Why Check a Sketch Against a Sign Chart?
A sketch can look plausible while contradicting what is known about a function’s second derivative. A curve labeled “concave down” on an interval, for example, cannot agree with a chart showing \(f''(x)>0\) throughout that interval. Checking such details is useful whether the sketch comes from a hand-drawn graph, a graphing program, or a set of verbal claims about a function.
The check is not an attempt to reconstruct the entire graph. A sign chart for \(f''\) gives information about concavity and the way slopes change. On its own, it does not give exact function values, the curve’s vertical position, or every local maximum and minimum. A careful check separates what the chart requires from what it leaves undetermined.
What a Sign Chart Requires of a Sketch
Read a sign chart interval by interval. On any interval where \(f''\) is positive, the slopes of \(f\) increase as \(x\) increases. The graph is concave up on that interval. Where \(f''\) is negative, the slopes decrease and the graph is concave down. A boundary between intervals is a place to inspect, not an automatic inflection point: the signs on its two sides must be compared.
In a sketch, concavity is about how the slope changes, not whether the curve itself rises or falls. A curve may be decreasing and concave up if its slopes are negative but becoming less negative. It may be increasing and concave down if its slopes are positive but becoming less positive. Keeping these two questions separate prevents many incorrect graph checks.
Use zeros of \(f''\), points where \(f''\) does not exist, and domain breaks to separate the sign chart into intervals.
Write “concave up” where \(f''>0\) and “concave down” where \(f''<0\).
Check the concavity on each interval, then check whether the slope behavior agrees with the sign of \(f''\).
There must be a change in concavity across the point, and the function must be continuous there under the usual definition.
Why the Signs Control the Slope Behavior
The link between the sign chart and a sketch follows from the Mean Value Theorem. If \(f\) is twice differentiable on an interval and \(f''>0\) there, then \(f'\) is strictly increasing. To see this, take any two points \(a<b\) in that interval. Apply the Mean Value Theorem to \(f'\) on \([a,b]\). For some \(\xi\) with \(a<\xi<b\),
Here \(f''(\xi)>0\) and \(b-a>0\), so \(f'(b)-f'(a)>0\), which means \(f'(b)>f'(a)\). Thus the slopes increase from left to right. If \(f''<0\) throughout the interval, the same equality gives \(f'(b)-f'(a)<0\), so the slopes decrease. This explains why a sketch with slopes decreasing across an interval cannot match a chart showing \(f''>0\) there.
Worked Example: A Concavity Change with the Wrong Labels
Worked Example: A Concavity Change with the Wrong Labels
Let \(f(x)=\frac{x^3}{6}\). A proposed sketch labels the graph concave down for \(x<0\), concave up for \(x>0\), and marks \(x=0\) as an inflection point. Check these claims.
First differentiate:
The sign of \(f''(x)=x\) is negative when \(x<0\), zero at \(x=0\), and positive when \(x>0\). Therefore the chart requires concave down on the left and concave up on the right. Those two concavity labels agree with the proposed sketch.
For the inflection-point check, \(f(0)=0\), and \(f\) is a polynomial, so it is continuous at \(0\). Its concavity changes from down to up there. Thus the point \((0,0)\) is an inflection point under the usual definition. The chart and the sketch agree on all three claims.
Check: The computed second derivative has exactly the signs used in the chart, and the proposed inflection point is on the graph because \(f(0)=0\).
A Zero Is Not Enough
A sign chart may show \(f''(c)=0\), but that fact alone does not establish an inflection point. The relevant question is whether concavity changes across \(c\). For instance, if \(f''\) is positive on both sides of a zero, the graph is concave up on both sides; a sketch that marks that zero as an inflection point conflicts with the chart.
Worked Example: A Zero Without a Change in Concavity
Let \(f(x)=\frac{x^4}{12}\). A sketch marks \(x=0\) as an inflection point because it is a zero of the second derivative. Check that mark.
Differentiate once and then again:
For \(x<0\), \(x^2>0\); for \(x>0\), \(x^2>0\); and \(f''(0)=0\). So the sign chart is positive on both sides of zero, with a zero at zero. The graph is concave up on both intervals and does not change concavity at \(0\). The proposed inflection mark is therefore inconsistent with the chart.
Also, \(f(0)=0\), so the function is continuous there. Continuity does not make up for the missing concavity change. Both parts of the usual check matter.
Check: Substituting \(x=-1\) and \(x=1\) into \(f''(x)=x^2\) gives \(1\) in both cases, confirming the positive signs on either side.
