Why the Domain Changes the Sketch
A formula does not determine a curve sketch by itself. The specified domain determines which parts of the curve are present, whether endpoint points are included, and where the graph must stop. A curve may approach a finite height at an excluded endpoint without containing the point at that height. Conversely, an endpoint included in the domain has an actual function value, even when the derivative is not defined there.
For a restricted-domain sketch, analyze the function on the interior of each interval in its domain, then examine every endpoint or excluded boundary separately. Derivative signs describe how the curve behaves between domain boundaries; endpoint values and one-sided limits determine how the graph meets or approaches those boundaries.
A Boundary Checklist
First identify the actual domain, not just the interval suggested by the formula. Mark the endpoints that belong to the domain with filled points; mark excluded endpoints or holes with open points if the corresponding finite limit exists. If a one-sided limit is infinite or does not exist, describe that behavior rather than assigning a finite endpoint height.
On each open interval in the domain, find where the function increases or decreases and where it is concave up or down. Critical points inside the domain are analyzed using the derivative information. An endpoint is not an interior critical point: it has no two-sided neighborhood within the domain. It still matters to the sketch and can be where an absolute maximum or minimum occurs.
Record included endpoints, excluded endpoints, and any excluded points inside the stated interval.
Use derivative signs and concavity information only where the function is defined and differentiable as required.
Calculate the function value at an included endpoint; calculate the appropriate one-sided limit at an excluded boundary.
Use filled points for actual function values and open points for finite limiting heights that are not attained.
Why Derivative Signs Still Control the Interior
The connection between derivative signs and the curve’s direction comes from the Mean Value Theorem. In particular, suppose \(f\) is continuous on a closed interval \([u,v]\) and differentiable on \((u,v)\). The theorem gives a number \(c\) with \(u<c<v\) such that
Since \(v-u>0\), if \(f'(x)>0\) throughout \((u,v)\), then \(f(v)-f(u)>0\). Thus \(f(v)>f(u)\), and the function increases across the interval. If \(f'(x)<0\) throughout, then \(f(v)<f(u)\), so the function decreases. The same reasoning applies when \(u\) or \(v\) is a domain endpoint, provided the continuity and differentiability conditions hold on the specified interval.
This is why an endpoint does not need a derivative in order for us to determine the curve’s direction up to that endpoint. For an excluded endpoint, take two inputs that are still inside the domain: the derivative sign orders their function values. A one-sided limit then describes the height the curve approaches as those inputs move toward the excluded boundary.
Worked Example: A Minimum Inside an Interval with an Open Endpoint
Worked Example: A Minimum Inside an Interval with an Open Endpoint
Sketch \(f(x)=x^2-4x\) on the restricted domain \([0,3)\), and identify its absolute extrema if they exist.
The interior of the domain is \((0,3)\). Differentiate:
Set the derivative equal to zero: \(2x-4=0\), so \(x=2\). For \(0<x<2\), \(f'(x)<0\), so \(f\) decreases. For \(2<x<3\), \(f'(x)>0\), so \(f\) increases. Also,
so the graph is concave up throughout the interior. The interior minimum occurs at \(x=2\), where \(f(2)=2^2-4(2)=4-8=-4\).
The left endpoint is included, so evaluate the function there: \(f(0)=0^2-4(0)=0\). The right endpoint \(3\) is excluded, so there is no function value \(f(3)\) in this restricted domain. Its left-hand limit is
The curve starts at the filled point \((0,0)\), decreases to \((2,-4)\), and then increases toward the open point \((3,-3)\). Its absolute minimum is \(-4\), attained at \(x=2\). Its absolute maximum is \(0\), attained at \(x=0\): on the increasing portion, the values approach \(-3\), which is still below \(0\).
Check: Substitution verifies the minimum value: \(f(2)=-4\). The derivative changes from negative to positive at \(2\), consistent with a minimum, and \(f(0)=0\) verifies the included left endpoint’s plotted value. The limiting height at the open endpoint is \(-3\), not an additional function value.
Endpoint Values, Limits, and Extrema
An endpoint value and a one-sided limit answer different questions. If \(a\) belongs to the domain, \(f(a)\) gives the height of an actual point on the graph. The limit \(\lim_{x\to a^+}f(x)\) describes nearby function values, and it equals \(f(a)\) when \(f\) is continuous from the right there. If \(a\) is excluded, \(f(a)\) is not defined on the restricted domain, even if the one-sided limit is finite.
For absolute extrema, a value must be attained by the function on its domain. A function may approach a number without ever reaching it. The largest value attained is an absolute maximum; the smallest value attained is an absolute minimum. A supremum is a least upper bound, and an infimum is a greatest lower bound; either may exist without being an attained maximum or minimum.
Worked Example: An Included Endpoint and an Unattained Upper Bound
Worked Example: An Included Endpoint and an Unattained Upper Bound
Analyze \(g(x)=\sqrt{x}\) on \([0,4)\), including its endpoint behavior and extrema.
