Why Gaps in the Domain Matter
A derivative sign chart records where a function is increasing or decreasing. When the domain has a gap, however, the intervals on either side of that gap are separate parts of the function’s domain. A change in derivative sign across the gap does not mean that the function turns at the missing input: there is no function value there to be a turning point.
This distinction matters both for identifying extrema and for describing monotonicity. A function can be decreasing on every component of its domain without being decreasing on the entire disconnected domain. Each conclusion must be stated on the set where it is actually justified.
Domain Components and Monotonicity
A domain is disconnected when it consists of separated pieces, rather than one interval. For example, \(\mathbb{R}\setminus\{1\}\) consists of the two intervals \((-\infty,1)\) and \((1,\infty)\). An excluded input such as \(1\) divides the domain into intervals on which derivative signs can be analyzed separately.
A function is increasing on a set if \(x_1<x_2\) in that set implies \(f(x_1)\leq f(x_2)\). It is decreasing if \(x_1<x_2\) implies \(f(x_1)\geq f(x_2)\). When we use derivative signs to describe monotonicity, we usually state the result on each interval in the domain. A conclusion for each interval does not automatically give the same conclusion on their union: comparing points from different intervals also requires checking their function values.
Worked Example: A Sign Change Across a Missing Input
Worked Example: The derivative changes sign at an excluded input
Consider \(f(x)=\ln|x|\), whose domain is \(\mathbb{R}\setminus\{0\}\). Find its derivative signs on the domain intervals and decide whether the sign change gives an extremum at \(0\).
For \(x\ne 0\), the derivative of \(\ln|x|\) is
The domain intervals are \((-\infty,0)\) and \((0,\infty)\). If \(x<0\), then \(1/x<0\), so \(f\) is decreasing on \((-\infty,0)\). If \(x>0\), then \(1/x>0\), so \(f\) is increasing on \((0,\infty)\).
The derivative signs change from negative to positive across \(0\). But \(f(0)\) is not defined, so \(0\) cannot be a local minimum. The signs describe a decreasing branch to the left and an increasing branch to the right, not a minimum at the gap.
As a check, \(f(-1)=\ln 1=0\) and \(f(1)=\ln 1=0\). Both values are defined, but neither supplies a value at \(0\). Differentiating \(\ln|x|\) on either interval gives \(1/x\), as used in the sign analysis.
Why the First Derivative Test Cannot Cross a Gap
The derivative sign test for a local extremum concerns behavior on the two sides of a point in the domain. The point itself must be present: the definition of a local extremum includes the value \(f(c)\). If \(c\) is excluded, that necessary value does not exist.
There is also a useful reason not to infer a comparison across a gap from derivative signs. The Mean Value Theorem applies to a function continuous on a closed interval and differentiable on its interior. If an excluded input lies between two points, the function is not defined throughout that closed interval. The theorem therefore cannot be applied across the gap to compare the endpoint values. Derivative signs can still establish monotonicity separately on each interval where the hypotheses hold.
Worked Example: Decreasing on Both Pieces, but Not on Their Union
Worked Example: Monotonicity on components does not settle global monotonicity
Let \(g(x)=\dfrac{1}{x-2}\), with domain \(\mathbb{R}\setminus\{2\}\). Determine its monotonicity on each domain interval, then check whether it is decreasing on the entire domain.
Differentiate:
For every \(x\ne 2\), the denominator \((x-2)^2\) is positive, so \(g'(x)<0\). Thus \(g\) is decreasing on \((-\infty,2)\) and on \((2,\infty)\).
To test decreasing behavior on the whole domain, compare points from opposite sides of the gap. Since \(1<3\), a decreasing function on the whole domain would have to satisfy \(g(1)\geq g(3)\). But
So \(g\) is not decreasing on the entire disconnected domain. Nor is it increasing there: \(2.5<3\), but \(g(2.5)=2\) and \(g(3)=1\), so the function value decreases for that pair. The derivative signs correctly establish decrease on each component; they do not by themselves determine comparisons across the gap.
As a check, differentiating \((x-2)^{-1}\) gives \(-1\cdot(x-2)^{-2}\cdot 1=-1/(x-2)^2\), which is negative on both intervals and undefined at the excluded input \(2\).
