Why Poles Need Special Attention
A rational function can be concave up on one side of a vertical asymptote and concave down on the other. A sign change in the second derivative may therefore occur across a pole. But a pole is not a point on the graph, so it cannot be an inflection point. To analyze concavity correctly, keep track of both the signs of the second derivative and the points where the function is not defined.
The method is to find the domain, compute and factor \(f''\), and divide the real line at every domain break and every zero of \(f''\). Test the sign on each resulting interval. A candidate is an inflection point only if the function is defined there and the concavity changes across it.
Poles, Holes, and Inflection Points
For example, canceling a factor can simplify a formula on its domain, but it does not change which inputs the original expression allows. A hole and a pole are both domain breaks. Neither can be an inflection point because neither has a graph point there.
For a twice-differentiable function on an interval, \(f''(x)>0\) indicates concavity up there, while \(f''(x)<0\) indicates concavity down. To identify the intervals for a rational function, use the second-derivative signs only on intervals contained in its domain. A zero of \(f''\) is a candidate for an inflection point, not proof of one; a domain break is never an inflection point.
A Sign-Chart Method
First record the original domain. Then find a simplified expression for \(f''\), while retaining every excluded input from the original function. The points that divide the sign chart are the domain breaks and the zeros of \(f''\). If factors in the formula for \(f''\) appear to create extra breaks, simplify them and check the original domain before treating them as genuine.
Find where the original denominator is zero. Keep canceled factors marked if the original expression excludes those inputs.
Differentiate on the domain, then factor the result so its signs can be tested.
Use the domain breaks and the zeros of \(f''\) as dividing points. Test the sign of \(f''\) within each domain interval.
A candidate must belong to the domain, and the signs must show a change in concavity across it.
Here is why the sign test works. On any one of these intervals, a rational function’s second derivative is continuous, because the interval contains no denominator zero of \(f''\). If \(f''\) had both positive and negative values on that interval, the intermediate value theorem would force \(f''\) to equal zero somewhere between them. But all zeros of \(f''\) have already been used to divide the intervals. Thus its sign is constant on each interval, and one test value determines the concavity there.
Worked Examples
Worked Example: A Sign Change Across a Pole
Determine the concavity intervals and inflection points of \(f(x)=1/(x-2)\).
The denominator is zero at \(x=2\), so the domain is \(\mathbb{R}\setminus\{2\}\). Write \(f(x)=(x-2)^{-1}\) and differentiate on this domain:
These derivatives can be checked by differentiating the preceding expressions: the derivative of \(-(x-2)^{-2}\) is \(2(x-2)^{-3}\). For \(x<2\), the denominator \((x-2)^3\) is negative, so \(f''(x)<0\). For \(x>2\), it is positive, so \(f''(x)>0\). The function is concave down on \((-\infty,2)\) and concave up on \((2,\infty)\).
The concavity changes across \(2\), but \(f(2)\) is undefined. Therefore, there is no inflection point at \(x=2\). The sign change occurs across a pole, not at a point on the graph.
Worked Example: Concavity Changes at a Point on the Graph
Determine the concavity intervals and inflection points of \(f(x)=x/(x^2-1)\).
The denominator factors as \((x-1)(x+1)\), so the domain excludes \(x=-1\) and \(x=1\). The quotient rule gives
Differentiate this expression using the product rule:
For a derivative check, differentiating \(-(x^2+1)(x^2-1)^{-2}\) gives the first line above. The factor \(x^2+3\) is positive for every real \(x\), so the sign of \(f''\) depends on \(x\) and \((x^2-1)^3\). The critical dividing inputs for this sign chart are \(-1,0,1\).
| Interval | Sign of \(x\) | Sign of \((x^2-1)^3\) | Sign of \(f''\) | Concavity |
|---|---|---|---|---|
| \((-\infty,-1)\) | Negative | Positive | Negative | Down |
| \((-1,0)\) | Negative | Negative | Positive | Up |
| \((0,1)\) | Positive | Negative | Negative | Down |
| \((1,\infty)\) | Positive | Positive | Positive | Up |
At \(x=0\), the function is defined and \(f(0)=0/(0-1)=0\). The concavity changes from up on \((-1,0)\) to down on \((0,1)\), so \((0,0)\) is an inflection point. At \(x=-1\) and \(x=1\), the function is undefined: the numerator is nonzero at both inputs, while the denominator is zero. These are poles, not inflection points.
Worked Example: A Canceled Factor Does Not Fill a Hole
Consider \(f(x)=(x^2-1)/(x-1)\), with the domain specified by the displayed expression. Find its concavity and determine whether the excluded input is an inflection point.
The original denominator excludes \(x=1\). Factoring the numerator and simplifying for \(x\ne1\) gives
On its domain, \(f'(x)=1\) and \(f''(x)=0\). These results check directly by differentiating \(x+1\), which is equal to the original function wherever the original function is defined. The function is linear on \((-\infty,1)\) and \((1,\infty)\), so it has no concavity change on either interval. At \(x=1\), its simplified formula would give \(2\), but the original expression does not define \(f(1)\). Thus the graph has a hole at \((1,2)\), not an inflection point.
A Common Mistake
A frequent error is to identify every change in the sign of \(f''\) as an inflection point. For \(f(x)=1/(x-2)\), one might reason incorrectly as follows:
The conclusion fails because \(f(2)\) does not exist. The corrected conclusion is that the function is concave down to the left of \(2\), concave up to the right, and has no inflection point at \(2\). A concavity change across a pole describes two separate pieces of the graph; it does not create a point joining them.
A second practical safeguard is to preserve the domain while canceling factors. Simplifying a formula is useful for differentiation, but it does not silently add a missing graph point. Mark the original domain breaks before simplifying, and check each potential inflection input against that domain.
Check Your Understanding
Use a sign chart where needed, and distinguish concavity changes from domain breaks.
- For \(f(x)=1/(x+3)\), find the concavity on either side of the pole. Is the pole an inflection point?
- For \(g(x)=x/(x^2-4)\), identify the domain breaks and the zeros of its second derivative.
- Why is a zero of \(f''\) a candidate for an inflection point rather than a guarantee?
- For \(h(x)=(x^2-9)/(x-3)\), with the domain of the displayed expression, explain why \(x=3\) is not an inflection input.
- State the condition, in addition to a change of concavity, that an input must satisfy to be an inflection point.
