Why a Zero of the Second Derivative Is Only a Candidate
A common step in analyzing a function is to find the values where its second derivative is zero. These values are useful: they divide the domain into intervals on which the sign of the second derivative can be checked. But the equation \(f''(c)=0\) alone does not tell us whether the function changes concavity at \(c\).
An inflection point requires a change in concavity. For example, the function must be concave down on one side of the point and concave up on the other. If it is concave up on both sides, there is no change in concavity, even if the second derivative equals zero at the point itself.
This distinction matters because a sign chart records behavior on intervals, not just the value of a derivative at one point. The value \(f''(c)=0\) gives information at \(c\); an inflection point is identified by comparing behavior on both sides.
How to Test a Candidate
Assume \(f\) is twice differentiable in a neighborhood of \(c\), and suppose \(f''(c)=0\). Since \(f\) is twice differentiable there, it is continuous at \(c\), so the point \((c,f(c))\) lies on its graph. To decide whether it is an inflection point, examine the sign of \(f''\) on the two nearby intervals.
Solve \(f''(x)=0\), then keep only solutions that lie in the domain under consideration.
Factor \(f''\) when possible, or choose test values from the intervals immediately to the left and right of the candidate.
If \(f''\) has opposite signs on the two sides, concavity changes. If the signs agree, it does not.
Here we use the second derivative concavity criterion: on an interval where \(f''(x)>0\), \(f\) is concave up; on an interval where \(f''(x)<0\), \(f\) is concave down. The criterion is applied on intervals, so a zero at the boundary between them is not itself enough to determine the outcome.
Worked Example: A Zero with No Change in Concavity
Worked Example: A Zero with No Change in Concavity
Determine whether \(x=0\) is an inflection point of \(f(x)=x^4\).
First differentiate:
The second derivative is zero when \(12x^2=0\), which gives the candidate \(x=0\). For every \(x\ne 0\), \(x^2>0\), so \(12x^2>0\). Thus \(f''\) is positive both immediately to the left and immediately to the right of zero, and \(f\) is concave up on both sides.
For a direct sign check, take \(x=-1\) and \(x=1\):
Although \(f''(0)=0\), the concavity does not change. Therefore \(x=0\) is not an inflection point. The graph point at the candidate is \((0,f(0))=(0,0)\), but it is not an inflection point. The checks agree with the factored expression \(12x^2\), which is positive on both sides of zero.
Why Even-Multiplicity Zeros Often Fail
For polynomial second derivatives, multiplicity can help explain what a sign chart will show. If a polynomial \(f''\) has a zero at \(c\) of multiplicity \(2k\), where \(k\) is a positive integer, it can be written near \(c\) in the form \((x-c)^{2k}g(x)\), where \(g\) is a polynomial and \(g(c)\ne 0\). Since \(g\) is continuous and nonzero at \(c\), it has the same sign on both sides of \(c\) sufficiently close to \(c\). Also, \((x-c)^{2k}>0\) whenever \(x\ne c\). So \(f''\) has the same sign on both sides: there is no change in concavity.
This reasoning concerns a zero of even multiplicity in a polynomial (or a factorization with the stated properties). It does not replace checking signs in general. In particular, when the factorization is not available or a factor also vanishes at \(c\), use a sign chart to determine the behavior on each side.
For contrast, a factor \((x-c)^{2k+1}\) has opposite signs on either side of \(c\). If the other factor is continuous and nonzero at \(c\), the product changes sign there. That pattern can produce an inflection point, but the conclusion still comes from verifying concavity on both sides—not from the equation \(f''(c)=0\) by itself.
Worked Example: A Sign Change at a Zero
Worked Example: A Sign Change at a Zero
Determine whether \(x=0\) is an inflection point of \(f(x)=x^5\).
Differentiate twice:
The only zero of \(f''\) is \(x=0\). For \(x<0\), \(x^3<0\), so \(20x^3<0\) and \(f\) is concave down. For \(x>0\), \(x^3>0\), so \(20x^3>0\) and \(f\) is concave up. A numerical check gives:
The concavity changes from down to up at zero, and \(f(0)=0\). Therefore \((0,0)\) is an inflection point. The signs follow directly from \(20x^3\), so the test values confirm the conclusion on the two sides.
A Candidate Can Be False Even When Another Is Genuine
It is especially important to test candidates separately when \(f''\) has several zeros. One zero may correspond to a change in concavity, while another does not. The following example has both outcomes.
Worked Example: One False Candidate and One Inflection Point
Find the inflection points of \(f(x)=x^4+x^5\).
Differentiate:
The factored second derivative is zero at \(x=0\) and at \(x=-\frac{3}{5}\). Its factors show the signs on the intervals separated by these values. For \(x<-\frac{3}{5}\), \(x^2>0\) and \(3+5x<0\), so \(f''(x)<0\). For \(-\frac{3}{5}<x<0\), both \(x^2>0\) and \(3+5x>0\), so \(f''(x)>0\). For \(x>0\), both factors are again positive, so \(f''(x)>0\).
For an explicit check, use one test value in each interval:
At \(x=-\frac35\), the sign changes from negative to positive, so the concavity changes from down to up. The function value is:
Thus \(\left(-\frac35,\frac{162}{3125}\right)\) is an inflection point. At \(x=0\), however, \(f''(0)=0\) but \(f''\) is positive on both sides. The concavity remains up, so \((0,0)\) is not an inflection point. The tested values confirm the signs in all three intervals, and the two candidates have been classified separately.
A Common Mistake
A frequent error is to treat every solution to \(f''(x)=0\) as an inflection point. For \(f(x)=x^4\), this incorrect calculation would be:
The conclusion does not follow. The equation shows only that \(x=0\) is a candidate. To correct the argument, compare the signs around zero: \(12x^2>0\) for every \(x\ne0\), on both sides. Therefore the function is concave up on both sides and has no inflection point there.
Check Your Understanding
For each question, focus on the signs of the second derivative on intervals to either side of the candidate.
- For \(f(x)=x^6\), find \(f''(x)\) and decide whether \(x=0\) is an inflection point.
- For \(f(x)=x^3\), compare the signs of \(f''\) on either side of zero. What does this imply about concavity?
- Why does a factor \((x-c)^4\) not change sign as \(x\) passes through \(c\)?
- For \(f(x)=x^4+x^5\), which zero of \(f''\) is an inflection candidate that does not become an inflection point?
- What additional check is required after finding a solution of \(f''(x)=0\)?
