From a Zero of f'' to an Inflection Point
The second derivative gives a useful way to classify concavity: where \(f''\) is positive, a twice-differentiable function is concave up; where \(f''\) is negative, it is concave down. This suggests a practical search for inflection points: first find where \(f''\) is zero, then check whether its sign changes on either side.
The second step matters. A zero of \(f''\) identifies a candidate, not automatically an inflection point. The question is whether concavity actually changes as the input passes through that value.
This tutorial considers candidates where \(f''(c)=0\), with \(f''\) defined on an interval around \(c\). It explains how to test the concavity on both sides. Other situations require separate analysis.
The Sign-Change Test
Suppose \(f''(c)=0\). Choose a small interval to the left of \(c\) and a small interval to the right. Determine the sign of \(f''\) on each interval. If \(f''\) is positive on one side and negative on the other, the concavity changes, so \((c,f(c))\) is an inflection point. If the signs agree on both sides, there is no concavity change at \(c\), so it is not an inflection point.
A sign chart makes the test systematic. List the zeros of \(f''\) in increasing order. These values divide the interval under consideration into subintervals. Find the sign of \(f''\) on each subinterval, then compare the signs on the two sides of every candidate.
Find \(f''(x)\) on the interval being studied.
Solve \(f''(x)=0\), keeping only values in the interior of the domain relevant to the test.
Use the ordered zeros to divide the interval and determine the sign of \(f''\) on each part.
Opposite signs mean a change in concavity; matching signs mean no change.
For each genuine inflection value \(c\), calculate \(f(c)\) to obtain \((c,f(c))\).
Why the Test Works
The second-derivative concavity criterion states that if \(f''(x)>0\) throughout an interval, then \(f\) is concave up there; if \(f''(x)<0\) throughout an interval, then \(f\) is concave down there. These are the consequences of the signs that the test compares.
Conversely, if the signs of \(f''\) are the same on both sides, the second-derivative criterion gives the same concavity on both sides. There is no concavity change at \(c\), so \(c\) is not an inflection value. Thus the comparison must use both sides, not just the value of \(f''\) at the candidate.
Worked Examples
Worked Example: A Sign Change from Negative to Positive
Find the inflection point of \(f(x)=x^3-3x\) by testing the zero of \(f''\).
First differentiate:
Differentiate again:
Set the second derivative equal to zero:
The value \(0\) divides the real line into \(x<0\) and \(x>0\). If \(x<0\), then \(6x<0\), so \(f''(x)<0\). If \(x>0\), then \(6x>0\), so \(f''(x)>0\). The sign changes from negative to positive, so the graph changes from concave down to concave up at \(x=0\).
Now find the corresponding function value:
The inflection point is \((0,0)\). As a check, \(f''(x)=6x\) is negative immediately to the left of \(0\) and positive immediately to the right, matching the concavity change.
Worked Example: A Zero Without a Sign Change
Determine whether \(f(x)=x^4\) has an inflection point at the zero of its second derivative.
Differentiate twice:
Solve \(f''(x)=0\):
For every \(x\ne 0\), \(x^2>0\), so \(12x^2>0\). In particular, \(f''\) is positive both immediately to the left and immediately to the right of \(0\). The signs match, so the function is concave up on both sides. There is no concavity change, and \((0,f(0))=(0,0)\) is not an inflection point.
Check the key values: \(f''(0)=12(0)^2=0\), while, for example, \(f''(-1)=12\) and \(f''(1)=12\). The zero at \(0\) is real, but it does not pass the sign-change test.
Worked Example: Testing Two Candidates
Find all inflection points of \(f(x)=x^4-2x^2\).
Differentiate:
Then differentiate again:
Set \(f''(x)=0\) and solve:
Write \(a=1/\sqrt{3}\). The candidates are \(-a\) and \(a\), which divide the real line into three intervals. The factor \(4\) is positive, so the sign of \(f''(x)\) is the sign of \(3x^2-1\). For \(x<-a\) or \(x>a\), \(x^2>1/3\), so \(3x^2-1>0\). For \(-a<x<a\), \(x^2<1/3\), so \(3x^2-1<0\). The sign chart is therefore:
At \(-a\), the sign changes from positive to negative. At \(a\), it changes from negative to positive. Both candidates are inflection values.
To find the coordinates, use \(a^2=1/3\) and \(a^4=1/9\):
Thus the inflection points are \(\left(-1/\sqrt{3},-5/9\right)\) and \(\left(1/\sqrt{3},-5/9\right)\). Substitution into the factored second derivative checks the candidates: \(f''(\pm a)=4(3(1/3)-1)=0\). The sign chart confirms that each zero has a concavity change.
Common Mistake: Stopping at f''(c) = 0
A frequent error is to solve \(f''(x)=0\) and label every solution an inflection point. For \(f(x)=x^4\), this mistaken argument would be:
The conclusion does not follow from the displayed calculation. It establishes only that \(0\) is a candidate. The correction is to inspect both sides: \(12x^2>0\) for \(x<0\) and for \(x>0\), so the concavity is up on both sides. Therefore \((0,0)\) is not an inflection point.
The reverse shortcut is also unsafe: the fact that a function has different values of \(f''\) at selected test points does not by itself establish a sign change at a particular candidate. Determine the sign throughout the intervals adjacent to that candidate, using algebra or an appropriate sign chart.
Using a Sign Chart Carefully
A sign chart is most useful when its intervals and signs are justified. For a polynomial second derivative, the real zeros divide the number line into intervals, and factoring often makes the sign on each interval clear. Alternatively, one can choose a test value in each interval and evaluate \(f''\) there, provided the sign is known to remain constant throughout that interval. Do not use a test value from one interval to infer the sign on another.
For several candidates, compare the signs on the adjacent intervals separately at each one. In the previous example, the sign pattern was positive, negative, positive. That pattern gives a sign change at each of the two zeros. In contrast, a pattern that stays positive on both sides of a zero indicates no inflection there.
Once a candidate passes the test, calculate \(f(c)\) to state the point on the graph. Reporting only the \(x\)-coordinate gives the location where concavity changes, but the inflection point itself is \((c,f(c))\).
Why the Test Is Useful
Finding zeros of \(f''\) narrows the search for changes in concavity to specific values that can be checked systematically. The sign chart then describes the concavity on whole intervals, rather than relying on the appearance of a formula or a few isolated values. It also prevents a common logical error: treating a necessary-looking algebraic condition as a sufficient conclusion.
Keep the distinction clear: solving \(f''(x)=0\) produces candidates; comparing the signs immediately to the left and right decides whether those candidates are inflection values. Finally, evaluating \(f\) gives the coordinates of the genuine inflection points.
Check Your Understanding
For each question, explain how the signs of the second derivative support your conclusion.
- For \(f(x)=x^3\), find the zero of \(f''\) and determine whether it is an inflection value.
- For \(f(x)=x^4\), why does the zero of \(f''\) fail the sign-change test?
- If \(f''\) is negative immediately to the left of \(c\) and positive immediately to the right, what happens to the concavity at \(c\)?
- For \(f(x)=x^4-2x^2\), state the signs of \(f''\) on the three intervals formed by its zeros.
- After identifying an inflection value \(c\), what additional calculation gives the coordinates of the point?
