Why Begin a Curve Sketch with Intervals?
A curve sketch brings together several kinds of information: where the function is defined, what happens near boundaries, and how its derivatives behave. If these facts are recorded without a clear interval structure, it is easy to overlook a break in the graph or to assign a sign to a region where the function is not defined.
The first step is to map the domain. Then mark any additional inputs needed to organize the analysis, such as zeros of a derivative or points where a derivative is undefined. These points divide the real line into intervals on which information can be examined consistently.
Domain Breaks and Analysis Boundaries
The domain of a function \(f\) is the set of inputs for which \(f(x)\) is defined. Domain restrictions can arise, for example, when a denominator is zero or when a real-valued square root has a negative radicand.
A domain break is an input excluded from the domain that separates parts of the domain. A sketch interval is an interval used to organize the analysis. Its endpoints may be domain breaks, included domain endpoints, zeros of a derivative, points where a derivative is undefined, or other inputs relevant to the sketch.
These terms describe different roles. Suppose \(c\) is excluded from the domain. The intervals to its left and right belong to separate pieces of the domain. By contrast, if \(f(c)\) is defined but \(f'(c)=0\), then \(c\) can be used to divide the derivative analysis even though the graph is not broken there.
Interval notation records whether an endpoint is included. For example, \([-2,1)\) includes \(-2\) but excludes \(1\), while \((1,\infty)\) contains inputs strictly greater than \(1\). If a point is included in the domain but is used to divide the analysis, record the point itself as well as the adjacent open intervals. Do not silently discard a defined endpoint or assume the function is defined at an excluded one.
A Reliable Partitioning Procedure
First determine the domain from the original function. Then collect the inputs that matter to the sketch and sort them from least to greatest. The resulting intervals are the places where you check one-sided behavior and record derivative or concavity signs. The process is an organizational tool: it does not, by itself, classify extrema or establish a complete graph.
Record every excluded input and every included endpoint. Do this from the original expression, before simplifying it.
Add inputs where a derivative or another quantity being studied is zero or undefined, when those inputs belong to the domain or split the relevant domain pieces.
Place them on the real line from left to right. The spaces between consecutive boundaries form the open intervals for the analysis.
For every interval, record whether \(f\) is defined there and the applicable one-sided behavior and derivative signs. Mark separately whether each boundary input is included in the domain.
Why Sign Information Is Organized by Intervals
An interval partition is useful because a continuous quantity cannot change sign on an interval without taking the value zero somewhere in between. This is a consequence of the Intermediate Value Theorem: if a continuous function has a positive value at one input and a negative value at another, it must equal zero at some input between them.
Apply this fact to \(q=f'\), but only on an interval where \(f'\) exists and is continuous. If \(f'\) has no zero on that interval, its sign cannot change there. The same reasoning applies to \(f''\) on an interval where \(f''\) is continuous and has no zero. Thus zeros and points where the relevant derivative is undefined are natural places to divide the sign analysis. This does not say that every derivative is continuous everywhere, or that every function has the same set of boundaries.
Worked Example: Simplification Does Not Restore an Excluded Input
Worked Example: A removable domain break
Consider \(f(x)=\frac{x^2-1}{x-1}\). Find its domain, simplify the formula where possible, and state the intervals for the domain analysis.
The denominator cannot be zero, so \(x-1\ne 0\), which gives \(x\ne 1\). Therefore the domain is \((-\infty,1)\cup(1,\infty)\).
Factor the numerator: \(x^2-1=(x-1)(x+1)\). For \(x\ne 1\), cancellation gives \(f(x)=x+1\). The condition \(x\ne 1\) remains in force: cancellation simplifies the values on the domain but does not add \(1\) to it.
The domain break at \(1\) gives two sketch intervals, \((-\infty,1)\) and \((1,\infty)\). On either interval, the simplified expression has derivative \(f'(x)=1\), so the derivative is positive on both intervals.
As a check on the domain distinction, substituting \(x=1\) into the original expression gives \(\frac{0}{0}\), which is undefined. The simplified expression would give \(2\), but that value belongs to an extension of the formula, not to the original function. The graph follows \(y=x+1\) on its domain and omits the point \((1,2)\).
Worked Example: Combining Derivative and Concavity Boundaries
Worked Example: A polynomial with several analysis intervals
For \(f(x)=x^3-3x\), find the boundaries needed to organize the signs of \(f'\) and \(f''\), then list the signs on each interval.
A polynomial is defined for every real number, so there are no domain breaks. Differentiate: \(f'(x)=3x^2-3=3(x-1)(x+1)\). The derivative is zero at \(x=-1\) and \(x=1\). Differentiate again: \(f''(x)=6x\), which is zero at \(x=0\).
