What a Restricted Domain Changes
A function’s formula does not by itself determine which points belong on its graph. The domain specifies which inputs are allowed. If the domain is restricted, the graph retains only the points whose inputs are in that restricted set.
This matters when a curve is visible only on certain intervals, when an interval endpoint is excluded, or when a point that might appear important in the full graph is removed. The derivative information and familiar shape of a curve can still help, but they must be applied only where the function is defined on the restricted domain.
Thus restricting the domain does not change the function’s output at an allowed input. It changes which inputs, and therefore which graph points, are included. If \(D\) consists of several separate intervals, the visible graph may consist of several separate pieces.
A Reliable Sketching Procedure
Start by identifying the domain of the formula, then intersect it with the stated restriction. This prevents a common error: drawing a point that the formula cannot define, or that the problem has explicitly excluded. Once the actual domain is clear, retain the corresponding parts of the curve and remove everything else.
Check the formula’s natural domain and apply the stated restriction. The resulting set is the domain of the graph being sketched.
For each allowed input \(x\), include the point \((x,f(x))\). Do not draw the curve at inputs outside the restricted domain.
If a boundary input belongs to the domain, plot its function value as an included point. If it does not belong, leave that point off the graph; when appropriate, show its location with an open circle.
Derivative signs, concavity, intercepts, and other known features describe the visible curve only where their inputs are allowed.
The open-circle convention is a visual aid, not an extra point on the graph. It indicates a value the curve would have at a boundary input if the formula were defined there, while making clear that the input is excluded. If the formula is undefined there, do not treat a simplified expression as permission to add the missing point.
Why Restriction Means Keeping a Subset of the Graph
The graph of \(f\) is the set of ordered pairs whose first coordinates belong to its domain. A restriction selects some of those pairs and discards the rest. In set notation, if \(D\subseteq A\), then
This identity justifies the sketching procedure. A point \((x,y)\) remains exactly when it was already on the original graph and its input \(x\) lies in \(D\). No new points can appear, and no retained input can acquire a different output. In particular, a turning point or intercept outside \(D\) is not a feature of the restricted graph, even though it may belong to the unrestricted curve.
Worked Example: A Line with a Removed Input
Worked Example: A Line with a Removed Input
Consider \(f(x)=\frac{x^2-1}{x-1}\) on the restricted domain \(D=[0,3]\setminus\{1\}\). Sketch the graph and mark its boundary points.
First, the denominator must be nonzero, so the original formula is defined when \(x\ne 1\). The specified set already excludes \(1\), and the resulting domain is \(D=[0,1)\cup(1,3]\).
For every allowed input, factor the numerator and simplify:
The graph therefore follows the line \(y=x+1\) on the two retained intervals. Its slope is \(1\), since \(\frac{d}{dx}(x+1)=1\), so each visible portion rises from left to right.
At the included left boundary, \(f(0)=0+1=1\), giving the included point \((0,1)\). At the excluded input \(x=1\), the simplified line would have value \(1+1=2\), so indicate the missing point with an open circle at \((1,2)\). At the included right boundary, \(f(3)=3+1=4\), giving the included point \((3,4)\).
Check the boundary calculations in the original formula where it is defined: \(f(0)=(-1)/(-1)=1\), and \(f(3)=8/2=4\). The formula is undefined at \(1\) because its denominator is zero. Thus the sketch is the line segment from \((0,1)\) up to, but not including, \((1,2)\), and the line segment from just after \((1,2)\) through the included point \((3,4)\).
Derivative Information Must Be Filtered Too
Earlier material on first and second derivatives describes how a curve behaves on intervals where the function is defined. Restriction does not invalidate that information, but it narrows where the information applies. If a derivative sign chart includes intervals beyond the restricted domain, those intervals cannot be used to describe visible portions of the graph.
For instance, if \(f'(x)<0\) on an allowed interval, the function decreases there. If that interval contains inputs removed from the domain, the graph has no visible portion over those inputs. A derivative zero outside \(D\) is likewise not a turning point on the restricted graph. Check the inputs against the domain before deciding whether any feature belongs in the sketch.
Worked Example: Two Separate Visible Pieces
Worked Example: Two Separate Visible Pieces
Let \(f(x)=x^2-4\) on \(D=[-2,0]\cup(1,3)\). Identify the visible pieces, their boundary points, and any intercept that remains.
The domain consists of two intervals. The first includes both \(-2\) and \(0\); the second excludes both \(1\) and \(3\). The derivative is \(f'(x)=2x\). On \([-2,0]\), it is negative for \(-2\le x<0\), so the first piece decreases as \(x\) moves toward \(0\). On \((1,3)\), it is positive, so the second piece increases.
