From Given Evidence to a Graph
A graph-analysis free-response question may give several kinds of evidence at once: selected values of \(f\), signs of \(f'\), signs of \(f''\), intercept information, and a restricted domain. The task is not to guess a familiar-looking curve. It is to translate each fact into a graph feature, then check that all the features fit together.
A good sketch is therefore a visual summary of justified conclusions. It should show where the graph exists, where it crosses or touches an axis, where it rises or falls, where it is concave up or down, and where extrema or inflection points are supported. It should not imply exact coordinates or smooth behavior that the information does not establish.
Translate Each Kind of Evidence
Before drawing, organize the information by role. A stated value \(f(a)=b\) gives the point \((a,b)\). A zero \(f(a)=0\) gives an \(x\)-intercept, while \(f(0)=b\) gives the \(y\)-intercept when \(0\) is in the domain. The domain specifies where the graph may exist; a coordinate excluded from the domain cannot be plotted as a point on the graph.
| Evidence | Graph conclusion |
|---|---|
| \(f(a)=b\) | The graph contains the point \((a,b)\). |
| \(f'(x)>0\) on an interval | \(f\) is increasing on that interval. |
| \(f'(x)<0\) on an interval | \(f\) is decreasing on that interval. |
| \(f''(x)>0\) on an interval | The graph is concave up on that interval. |
| \(f''(x)<0\) on an interval | The graph is concave down on that interval. |
| \(f'\) changes from positive to negative at an interior point | There is a local maximum there, provided the function is defined and differentiable as required by the test. |
| \(f'\) changes from negative to positive at an interior point | There is a local minimum there, provided the function is defined and differentiable as required by the test. |
| Concavity changes at \(x=c\) | An inflection point requires a graph point at \(x=c\); a concavity change across a missing coordinate is not enough. |
For extrema, the sign change matters. A zero of \(f'\) alone does not establish a maximum or minimum; the derivative may keep the same sign on both sides. Also distinguish an interior local extremum from an absolute extremum on a closed interval: endpoint values may affect the absolute classification.
A Reliable Sketching Sequence
Mark the domain and the given points first. Then record the derivative signs on intervals separated by critical numbers or domain breaks. Use those signs to draw the overall direction of the graph. Add extrema only where the relevant sign changes support them. Next use second-derivative signs to indicate concavity, and mark an inflection point only when the graph exists at the coordinate and its concavity changes there. Finally place known intercepts and any roots whose existence follows from the evidence.
Mark endpoints, excluded coordinates, and every supplied point \((a,f(a))\). Put stated intercepts on the appropriate axes.
Separate intervals at zeros of \(f'\) and at domain breaks. Record increasing or decreasing behavior and classify supported extrema.
Mark concavity on each interval. Investigate a possible inflection coordinate only if the function has a graph point there.
Use supplied zeros or continuity and monotonicity to justify any additional root claims. Compare every feature with the stated values and domain.
Why Derivative Signs Determine the Direction
The connection between derivative signs and graph direction follows from the Mean Value Theorem. If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then for some \(c\) between \(a\) and \(b\),
If \(f'(x)>0\) throughout an interval, take any \(a<b\) in that interval. The theorem gives \(f(b)-f(a)=f'(c)(b-a)>0\), so \(f(b)>f(a)\). Thus \(f\) is increasing. If \(f'(x)<0\), the same reasoning gives \(f(b)<f(a)\), so \(f\) is decreasing. This is why a sign chart controls the broad shape of a sketch, rather than merely indicating isolated tangent slopes.
Worked Example: Sketching a Cubic from Its Features
Worked Example: Sketching a Cubic from Its Features
Suppose \(f(x)=x^3-3x\) on \([-2,2]\). Determine its intercepts, extrema, and concavity, then describe a consistent sketch.
First evaluate the function at the endpoints and at convenient interior values:
To find the zeros, factor the function:
Thus the \(x\)-intercepts are \((-\sqrt{3},0)\), \((0,0)\), and \((\sqrt{3},0)\); the \(y\)-intercept is \((0,0)\) as well.
Differentiate and factor:
The critical numbers are \(x=-1\) and \(x=1\). The sign of \(f'\) is positive for \(x<-1\), negative for \(-1<x<1\), and positive for \(x>1\). Therefore the graph increases, then decreases, then increases. At \(x=-1\), the sign changes from positive to negative, so \(f(-1)=2\) is a local maximum. At \(x=1\), it changes from negative to positive, so \(f(1)=-2\) is a local minimum.
Now calculate the second derivative:
For \(x<0\), \(f''(x)<0\), so the graph is concave down. For \(x>0\), \(f''(x)>0\), so it is concave up. The function is defined at \(x=0\), and \(f(0)=0\), so the concavity change gives an inflection point at \((0,0)\).
The sketch must pass through the listed points, cross the \(x\)-axis at the three zeros, rise to the local maximum \((-1,2)\), fall to the local minimum \((1,-2)\), and change concavity at \((0,0)\). The endpoints are \((-2,-2)\) and \((2,2)\). As a check, substituting \(x=-1\) and \(x=1\) into \(f\) gives the stated extremum heights, and differentiating \(f\) gives the derivative signs used above.
Inflection Points Require a Point on the Graph
A change in concavity is important evidence, but it does not by itself guarantee an inflection point at the dividing coordinate. In the usual AP setting, an inflection point is a point on the graph where the function changes concavity. If \(f''\) changes sign across \(c\), but \(f(c)\) does not exist, there is no graph point at \(x=c\) to call an inflection point. If \(f(c)\) exists but its value is not supplied, you may mark the horizontal coordinate of the change, but you cannot assign an unsupported vertical coordinate.
