Why Poles Require Care
A rational function can increase on one part of its domain and decrease on another, with a pole separating those parts. A derivative sign chart can reveal where the function has local maxima or minima, but it must respect the function’s domain: a pole is not a point where the function has a value, so it cannot be a local extremum.
The key is to analyze each interval of the domain separately. First locate the poles and critical numbers. Then determine the sign of the derivative on every interval they create. For a nonconstant rational function, a change from positive to negative identifies a local maximum, while a change from negative to positive identifies a local minimum. Constant rational functions are an exception to this sign-change rule and will be treated separately.
Domain Points, Poles, and Local Extrema
A rational function has the form \(f(x)=P(x)/Q(x)\), where \(P\) and \(Q\) are polynomials and \(Q(x)\ne 0\) at every point in the function’s domain. Its domain excludes the zeros of the denominator in its defining expression. If a zero of the denominator is a pole, the function is not defined there. A pole can divide the domain into separate intervals, and the derivative signs on one side of a pole do not by themselves determine the signs on the other side.
Critical numbers are domain points where the derivative is zero or does not exist. Poles are excluded from the domain, so they are not critical numbers under this definition. They can still be important in a sign chart because they mark boundaries between domain intervals.
The Sign-Change Rule
Suppose \(c\) is a domain point where a rational function is differentiable. If \(f'(x)>0\) immediately to the left of \(c\), then \(f\) is increasing there. If \(f'(x)<0\) immediately to the right, then \(f\) is decreasing there. This change from increasing to decreasing gives a local maximum at \(c\). The reverse change gives a local minimum.
The signs here are required to hold throughout some sufficiently small interval on each side of \(c\), with no pole or other domain break in those intervals. The theorem classifies stationary points; it does not make an excluded pole into a candidate. For rational functions, the derivative exists at every domain point, so the critical numbers to test are the domain points where the derivative is zero.
Why the Rule Works
On any interval where a differentiable function has positive derivative, the mean value theorem implies that it is increasing. Indeed, for \(x_1<x_2\) in that interval, the theorem gives a point \(\xi\) between them such that \(f(x_2)-f(x_1)=f'(\xi)(x_2-x_1)\). If \(f'(\xi)>0\), then \(f(x_2)>f(x_1)\). A negative derivative gives the reverse inequality and hence decreasing behavior. Applying this reasoning on either side of \(c\) proves the two sign-change conclusions.
There is also a reason to qualify the rule for nonconstant rational functions. The derivative of a rational function is rational. If the function is nonconstant, its derivative cannot be identically zero; when written as a quotient of polynomials, its numerator is therefore a nonzero polynomial. A nonzero polynomial has only finitely many zeros. Consequently, near any particular critical number \(c\), we can choose a small interval on each side with no other derivative zero and no pole. The derivative then has a fixed sign on each of those intervals. If the signs are the same, the function moves in the same direction on both sides, so \(c\) is not a local extremum. If they are opposite, the function changes direction and \(c\) is an extremum.
For completeness, if a differentiable function has a local extremum at an interior point where its derivative exists, Fermat’s theorem gives \(f'(c)=0\). Thus, for a nonconstant rational function, checking its domain critical numbers and the derivative signs around them identifies its local extrema. Constant rational functions must be handled separately: their derivative is zero everywhere, so there is no sign change to test.
A Reliable Sign-Chart Procedure
Locate the poles and any other points excluded from the function’s domain. These points divide the domain into intervals.
Find where \(f'(x)=0\) among domain points. Do not list a pole as a critical number or an extremum candidate.
Use a test value from each resulting interval, keeping the intervals on opposite sides of a pole separate.
For a nonconstant rational function, positive-to-negative gives a local maximum and negative-to-positive gives a local minimum. No sign change gives no local extremum.
Worked Examples
Worked Example: Two Extrema Separated by a Pole
Classify the local extrema of \(f(x)=\dfrac{x^2+1}{x-1}\).
The denominator is zero at \(x=1\), so the domain is \(\mathbb{R}\setminus\{1\}\). Differentiate using the quotient rule:
Set the numerator equal to zero and solve:
These are both in the domain. The denominator \((x-1)^2\) is positive everywhere in the domain, so the sign of \(f'\) is determined by its numerator. Since that numerator is an upward-opening quadratic, it is positive outside its roots and negative between them. The pole at \(1\) divides the negative-sign region into two domain intervals:
At \(x=1-\sqrt{2}\), the derivative changes from positive to negative, so there is a local maximum. At \(x=1+\sqrt{2}\), it changes from negative to positive, so there is a local minimum. To compute their values, rewrite the function by division:
At \(x=1-\sqrt{2}\), we have \(x-1=-\sqrt{2}\), and therefore
At \(x=1+\sqrt{2}\), we have \(x-1=\sqrt{2}\), and therefore
Thus the local maximum is \((1-\sqrt{2},\,2-2\sqrt{2})\), and the local minimum is \((1+\sqrt{2},\,2+2\sqrt{2})\). Check the locations: \(1-\sqrt{2}<1\) and \(1+\sqrt{2}>1\), so each extremum is on its stated side of the pole. The derivative’s signs also verify the two classifications.