Evaluate the endpoints that belong to the domain:
Thus the first piece runs from the included point \((-2,0)\) to the included point \((0,-4)\). On the second interval, the boundary values suggested by the formula are \(f(1)=1-4=-3\) and \(f(3)=9-4=5\), but neither input belongs to \(D\). Show open circles at \((1,-3)\) and \((3,5)\).
The equation \(f(x)=0\) gives \(x^2-4=0\), so \((x-2)(x+2)=0\) and \(x=-2\) or \(x=2\). Both inputs belong to \(D\): the first gives the included point \((-2,0)\), and the second gives \((2,0)\) on the increasing second piece. Direct substitution checks both: \(f(-2)=0\) and \(f(2)=4-4=0\). There is no graph between the two intervals, because no inputs in \((0,1]\) belong to the domain.
Included and Excluded Points Are Different
When an interval endpoint is included, the function value at that input is part of the graph. When the endpoint is excluded, it is not. This remains true even if the formula gives a perfectly ordinary value at the excluded input. The domain, not the appearance of the formula, decides whether the point is included.
Conversely, a formula may be undefined at an input even when a simplified expression appears to give a value there. In the first example, the expression \(x+1\) was equal to the original quotient only for \(x\ne 1\). The original function therefore did not have a value at \(1\), and simplification did not fill the missing point.
Worked Example: A Turning Point Outside the Domain
Worked Example: A Turning Point Outside the Domain
Consider \(g(x)=(x-2)^2\) with domain \(D=[-1,1)\). Describe the visible curve and explain why its vertex is absent.
The full parabola has derivative \(g'(x)=2(x-2)\). For every \(x\in[-1,1)\), we have \(x-2<0\), so \(g'(x)<0\) throughout the retained interval. The visible graph is therefore decreasing over its domain.
The included left boundary gives \(g(-1)=(-1-2)^2=(-3)^2=9\), so \((-1,9)\) is included. At the excluded boundary \(x=1\), the formula gives \(g(1)=(1-2)^2=1\). Mark \((1,1)\) as an open endpoint. The vertex of the full parabola is at \(x=2\), because \(g'(2)=2(2-2)=0\), but \(2\notin D\). It is not on the restricted graph.
As a check on the endpoint and derivative calculations, \(g'(-1)=2(-3)=-6<0\), consistent with decreasing behavior, and the value at the included point is \(g(-1)=9\). The sketch is only the decreasing portion from \((-1,9)\) toward the excluded point \((1,1)\); it does not continue to the vertex.
A Common Mistake: Drawing the Entire Curve
A frequent error is to find an important feature of the unrestricted curve and automatically include it in the restricted sketch. For \(g(x)=(x-2)^2\) on \([-1,1)\), one might calculate \(g'(2)=0\), correctly locate the full parabola’s vertex at \((2,0)\), and then incorrectly draw that vertex as part of the answer.
The correction is to check domain membership: \(2\notin[-1,1)\), so \((2,0)\) is not on the graph of the restricted function. The derivative calculation is correct, but it concerns an input that the restricted function does not allow. Keep the portion of the curve for \(-1\le x<1\), include \((-1,9)\), and leave \((1,1)\) open.
Why Restricted-Domain Sketches Matter
Many mathematical models apply only over a stated range of inputs. A quantity may be meaningful only for nonnegative values, a formula may be used only on a specified interval, or a problem may deliberately remove certain inputs. In each case, the graph must represent the function that is actually defined, not a larger curve suggested by the same formula.
Accurate boundary marks and separate graph pieces also prevent false conclusions. An excluded intercept is not a zero of the restricted function, and an excluded turning point is not a turning point on its graph. The method is straightforward: establish the domain, retain precisely the points above it, and use the known curve shape only on those retained inputs.
Check Your Understanding
For each question, pay attention to the domain before deciding what the graph contains.
- For \(f(x)=x^2\) on \([-2,1)\), which boundary point is included, and where should the excluded boundary be marked?
- If a derivative is zero at an input outside the restricted domain, does that input give a turning point on the restricted graph? Explain.
- For \(f(x)=\frac{x^2-4}{x-2}\) on \([0,3]\setminus\{2\}\), what point is missing from the line obtained by simplifying the formula?
- Why can a restricted domain consisting of two disjoint intervals produce two separate pieces of graph?
- What determines whether a boundary point is included: the value suggested by the formula, or membership of its input in the domain?
