From a Feature List to a Consistent Sketch
A complete curve sketch is not a collection of separate facts. It is a single picture constrained by the domain, intercepts, derivative signs, concavity, and end behavior. When features compete—for example, a local maximum lies on a branch that approaches a vertical asymptote—the sketch must respect all of them at once.
A reliable approach is to analyze the function in order, dividing its domain wherever the behavior could change. Then use the resulting intervals to assemble the graph. The goal is not to make a drawing look plausible; it is to make every part of the drawing follow from the analysis.
A Reliable Order of Analysis
For a function with several potentially competing features, begin by marking the domain breaks and any included endpoints. Find intercepts and symmetry where they help locate the graph. Then calculate \(f'\) and \(f''\), and make sign charts using critical numbers, possible inflection-point inputs, and domain breaks as interval boundaries.
Separate the real line at excluded inputs and show any restricted-domain endpoints. No branch can cross a point outside the domain.
Find intercepts, symmetry, critical numbers, and points where \(f''\) is zero or undefined, while checking that each input belongs to the domain.
Use the signs of \(f'\) for increasing or decreasing behavior and the signs of \(f''\) for concavity. Test each interval, not just the boundary values.
Calculate limits at domain breaks and at the ends of the domain. These determine how the branches approach their boundaries.
Place the points and branches so that every interval trend, concavity, asymptote, and endpoint condition is satisfied.
Why Derivative Signs Constrain the Graph
The connection between derivative signs and shape follows from the Mean Value Theorem. If \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f'(x)>0\) throughout \((a,b)\), then for some \(c\in(a,b)\),
Thus \(f(b)>f(a)\): the function is increasing on that interval. A negative derivative gives the corresponding decreasing behavior. Similarly, if \(f''(x)>0\) throughout an interval, apply the Mean Value Theorem to \(f'\): for \(a<b\), there is a \(c\in(a,b)\) such that
So \(f'\) is increasing there, which gives concavity up. If \(f''<0\), then \(f'\) is decreasing and the graph is concave down. These implications explain why a sign chart determines the direction and bending of each branch, rather than merely suggesting a shape.
Worked Example: A Quartic with Two Minima and Two Inflection Points
Worked Example: A Quartic with Two Minima and Two Inflection Points
Analyze \(f(x)=\frac{x^4}{4}-x^2\) and describe the complete sketch.
Domain, symmetry, and intercepts. This polynomial is defined for every real \(x\), and it is even because \(f(-x)=f(x)\). Its intercepts satisfy
Therefore the \(x\)-intercepts are \((-2,0)\), \((0,0)\), and \((2,0)\); the \(y\)-intercept is also \((0,0)\).
First derivative and critical numbers.
The critical numbers are \(-\sqrt2\), \(0\), and \(\sqrt2\). The sign of \(f'\), in order across the real line, is negative, positive, negative, positive. Therefore \(f\) decreases on \((-\infty,-\sqrt2)\), increases on \((-\sqrt2,0)\), decreases on \((0,\sqrt2)\), and increases on \((\sqrt2,\infty)\).
The function values at those numbers are
So there are local minima at \((-\sqrt2,-1)\) and \((\sqrt2,-1)\), and a local maximum at \((0,0)\). The derivative signs on both sides of each point confirm these classifications.
Second derivative and inflection points.
Setting \(f''(x)=0\) gives \(x=\pm\sqrt{2/3}\). The second derivative is positive when \(|x|>\sqrt{2/3}\) and negative when \(|x|<\sqrt{2/3}\). Its sign changes at both values, so both are inflection-point inputs. Since \(x^2=2/3\) there, \(x^4=4/9\), and the corresponding function value is
The inflection points are \(\left(-\sqrt{2/3},-5/9\right)\) and \(\left(\sqrt{2/3},-5/9\right)\). The graph is concave up outside these inputs and concave down between them.
End behavior and sketch check. The leading term is \(x^4/4\), so \(f(x)\to\infty\) as \(x\to\infty\) and as \(x\to-\infty\). The even symmetry makes the two halves mirror images. Each side descends from the upper end, passes through its intercept at \(x=-2\) or \(x=2\), reaches a minimum at \(x=\pm\sqrt2\), and rises to the central maximum at the origin. The concavity changes on each inner slope at the stated inflection point. Differentiating the original function gives \(x^3-2x\) and then \(3x^2-2\), confirming both derivatives used in the analysis.
Worked Example: Two Branches, a Vertical Asymptote, and a Slant Asymptote
Worked Example: Two Branches, a Vertical Asymptote, and a Slant Asymptote
Analyze \(g(x)=x+\frac{1}{x}\), with its domain split at the excluded input \(x=0\).