Worked Example: A Concavity Change at an Excluded Coordinate
Let \(f(x)=\frac{1}{x-1}\) on the domain \([-1,2]\setminus\{1\}\). Determine whether \(x=1\) is an inflection point.
The domain excludes \(1\), so \(f(1)\) is undefined. Differentiate on the domain:
For \(x<1\), \((x-1)^3<0\), so \(f''(x)<0\) and the graph is concave down. For \(x>1\), \((x-1)^3>0\), so \(f''(x)>0\) and the graph is concave up. Concavity changes across \(x=1\), but \(f(1)\) does not exist. Therefore \(x=1\) is not an inflection point.
There is no \(x\)-intercept because \(\frac{1}{x-1}=0\) has no solution. Since \(0\) is in the domain, \(f(0)=1/(0-1)=-1\), so the graph contains the \(y\)-intercept \((0,-1)\). Also \(f'(x)<0\) on both parts of the domain, so the graph decreases on \([-1,1)\) and on \((1,2]\). It has two separate branches, not a point at \(x=1\). The key correction is to show the concavity change on either side of the excluded coordinate without placing an inflection point there.
Worked Example: Combining Values, Extrema, Roots, and Concavity
Worked Example: Combining Values, Extrema, Roots, and Concavity
Suppose \(f\) is continuous on \([-3,3]\), differentiable on \((-3,3)\), and the following information is given:
- \(f(-3)=-2,\ f(-1)=3,\ f(0)=1,\ f(2)=-1,\ f(3)=2\).
- \(f'(x)>0\) on \((-3,-1)\), \(f'(x)<0\) on \((-1,2)\), and \(f'(x)>0\) on \((2,3)\).
- \(f''(x)<0\) on \((-3,0)\) and \(f''(x)>0\) on \((0,3)\).
At \(x=-1\), \(f'\) changes from positive to negative, so there is a local maximum at the given point \((-1,3)\). At \(x=2\), \(f'\) changes from negative to positive, so there is a local minimum at \((2,-1)\). The graph increases from \((-3,-2)\) to \((-1,3)\), decreases from \((-1,3)\) to \((2,-1)\), and increases from \((2,-1)\) to \((3,2)\).
Because \(f\) is continuous and changes from negative at \(-3\) to positive at \(-1\), the Intermediate Value Theorem guarantees at least one zero in \((-3,-1)\). Strict increase there guarantees at most one zero, so there is exactly one. Between \(-1\) and \(2\), the values decrease from \(3\) to \(-1\); continuity guarantees a zero, and strict decrease makes it unique. Between \(2\) and \(3\), the values increase from \(-1\) to \(2\), so there is exactly one zero there as well. The evidence establishes three roots, but not their exact coordinates.
The graph is concave down on \((-3,0)\) and concave up on \((0,3)\). Since the supplied value \(f(0)=1\) places the point \((0,1)\) on the graph, the concavity change gives an inflection point at \((0,1)\). A coherent sketch therefore includes the three monotonic branches, the two extrema, the inflection point, the five supplied points, and one root in each of the three intervals just identified.
As a consistency check, the given values agree with the required directions: \(-2<3\) on the first increasing interval, \(3>-1\) on the decreasing interval, and \(-1<2\) on the last increasing interval. The concavity signs also change at \(0\), where the function is defined. No exact root locations or additional extrema follow from the supplied evidence.
Common Mistake: Drawing More Than the Evidence Allows
A common error is to see \(f''\) change sign at \(x=c\) and automatically label \(c\) an inflection point. For example, a student might write, “\(f''\) changes from negative to positive at \(x=1\), so the inflection point is \((1,0)\).” That conclusion is unsupported: the sign change does not establish that \(f(1)\) exists, much less that it equals zero.
The correction is to check the domain and function value. If \(f(1)\) is undefined, there is no inflection point there. If \(f(1)\) exists but is unknown, you may report a change in concavity at \(x=1\) and identify the point as \((1,f(1))\), but you must not invent its height. More generally, a sketch should distinguish exact facts from qualitative conclusions. A guaranteed root in an interval is not an exact root coordinate, and a critical number is not automatically a local extremum.
Why a Coherent Sketch Matters
Graph-analysis questions assess whether you can connect mathematical evidence rather than treat each fact separately. A graph that marks all intercepts but ignores a domain restriction is not correct. A curve that rises through an interval where \(f'<0\) contradicts the derivative data. A claimed inflection point at a hole is not a point on the graph. These inconsistencies can invalidate an otherwise polished sketch.
Before finalizing, check that every segment moves in the direction required by \(f'\), each bend matches the sign of \(f''\), extrema occur only where supported, supplied values are plotted accurately, and no point is drawn outside the domain. When exact coordinates are unavailable, keep the sketch qualitative and label only what the information justifies.
Check Your Understanding
Use the principles in this tutorial to answer the following questions.
- If \(f'\) is negative on both sides of a critical number, what can you conclude about increasing or decreasing behavior there? Does the critical number necessarily give an extremum?
- What additional fact is needed before a concavity change across \(x=c\) can identify an inflection point?
- If \(f\) is continuous and strictly increasing on \((a,b)\), with \(f(a)<0<f(b)\), what can you conclude about roots in \((a,b)\)?
- Why should a graph-analysis sketch show separate branches on either side of an excluded domain value?
- If \(f(c)\) is known to exist but its value is not supplied, what can a concavity change at \(c\) tell you—and what can it not tell you?
