When the Second Derivative Fails at a Candidate
A common way to locate possible inflection points is to find where the second derivative is zero or undefined, then check whether concavity changes. But these two kinds of candidates are not equivalent: a point can be an inflection point even when \(f''\) does not exist there.
The key question is not whether \(f''(c)\) exists. It is whether the graph changes from concave down to concave up, or from concave up to concave down, as it passes through \(x=c\). To answer that question, examine concavity on intervals immediately to the left and right.
The requirement that the function be continuous ensures that the graph passes through \((c,f(c))\). A point where the function is not defined cannot be an inflection point under this definition, even if concavity differs on separate branches.
Testing Concavity on Both Sides
Suppose \(f''(x)\) exists for \(x\) in an interval immediately to the left of \(c\), and also on an interval immediately to the right of \(c\). Its sign on each interval determines the concavity there: positive second derivative means concave up, and negative second derivative means concave down.
This is a local test. The intervals need not extend far from \(c\), but the sign must hold throughout some interval on each side—not just at one selected point. If the signs agree, there is no change of concavity across \(c\), so \(c\) is not an inflection point.
Confirm that \(f(c)\) exists and that \(f\) is continuous at \(c\).
Determine where \(f''\) exists near \(c\), and find its sign on each side separately.
Opposite signs mean concavity changes; matching signs mean it does not.
If concavity changes and the function is continuous at \(c\), identify the inflection point as \((c,f(c))\).
Why the Sign of \(f''\) Gives Concavity
The connection between the second derivative and concavity comes from how the first derivative changes. On any interval where \(f''>0\), the Mean Value Theorem applied to \(f'\) shows that \(f'\) is increasing there. Indeed, for any \(u<v\) in that interval, there is some \(w\) between them such that
Since \(v-u>0\), this implies \(f'(v)>f'(u)\). Increasing tangent slopes correspond to a graph that bends upward, or is concave up. If \(f''<0\), the same argument gives \(f'(v)<f'(u)\), so the graph is concave down. This reasoning is applied separately on each side of \(c\); it does not require \(f''(c)\) to exist.
A Point with an Undefined Second Derivative Can Be an Inflection Point
Worked Example: A Change in Concavity at a Piecewise Join
Consider \(f(x)=x|x|\). Determine whether \(x=0\) is an inflection point, even though the second derivative there may not exist.
For \(x<0\), \(|x|=-x\), so \(f(x)=-x^2\). For \(x>0\), \(|x|=x\), so \(f(x)=x^2\). At \(x=0\), \(f(0)=0\). The two formulas both approach \(0\), so \(f\) is continuous at \(0\).
First check the derivative at the join using its definition:
For \(x<0\), differentiating \(-x^2\) gives \(f'(x)=-2x\). For \(x>0\), differentiating \(x^2\) gives \(f'(x)=2x\). Both derivative formulas approach \(0=f'(0)\), consistent with the derivative found from the limit. Differentiating again gives
At \(x=0\), the derivative of \(f'\) from the left is \(-2\), while from the right it is \(2\). For example, the difference quotient for \(f'\) is \(-2\) when \(h<0\) and \(2\) when \(h>0\), so \(f''(0)\) does not exist. On the left, \(f''=-2<0\), so the graph is concave down; on the right, \(f''=2>0\), so it is concave up. The concavity changes, and the function is continuous at \(0\). Therefore \((0,0)\) is an inflection point.
As a check, the formulas \(-x^2\) and \(x^2\) give second derivatives \(-2\) and \(2\), respectively, confirming the signs on the two sides. The differing one-sided difference quotients confirm that there is no second derivative at the join.
Vertical Tangents and Inflection Points
An undefined derivative does not prevent a concavity test. For instance, a vertical tangent can occur where the first derivative becomes unbounded. If the second derivative exists on both sides and has opposite signs there, the point can still be an inflection point.
Worked Example: A Vertical Tangent with Changing Concavity
Let \(f(x)=\sqrt[3]{x}\), and test \(x=0\). The function is continuous at \(0\), with \(f(0)=0\). For \(x\ne 0\), the power rule gives
For \(x<0\), \(x^{5/3}<0\), so \(-2/(9x^{5/3})>0\). The graph is concave up on the left. For \(x>0\), \(x^{5/3}>0\), so \(-2/(9x^{5/3})<0\). The graph is concave down on the right.
The first derivative is positive on both sides and grows without bound as \(x\) approaches \(0\), since \(x^{2/3}\) approaches \(0\) through positive values. Thus the tangent becomes vertical at the origin. Meanwhile, the opposite signs of \(f''\) show that concavity changes there. Since the function is continuous, \((0,0)\) is an inflection point.
For a derivative check, differentiating \(f'(x)=\frac13x^{-2/3}\) gives \(\frac13(-\frac23)x^{-5/3}=-\frac29x^{-5/3}\), as stated. The sign check on each side confirms the concavity conclusion.
A Cusp Does Not Automatically Mean an Inflection Point
A sharp point or cusp can make a derivative undefined, but that fact alone tells us nothing about whether concavity changes. In particular, a cusp may be concave the same way on both sides.
Worked Example: A Cusp with No Change in Concavity
Consider \(f(x)=|x|^{2/3}\) at \(x=0\). The function is continuous and \(f(0)=0\). For \(x<0\), \(f(x)=(-x)^{2/3}\); for \(x>0\), \(f(x)=x^{2/3}\). Differentiating each formula gives
The left derivative tends to negative infinity and the right derivative tends to positive infinity as \(x\) approaches \(0\). The slopes therefore do not approach a single finite derivative; the graph has a cusp at the origin. Differentiate once more on each side:
Both expressions are negative throughout their respective intervals. Thus the graph is concave down both to the left and to the right of \(0\). The concavity does not change, so \((0,0)\) is not an inflection point. This conclusion holds even though \(f''(0)\) is undefined and the graph has a cusp.
As a check, differentiating the left first-derivative formula gives \(-\frac{2}{9}(-x)^{-4/3}\), and differentiating the right formula gives \(-\frac{2}{9}x^{-4/3}\). Since both are negative, the signs agree and there is no concavity change.
Common Pitfall: Treating Undefined as Proof
A frequent error is to declare every point where \(f''\) is undefined an inflection point. For the cusp \(f(x)=|x|^{2/3}\), this incorrect calculation would be: “\(f''(0)\) is undefined, therefore \((0,0)\) is an inflection point.” The conclusion does not follow. Undefinedness only makes \(x=0\) a candidate to investigate; the second derivative is negative on both sides, so the concavity does not change.
The opposite mistake is to exclude a point because the second derivative does not exist there. The example \(f(x)=x|x|\) shows why that is wrong: \(f''(0)\) is undefined, yet the concavity changes from down to up. In both cases, compare the two sides rather than using the status of \(f''(c)\) as the verdict.
Check Your Understanding
For each question, focus on the concavity on both sides of the candidate.
- What two conditions establish an inflection point under the definition used here?
- If \(f''\) is negative on both sides of a continuous point \(c\), can \(c\) be an inflection point? Explain.
- For \(f(x)=x|x|\), what are the signs of \(f''\) immediately to the left and right of \(0\)?
- Why is the origin not an inflection point of \(f(x)=|x|^{2/3}\), even though the second derivative is undefined there?
- For \(f(x)=\sqrt[3]{x}\), what feature of the first derivative indicates a vertical tangent at the origin, and what feature of the second derivative establishes an inflection point?