Domain, intercepts, and asymptotes. The domain is \((-\infty,0)\cup(0,\infty)\). There is no \(y\)-intercept because \(g(0)\) is undefined. For an \(x\)-intercept, \(x+1/x=0\) would imply \(x^2+1=0\), which has no real solution. Also, \(g(x)-x=1/x\to0\) as \(x\to\pm\infty\), so \(y=x\) is a slant asymptote. At the domain break,
so \(x=0\) is a vertical asymptote, with different directions on its two sides.
Monotonicity and critical points.
The derivative is zero at \(x=-1\) and \(x=1\). Since \(x^2\) is positive on the domain, the derivative is positive when \(|x|>1\) and negative when \(0<|x|<1\). Thus \(g\) increases on \((-\infty,-1)\), decreases on \((-1,0)\) and \((0,1)\), and increases on \((1,\infty)\). The critical values are
The point \((-1,-2)\) is a local maximum; \((1,2)\) is a local minimum.
Concavity and branch behavior.
The negative branch is concave down, while the positive branch is concave up. There is no inflection point at \(x=0\): the function is not defined or continuous there, and the two branches are separate. On the left, the branch approaches the slant asymptote as \(x\to-\infty\), rises to \((-1,-2)\), and then falls toward \(-\infty\) as \(x\to0^-\). On the right, it falls from \(+\infty\) as \(x\to0^+\), reaches \((1,2)\), and rises toward the slant asymptote as \(x\to\infty\). Differentiating \(g'\) gives \(2/x^3\), confirming the concavity calculation.
Worked Example: A Restricted Domain with an Endpoint and an Inflection
Worked Example: A Restricted Domain with an Endpoint and an Inflection
Sketch \(h(x)=x^3-3x\) on the restricted domain \([-2,1]\). This example shows why an endpoint and an interior critical number must be treated differently.
Reference points. The \(x\)-intercepts solve \(x(x^2-3)=0\), giving \(x=0\) and \(x=\pm\sqrt3\). Only \(-\sqrt3\) and \(0\) belong to \([-2,1]\). The endpoints have values
Increasing and decreasing behavior.
The interior critical number is \(x=-1\), where \(h(-1)=-1+3=2\). The derivative is positive on \((-2,-1)\) and negative on \((-1,1)\), so the graph rises from the left endpoint to \((-1,2)\), then falls toward the right endpoint. Although \(h'(1)=0\), \(1\) is an endpoint, not an interior critical number under the definition used here.
Concavity and inflection.
For \(-2<x<0\), \(h''(x)<0\), so the graph is concave down. For \(0<x<1\), \(h''(x)>0\), so it is concave up. The concavity changes at \(x=0\), and \(h(0)=0\), so \((0,0)\) is an inflection point. The completed sketch begins at the closed endpoint \((-2,-2)\), rises through \((-\sqrt3,0)\) to \((-1,2)\), then falls through \((0,0)\) to the closed endpoint \((1,-2)\), changing concavity at the origin. Differentiation verifies \(h'=3x^2-3\) and \(h''=6x\), as used above.
When Features Seem to Compete
Several common conflicts are resolved by respecting the interval structure. A local maximum on one branch does not determine what happens on another branch separated by a vertical asymptote. A zero of \(f''\) is only a candidate for an inflection point: the concavity must actually change, and the point must belong to the graph. A derivative that vanishes at an included endpoint does not make that endpoint an interior critical number. Finally, a branch approaching an asymptote must still follow the derivative and concavity signs on its way there.
Why the Final Audit Matters
A sketch is complete only when the features fit together. For each interval, check that the direction agrees with the sign of \(f'\), the bending agrees with the sign of \(f''\), and the branch approaches any asymptote in the required direction. Also check that intercepts and extrema have the calculated coordinates, that restricted endpoints are shown, and that no branch crosses a gap in the domain.
This audit catches errors that a list of correct calculations can hide. For instance, a branch drawn rising toward an asymptote when the derivative is negative cannot be right, even if the asymptote and critical point are both correctly marked. The interval-by-interval analysis is what turns separate calculations into one justified curve.
Check Your Understanding
Use the examples and the interval-based method to answer these questions.
- For \(f(x)=x^4/4-x^2\), why are \(x=\pm\sqrt{2/3}\) inflection-point inputs rather than merely candidates?
- Why does \(g(x)=x+1/x\) have no inflection point at \(x=0\), even though its concavity signs differ on the two branches?
- For \(h(x)=x^3-3x\) on \([-2,1]\), which point is an interior critical point, and why is \(x=1\) classified differently?
- What does a positive sign for \(f'\) throughout an interval tell you about the graph there?
- Name two checks that can reveal an inconsistency in a completed curve sketch.
