Zero Is a Clue, Not a Conclusion
A zero second derivative often draws attention to a point when you are investigating concavity. But the value \(f''(c)=0\) at a single point does not tell you whether the graph changes concavity there. To decide that, you must examine the second derivative on both sides of \(c\).
The difference is easy to see by comparing \(f(x)=x^4\) with \(g(x)=x^3\). Both functions have a zero second derivative at \(x=0\). Yet \(x^4\) is concave up on both sides of zero, while \(x^3\) changes from concave down to concave up. Only the second function has an inflection point there.
What Must Change at an Inflection Point?
A function is concave up on an interval when its slopes are increasing there, and concave down when its slopes are decreasing. The second derivative gives a sign test: on an interval where \(f''(x)>0\), the function is concave up; where \(f''(x)<0\), it is concave down.
In this lesson, an inflection point is a point on the graph where the function changes concavity. Thus, to establish an inflection point at \(x=c\), check whether the concavity on one side of \(c\) differs from the concavity on the other side. When \(f''\) is available, its signs on those intervals provide the test.
The condition \(f''(c)=0\) describes the second derivative at one input. It does not specify whether \(f''\) is positive or negative immediately to the left and right. But those neighboring signs are what determine whether concavity changes.
Worked Example: \(x^4\) Has No Inflection Point at Zero
Worked Example: \(x^4\) Has No Inflection Point at Zero
Determine whether \(f(x)=x^4\) has an inflection point at \(x=0\).
First calculate the derivatives:
At the proposed point,
This establishes that the second derivative is zero at zero, but it does not settle the inflection question. Check its sign on either side. For every \(x<0\), \(x^2>0\), so \(12x^2>0\). For every \(x>0\), \(x^2>0\), so again \(12x^2>0\).
The concavity is up on both sides of zero; it does not change. Therefore, \((0,f(0))=(0,0)\) is not an inflection point.
Check: The derivative calculation gives \(f''(x)=12x^2\), and substituting any nonzero value on either side, such as \(x=-1\) or \(x=1\), gives \(f''(-1)=12\) and \(f''(1)=12\). Both signs are positive, confirming the same concavity on each side.
Worked Example: \(x^3\) Does Have an Inflection Point at Zero
Worked Example: \(x^3\) Does Have an Inflection Point at Zero
Determine whether \(g(x)=x^3\) has an inflection point at \(x=0\).
Differentiate twice:
At zero, \(g''(0)=6(0)=0\). Now check the signs on either side. If \(x<0\), then \(6x<0\), so the function is concave down. If \(x>0\), then \(6x>0\), so it is concave up.
The concavity changes from down to up at zero. Since \(g(0)=0\), the inflection point is \((0,0)\).
Check: For test values \(-1\) and \(1\), \(g''(-1)=-6<0\) and \(g''(1)=6>0\), confirming the sign change. The point lies on the graph because \(g(0)=0\).
Why the Sign Check Works
The second-derivative concavity theorem connects signs of \(f''\) on an interval with the shape of the function on that interval. In particular, if \(f''>0\) throughout an interval, the function is concave up there; if \(f''<0\) throughout an interval, it is concave down there. These are interval statements, not conclusions drawn from one value.
Suppose \(f''\) is defined on intervals immediately to the left and right of \(c\). If \(f''\) is positive on both, the concavity is up on both. If it is negative on both, the concavity is down on both. Neither situation gives a change. A change in sign from one side to the other gives different concavity on the two intervals, which is the needed evidence for an inflection point.
This reasoning also explains why evaluating \(f''(c)\) alone cannot answer the question. A single value is not a sign analysis on either neighboring interval. The examples \(x^4\) and \(x^3\) have the same second-derivative value at zero, but their neighboring signs differ.
Worked Example: Finding Inflection Points from a Factored Second Derivative
Worked Example: Finding Inflection Points from a Factored Second Derivative
Find the inflection points of \(h(x)=x^4-2x^2\), using the second derivative to test candidates.
Differentiate:
Set the second derivative equal to zero to locate possible sign changes:
Factor the second derivative to make its sign easier to inspect:
The candidate values divide the real line into three intervals. If \(x<-\frac{1}{\sqrt{3}}\), then \(x^2>\frac13\), so \(12x^2-4>0\). If \(-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\), then \(x^2<\frac13\), so \(12x^2-4<0\). If \(x>\frac{1}{\sqrt{3}}\), then \(x^2>\frac13\), so \(12x^2-4>0\).
The signs change at both candidates, so both are inflection points. To find their \(y\)-coordinates, use \(x^2=\frac13\) and \(x^4=\frac19\):
Thus the inflection points are \(\left(-\frac{1}{\sqrt{3}},-\frac59\right)\) and \(\left(\frac{1}{\sqrt{3}},-\frac59\right)\).
Check: Substitution into \(h''(x)=12x^2-4\) gives \(12(\frac13)-4=4-4=0\) at each candidate. The interval signs \(+,-,+\) verify that each zero corresponds to a concavity change, rather than relying on the zero value alone.
A Common Mistake and Its Correction
A tempting shortcut is to find where \(f''(x)=0\) and label every resulting point an inflection point. For \(f(x)=x^4\), that shortcut produces the following incorrect conclusion:
The implication is not justified. The calculation identifies a candidate, but \(12x^2>0\) both for \(x<0\) and for \(x>0\). Therefore the function is concave up on both sides, so there is no inflection point at zero.
A Reliable Decision Process
When a zero of the second derivative appears, use a sign chart rather than treating the zero as a verdict. The process is short, but each step has a distinct purpose.
Solve \(f''(x)=0\) and record the resulting \(x\)-values.
Determine the sign of \(f''\) on intervals immediately to the left and right of each candidate.
Opposite signs mean the concavity changes; matching signs mean it does not.
For each confirmed change, calculate \(f(c)\) and report \((c,f(c))\).
Remember that the sign chart concerns intervals, not just the candidate values. A zero of \(f''\) is useful because it can mark a boundary between sign intervals. It earns the label “inflection point” only when the concavity on those intervals differs.
Why This Distinction Matters
Inflection points describe a change in the way a graph bends, so correctly identifying them requires evidence about the graph on both sides of the proposed point. Confusing a necessary-looking algebraic clue with the definition can lead to false conclusions, as \(x^4\) demonstrates. Conversely, a sign-change test confirms the geometric feature directly, as \(x^3\) demonstrates.
For AP Calculus problems, make the evidence explicit: state the second derivative, show its sign on each relevant interval, translate those signs into concavity, and then give the point’s coordinates when a change occurs. This makes the conclusion both accurate and justified.
Check Your Understanding
Use second-derivative signs to answer each question. For any claimed inflection point, identify the concavity on both sides.
- For \(p(x)=x^6\), calculate \(p''(x)\). Is \(x=0\) an inflection point? Explain using the signs on either side.
- For \(q(x)=x^5\), calculate \(q''(x)\). Does the concavity change at zero?
- Why does \(f''(c)=0\) not, by itself, establish that \((c,f(c))\) is an inflection point?
- If \(f''\) is negative immediately to the left of \(c\) and positive immediately to the right, what happens to the concavity at \(c\)?
- For \(r(x)=x^4-2x^2\), what sign pattern does \(r''\) have across its two zeros, and what does that imply?