Continuity Matters at a Proposed Inflection Point
A change in concavity on the two sides of a point does not by itself make that point an inflection point under the usual definition. The function must also be continuous there. This matters when a sketch has a hole, a jump, or another break at the location where the concavity changes. A sign chart may describe the curvature on each side, but it cannot remove a discontinuity in the function.
Worked Example: Concavity Changes Across a Discontinuity
Define \(f(x)=x^3\) when \(x\ne 0\), and define \(f(0)=1\). A proposed sketch marks the point \((0,1)\) as an inflection point because the concavity changes across \(x=0\). Test the claim.
For \(x\ne 0\), the formula is \(x^3\), so on both open intervals on either side of zero,
When \(x<0\), \(6x<0\), so the graph is concave down. When \(x>0\), \(6x>0\), so it is concave up. There is a change in concavity across zero.
Now check continuity at zero. For \(x\ne0\), \(f(x)=x^3\), and therefore \(\lim_{x\to0}f(x)=0\). But \(f(0)=1\), so
The function is not continuous at zero. Consequently, \((0,1)\) is not an inflection point under the usual definition, despite the change in concavity on either side. The proposed sketch is wrong if it labels that point as an inflection point.
Check: The side-by-side second-derivative signs are indeed negative and positive, but the limit test shows that the required continuity condition fails.
Check Slope Claims, Not Just Shape Labels
A sign chart also constrains how slopes can compare at different points. If \(f''>0\) throughout an interval, a sketch cannot show a smaller derivative value at a point to the right than at a point to the left. The Mean Value Theorem argument above gives a direct test. This can uncover a contradiction even when a drawing’s concavity is difficult to judge by eye.
Worked Example: Slope Values That Cannot Match the Chart
Suppose \(f\) is twice differentiable on an interval containing \([-1,1]\), and its sign chart says \(f''(x)>0\) throughout that interval. A proposed sketch also claims \(f'(-\frac12)=2\) and \(f'(\frac12)=-2\). Check whether both slope claims can be true.
Apply the Mean Value Theorem to \(f'\) between \(a=-\frac12\) and \(b=\frac12\). It gives some \(\xi\) between these values such that
Using the proposed slope values, the left side is \(-2-2=-4\), while the distance factor on the right is \(\frac12+\frac12=1\). Thus the equation would require \(-4=f''(\xi)\). But the chart states \(f''(\xi)>0\). A negative number cannot equal a positive number, so the two proposed slope values are incompatible with the chart.
Check: The chart requires \(f'\) to increase from left to right, whereas the proposed values decrease from \(2\) to \(-2\). The contradiction is confirmed by the Mean Value Theorem calculation.
What the Chart Cannot Tell You
A second-derivative sign chart is a consistency test, not a complete description of a graph. Unless additional information is given, it does not specify:
- the value \(f(c)\) at any particular input;
- whether the graph is above or below the horizontal axis;
- the exact steepness of the curve at a point; or
- the location of every local maximum or minimum.
For example, \(f''>0\) says the slopes increase, but the slopes could still be negative, positive, or change from negative to positive. To determine whether the function is rising or falling at a particular point, information about \(f'\) is needed. Do not reject a sketch simply because it rises where you expected a fall unless the available information actually requires it.
A Practical Consistency Checklist
When a sign chart and a proposed sketch are presented together, check the claims in a fixed order. First confirm that the stated intervals match the signs of \(f''\). Then compare the corresponding concavity labels. At each boundary, compare the signs on the left and right rather than treating a zero or an undefined value as an automatic change. If an inflection point is claimed, check continuity there as well as the change in concavity. Finally, use slope information only when the chart or other data provides a basis for checking it.
Check Your Understanding
Use the sign chart and the stated sketch information to decide which claims are consistent.
- If \(f''<0\) throughout an interval, what must happen to the slopes of \(f\) as \(x\) increases?
- A sign chart shows \(f''>0\) on both sides of \(c\), with \(f''(c)=0\). Does this establish an inflection point at \(c\)? Explain.
- A proposed sketch shows a concavity change at \(c\), but the function is discontinuous there. Is \(c\) an inflection point under the usual definition?
- Suppose \(f''>0\) on an interval and a sketch claims \(f'(a)>f'(b)\) for \(a<b\) in that interval. What is the contradiction?
- Does a chart showing \(f''>0\) determine whether \(f\) is increasing? State what the chart does determine.