Constructing a Sign Chart Around an Excluded Input
To read a chart correctly, distinguish three kinds of locations: zeros of the derivative that lie in the domain, points in the domain where the derivative does not exist, and inputs excluded from the domain. The first two may divide an interval into regions of constant derivative sign. The third divides the domain itself, so the intervals on either side must be treated as separate pieces.
Mark every excluded input. These inputs divide the domain into its separate intervals.
Include only points that lie in the relevant domain interval, and keep excluded inputs marked as gaps.
Use a valid sign analysis or a test value from each interval; do not combine intervals across a gap.
Record where the function increases or decreases, then separately check any claim about comparisons across different components.
A change in signs across a gap does not create an extremum at the missing input.
Worked Example: Domain Gap and Genuine Extrema
Worked Example: Classifying points on either side of a gap
Consider \(h(x)=\dfrac{(x-2)^2}{x-1}\), with domain \(\mathbb{R}\setminus\{1\}\). Find the derivative signs and classify any local extrema at domain points.
By the quotient rule,
The denominator is positive wherever \(x\ne 1\). The derivative is zero at \(x=0\) and \(x=2\), and undefined at the excluded input \(x=1\). The relevant intervals are therefore \((-\infty,0)\), \((0,1)\), \((1,2)\), and \((2,\infty)\).
On \((-\infty,0)\), both \(x\) and \(x-2\) are negative, so their product is positive and \(h'(x)>0\). On \((0,1)\), \(x>0\) and \(x-2<0\), so \(h'(x)<0\). On \((1,2)\), the same factor signs give \(h'(x)<0\). On \((2,\infty)\), both factors are positive, so \(h'(x)>0\).
Thus \(h\) increases on \((-\infty,0)\), decreases on \((0,1)\) and \((1,2)\), and increases on \((2,\infty)\). At \(x=0\), the derivative sign changes from positive to negative, and \(0\) is in the domain, so \(h\) has a local maximum there. Its value is
At \(x=2\), the derivative sign changes from negative to positive, and \(2\) is in the domain, so \(h\) has a local minimum there. Its value is
The excluded input \(1\) is not a third extremum. The derivative is negative on both sides of \(1\), so the function decreases on the intervals adjacent to the gap; the gap itself has no function value. Finally, the derivative calculation is checked by differentiating the quotient directly: its numerator simplifies to \(x(x-2)\), while its denominator is \((x-1)^2\), exactly as above.
Common Mistake: Treating a Gap Like a Critical Point
A common error is to put an excluded input into the chart as if it were a point of the function, observe a sign change, and label it a local maximum or minimum. For example, for \(f(x)=\ln|x|\), one might write “\(f'\) changes from negative to positive at \(0\), so \(f\) has a minimum at \(0\).” This conclusion is invalid: \(f(0)\) is undefined, and a local minimum requires a function value at the candidate point.
The correction is to write the two intervals separately: \(f\) decreases on \((-\infty,0)\) and increases on \((0,\infty)\). Then state that \(0\) is excluded and is not an extremum. A sign chart is a record of derivative behavior on the domain, not a way to fill in missing inputs.
Why This Distinction Matters
Keeping domain components separate prevents two kinds of overclaiming. First, it prevents labeling a missing input as an extremum. Second, it prevents concluding that behavior on each component determines monotonicity on the entire domain. When the domain has gaps, claims about the function as a whole may require direct comparisons between values in different components.
For every sign chart, make the domain visible and label excluded inputs distinctly from domain points where the derivative is zero or undefined. That small bookkeeping step makes the conclusions more precise: monotonicity is reported on the correct intervals, and extrema are considered only where the function exists.
Check Your Understanding
Use the definitions and sign-chart principles above to answer these questions.
- Why does a change in derivative sign across an excluded input not establish a local extremum there?
- If a function is decreasing on each of two domain intervals separated by a gap, what additional issue must be checked before calling it decreasing on the union?
- For \(f(x)=\ln|x|\), state the derivative sign and monotonicity on each domain interval.
- What condition in the Mean Value Theorem fails when an excluded input lies inside the interval being considered?
- For \(h(x)=\dfrac{(x-2)^2}{x-1}\), which domain points are local extrema, and what happens at the excluded input?