For \(0<x<4\),
Thus \(g\) increases and is concave down on the interior. The left endpoint is included, and \(g(0)=\sqrt{0}=0\), so the graph contains \((0,0)\). The right endpoint is excluded. Its one-sided limit is
Therefore the curve approaches the open point \((4,2)\), but that point is not on the graph. Because \(g\) is increasing from \(0\), its absolute minimum is \(0\), attained at \(x=0\). Its values approach \(2\) from below, so \(2\) is the supremum, but there is no absolute maximum: \(g(x)=2\) would require \(x=4\), which is excluded.
The derivative formula is not defined at \(x=0\), but that does not remove \((0,0)\) from the graph. In fact, the slope of a secant from \((0,0)\) to \((x,\sqrt{x})\) is \(\sqrt{x}/x=1/\sqrt{x}\), which grows without bound as \(x\to0^+\). The curve therefore rises steeply from the included endpoint.
Check: Squaring \(g(x)=2\) gives \(x=4\), confirming that no input in \([0,4)\) attains the limiting height. Also \(g(0)=0\), verifying the included endpoint value and the minimum.
An Excluded Point Inside the Stated Interval
A restricted domain need not be one uninterrupted interval. A formula may be undefined at an interior point, splitting the graph into separate pieces. Analyze each piece with its own one-sided behavior at that excluded point. If both one-sided limits agree at a finite height, the graph has a hole there, not a filled point, unless the function’s domain explicitly includes a value at that input.
Worked Example: A Hole Between Two Domain Pieces
Analyze \(h(x)=\dfrac{x^2-1}{x-1}\) on the domain \([0,2]\) with \(x=1\) excluded.
For every domain input \(x\ne1\), factor the numerator and simplify:
The cancellation is valid only for \(x\ne1\); it does not define the original function at \(x=1\). Thus the domain is \([0,1)\cup(1,2]\). On each piece, \(h'(x)=1>0\) and \(h''(x)=0\), so each piece is increasing and straight. At the included endpoints, \(h(0)=1\) and \(h(2)=3\).
Approaching the excluded point from either side gives
The sketch is the line \(y=x+1\) over the specified domain, with filled points at \((0,1)\) and \((2,3)\), and an open point at \((1,2)\). The absolute minimum is \(1\), attained at \(x=0\), and the absolute maximum is \(3\), attained at \(x=2\). The hole does not prevent either endpoint value from being attained.
Check: For \(x=0\), the original expression gives \((0^2-1)/(0-1)=(-1)/(-1)=1\). For \(x=2\), it gives \((4-1)/(2-1)=3\). At \(x=1\), the original denominator is zero, so the point \((1,2)\) is correctly left open even though the simplified expression approaches \(2\).
Common Pitfall: Plotting a Limit as a Function Value
A frequent error is to substitute an excluded endpoint into a simplified or extended formula and treat the result as an actual endpoint value. For example, for \(g(x)=\sqrt{x}\) on \([0,4)\), the incorrect statement is: “\(g(4)=2\), so the maximum is \(2\).” This both assigns a value outside the domain and confuses a limit with an attained maximum.
The correction is to keep the domain attached to every conclusion: \(4\notin[0,4)\), so \(g(4)\) is not a value of the restricted function. Instead, \(\lim_{x\to4^-}g(x)=2\). Since \(g(x)<2\) for every \(x\) in the domain, \(2\) is a supremum, not a maximum. On a sketch, the correct representation is an open point at \((4,2)\).
Putting the Pieces Together
For a restricted-domain curve sketch, interior derivative and concavity analysis tell you how each piece moves and bends. Boundary analysis tells you where each piece begins or ends, whether those ends belong to the graph, and what heights the curve approaches. These are separate tasks: neither a derivative sign nor a finite limit can turn an excluded input into a domain point.
When comparing candidate heights, include actual endpoint values and interior extrema. Also inspect limits at excluded boundaries, because they can determine whether values are bounded even when no endpoint value exists. A limiting height can be a supremum or infimum without being an extremum; state whether the corresponding value is attained.
Check Your Understanding
For each question, distinguish actual function values from one-sided limits.
- For \(f(x)=x^2-4x\) on \([0,3)\), what are the function value at the included left endpoint and the limit at the excluded right endpoint?
- For \(g(x)=\sqrt{x}\) on \([0,4)\), why is \(2\) a supremum but not a maximum?
- What conditions on a function allow the Mean Value Theorem to establish that a positive derivative implies an increase across an interval?
- For \(h(x)=\dfrac{x^2-1}{x-1}\) with \(x=1\) excluded, why is \((1,2)\) an open point rather than a filled point?
- On a restricted domain, can an endpoint be an absolute extremum even if the derivative is undefined there? Explain briefly.
