakpoints before deciding what a test value can tell you.
- For \(f(x)=x^3-3x\), find the zeros of \(f''\), partition the real line, and classify concavity on each interval.
- For \(g(x)=\frac{1}{x^2}\), what points must be excluded from the domain, and what test values would you use to classify concavity on the resulting intervals?
- Why is it important to distinguish a point where \(f\) is undefined from a point where \(f\) is defined but \(f''\) is undefined?
- State the condition that allows one test value to determine the sign of \(f''\) throughout an interval.
- What is wrong with using one test value on an interval that contains a zero or an undefined point of \(f''\)?