Combining those boundaries and ordering them gives \(-1,0,1\). The intervals to check are \((-\infty,-1)\), \((-1,0)\), \((0,1)\), and \((1,\infty)\). The sign of \(f'\) follows from \(3(x-1)(x+1)\); the sign of \(f''\) follows from \(6x\).
| Interval | Sign of \(f'\) | Sign of \(f''\) |
|---|---|---|
| \((-\infty,-1)\) | Positive | Negative |
| \((-1,0)\) | Negative | Negative |
| \((0,1)\) | Negative | Positive |
| \((1,\infty)\) | Positive | Positive |
For example, on \((-1,0)\), \(x-1<0\) and \(x+1>0\), so \(f'(x)<0\); also \(x<0\), so \(f''(x)<0\). The other rows follow by checking the factors on each interval. The function remains defined at all three boundary inputs. The boundaries organize the sign information; they are not domain breaks.
As a check, evaluate the derivatives at the boundaries: \(f'(-1)=0\), \(f'(1)=0\), and \(f''(0)=0\), as required by the equations used to select them.
Worked Example: A Domain Endpoint and an Excluded Input
Worked Example: A square root and a denominator restriction
Consider \(g(x)=\frac{\sqrt{x+2}}{x-1}\). Determine its domain, intervals, and derivative signs on the interiors of those intervals.
For a real square root, \(x+2\ge 0\), so \(x\ge -2\). The denominator also requires \(x\ne 1\). Thus the domain is \([-2,1)\cup(1,\infty)\). The input \(-2\) is included, while \(1\) is excluded.
For \(x>-2\) and \(x\ne 1\), differentiate using the product rule, writing \(g(x)=(x+2)^{1/2}(x-1)^{-1}\): $$ g'(x)=\frac{1}{2\sqrt{x+2}(x-1)}-\frac{\sqrt{x+2}}{(x-1)^2} =\frac{(x-1)-2(x+2)}{2\sqrt{x+2}(x-1)^2} =\frac{-x-5}{2\sqrt{x+2}(x-1)^2}. $$ On the derivative’s domain, \(x>-2\), so \(-x-5<-3<0\). The denominator is positive because \(\sqrt{x+2}>0\) and \((x-1)^2>0\). Therefore \(g'(x)<0\) on both \((-2,1)\) and \((1,\infty)\).
The endpoint value is \(g(-2)=\frac{0}{-3}=0\). The derivative formula does not apply at \(-2\), since its denominator contains \(\sqrt{x+2}\), which is zero there. The sketch intervals are \([-2,1)\) and \((1,\infty)\), with derivative signs recorded on their interiors. As a check, substituting any allowed input into the derivative formula preserves a negative numerator and a positive denominator, confirming the sign conclusion.
One-Sided Behavior Belongs to Each Side
At an excluded input, the behavior from the left and right must be recorded separately whenever both sides are relevant. The two sides may approach different finite values, the same value, or unbounded values. At an included domain endpoint, only the side that lies within the domain is relevant. This is why interval labels and endpoint inclusion matter: they identify which one-sided questions can be asked.
For example, in the first worked example the simplified formula gives \(x+1\) on both sides of \(1\), so both one-sided limits are \(2\), even though \(f(1)\) is undefined. In the square-root example, the domain begins at \(-2\), so there is no part of the domain immediately to the left of that endpoint. The interval map prevents either distinction from being hidden in a single broad sign chart.
Common Mistake: Treating Every Boundary as a Domain Break
A point where \(f'(c)=0\) is a useful boundary for organizing derivative signs, but it is not automatically excluded from the domain. In the polynomial example, \(f\) is defined at \(-1\), \(0\), and \(1\). Calling these domain breaks would incorrectly suggest that the graph is missing those points.
The reverse error is also common: canceling a factor and then treating a formerly excluded input as defined. In the rational example, the original denominator still vanishes at \(1\), so the original function remains undefined there. Keep the domain record separate from any simplified expression.
Using the Interval Map in a Sketch
Once the intervals are established, maintain a consistent record for each one. Note whether the function is defined throughout it, record the applicable derivative signs, and examine one-sided behavior at its boundaries. Then combine this information with other known features of the curve. A complete sketch should preserve domain breaks and included endpoints rather than joining pieces just because their formulas look similar.
The interval map is not a substitute for computing limits or derivatives. Its purpose is to make clear where each calculation applies, which side of a boundary it describes, and whether that boundary is part of the graph. This simple discipline helps prevent gaps in the reasoning as well as gaps or false connections in the final sketch.
Check Your Understanding
For each question, distinguish carefully between the domain and the intervals used to organize analysis.
- Why does canceling a factor in a rational expression not necessarily add the canceled input to the original function’s domain?
- For \(h(x)=\sqrt{x-3}\), state the domain and identify the included endpoint.
- If \(f'(c)=0\) and \(f(c)\) is defined, is \(c\) necessarily a domain break? Explain.
- What continuity condition allows you to conclude that a quantity cannot change sign on an interval without being zero there?
- Why should one-sided behavior at an excluded input be recorded separately on each side?