Worked Example: A Local Maximum Between Two Poles
Classify the local extrema of \(g(x)=\dfrac{1}{x^2-4}\).
The denominator vanishes at \(x=-2\) and \(x=2\), so the domain consists of \((-\infty,-2)\), \((-2,2)\), and \((2,\infty)\). Differentiate:
The denominator of \(g'\) is positive on the domain. Thus \(g'(x)>0\) when \(x<0\) and \(g'(x)<0\) when \(x>0\). The only domain critical number is \(x=0\), which lies between the two poles. Within that interval the signs change from positive to negative, so \(x=0\) is a local maximum. Its value is
There is no extremum at \(x=-2\) or \(x=2\): neither point belongs to the domain. The classification is consistent with the derivative signs on the interval around zero, and \(g'(0)=0\) confirms that zero is a critical number.
Worked Example: A Pole Is Not an Extremum
Consider \(h(x)=\dfrac{1}{x-1}\). Its domain is \(\mathbb{R}\setminus\{1\}\), and
The function is decreasing on each of the two domain intervals \((-\infty,1)\) and \((1,\infty)\). There are no domain critical numbers because the derivative is never zero. Although \(x=1\) separates the intervals, it is excluded from the domain, and \(h(1)\) does not exist. Therefore \(x=1\) is not a local extremum. The derivative is negative on both sides, but this comparison concerns two distinct intervals and does not turn the pole into a point at which the function is defined.
As a check, differentiating \(1/(x-1)\) gives \(-1/(x-1)^2\), which is indeed negative for every domain point.
Worked Example: The Constant-Function Exception
Let \(r(x)=\dfrac{x^2+1}{x^2+1}\). Since \(x^2+1>0\) for all real \(x\), the domain is all of \(\mathbb{R}\), and \(r(x)=1\) everywhere. Differentiating the quotient confirms the same result:
For any real \(c\) and any \(x\), \(r(x)=1=r(c)\). Hence both \(r(x)\leq r(c)\) and \(r(x)\geq r(c)\) hold for all nearby \(x\). Every point is therefore both a local maximum and a local minimum under the non-strict definitions. There is no derivative sign change because the derivative is zero everywhere. This is why the sign-change classification was stated for nonconstant rational functions.
Check: substituting any real number, such as \(x=0\), gives \(r(0)=1/1=1\), as required.
Common Pitfalls
A frequent error is to mark a pole as an extremum because the derivative sign appears to change across it. For example, the signs around a vertical asymptote might be positive on one side and negative on the other. That does not establish a local maximum: the definition compares function values near a domain point \(c\), and the pole is not a domain point at all.
Another error is to use one test point for an entire region that actually contains a pole. A derivative sign chart must split at each pole. In the first worked example, \(f'\) is negative both on \((1-\sqrt{2},1)\) and on \((1,1+\sqrt{2})\), but those are separate intervals because \(f\) is undefined at \(1\).
Finally, do not apply the sign-change rule blindly to a constant function. A statement such as “a local extremum must have a derivative sign change” is false for constants: every domain point of a constant function is both a local maximum and a local minimum. State the nonconstant condition, or check constant functions separately.
Why These Classifications Matter
Local extrema describe changes in a function’s behavior near particular domain points. In rational functions, poles make the domain disconnected, so a complete analysis must combine algebraic work with attention to where the function exists. Correctly separating intervals prevents false comparisons across excluded points and ensures that a local extremum is assigned only where the function actually has a value.
The practical sequence is straightforward: establish the domain, split at poles, find domain critical numbers, and test derivative signs on each resulting interval. The derivative signs classify the direction of change; the domain determines which points are eligible to be extrema. Constant rational functions are the special case in which every domain point is both types of local extremum, irrespective of sign changes.
Check Your Understanding
Use the domain and derivative signs carefully in each question.
- For \(f(x)=\dfrac{x^2+1}{x-1}\), why must the sign chart be split at \(x=1\), even though \(f'\) is negative on both sides of that pole between the two critical numbers?
- For \(g(x)=\dfrac{1}{x^2-4}\), identify the domain critical number and state the derivative sign change there.
- Why can a pole never itself be a local maximum or minimum?
- What derivative sign changes indicate a local maximum and a local minimum for a nonconstant rational function?
- Why does the derivative sign-change rule need a separate qualification for constant rational functions?
