When the Second Derivative Is Undefined
A point where \(f''\) does not exist may still be important when studying a graph’s concavity. The missing derivative does not, by itself, tell us whether the graph changes concavity there. Instead, we examine the concavity on the two sides of the point.
This is especially useful for functions with cusps, vertical tangents, or fractional powers. Such functions can be defined and continuous at a point even though their derivatives—or their second derivatives—fail to exist there.
The point must belong to the graph. A location where the function is not defined cannot be an inflection point under this definition. When \(f''(c)\) is undefined, the test focuses on the intervals immediately to the left and right of \(c\), not on the derivative’s value at \(c\).
Using Concavity on Each Side
For a twice-differentiable function on an interval, the second-derivative criterion says that \(f''(x)>0\) throughout the interval implies that \(f\) is concave up there, while \(f''(x)<0\) throughout the interval implies that \(f\) is concave down there. This criterion applies on the intervals where the second derivative exists; it does not require \(f''\) to exist at an endpoint.
To test a point \(c\) where \(f''(c)\) is undefined, first check that \(f(c)\) exists and that \(f\) is continuous at \(c\). Then determine the sign of \(f''\) on a nearby interval to the left and on a nearby interval to the right. If these signs establish opposite concavities, the graph changes concavity at \(c\).
This test works because concavity is determined locally on intervals, not by the existence of a second derivative at one isolated point. If the signs are established on both sides, the second-derivative criterion identifies the concavity on each side. Comparing those two concavities then answers whether there is a change at \(c\).
Worked Example: A Cusp Without an Inflection Point
Worked Example: Testing a Cusp
Consider \(f(x)=x^{2/3}\) at \(x=0\). The function is defined at zero and \(f(0)=0\). It is continuous there, so \((0,0)\) belongs to the graph and is eligible for an inflection-point test.
For \(x\ne 0\), the power rule gives
The expression for \(f''\) is undefined at zero. For every \(x\ne 0\), however, \(x^{4/3}=(x^{1/3})^4>0\), so \(x^{-4/3}>0\). Therefore
The function is concave down on both sides. For a sign check, at \(x=-1\) and \(x=1\), \(f''(x)=-2/9\) in each case. There is no change in concavity across zero, so \((0,0)\) is not an inflection point. The cusp and the undefined value of \(f''(0)\) do not alter that conclusion.
Check: The function value at the proposed point is \(f(0)=0\), and the second derivative is negative on both adjacent sides. Both checks agree with the conclusion that this point is not an inflection point.
Worked Example: A Vertical Tangent with an Inflection Point
Worked Example: A Fractional Power with Opposite Concavities
Consider \(f(x)=x^{1/3}\) at \(x=0\). The function is defined and continuous at zero, with \(f(0)=0\). For \(x\ne0\), differentiate twice:
When \(x<0\), \(x^{1/3}<0\), so \(x^{5/3}=(x^{1/3})^5<0\). Thus \(x^{-5/3}<0\) and \(f''(x)>0\). When \(x>0\), \(x^{5/3}>0\), so \(x^{-5/3}>0\) and \(f''(x)<0\). Hence
The graph is concave up to the left of zero and concave down to the right. Therefore \((0,0)\) is an inflection point.
To check the tangent behavior, compute the difference quotient at zero:
For either sign of \(h\), this quotient tends to \(+\infty\) as \(h\to0\), so the derivative at zero is not finite. The graph has a vertical tangent there. This does not prevent the point from being an inflection point: the change in concavity, not the existence of a finite tangent slope, decides the test.
Check: At \(x=-1\), \(f''(-1)=2/9>0\); at \(x=1\), \(f''(1)=-2/9<0\). Also, \(f(0)=0\). These values confirm both the concavity change and the point on the graph.
Worked Example: An Inflection Point with a Horizontal Tangent
Worked Example: A Fractional Power with a Defined First Derivative
Let \(f(x)=x^{5/3}\) and test \(x=0\). The function is defined and continuous there, and \(f(0)=0\). For \(x\ne0\),
The first derivative also exists at zero. By the definition of derivative,
But \(f''(0)\) does not exist as a finite derivative. Using \(f'(0)=0\), its difference quotient would be
which is unbounded as \(h\to0\). So zero is a point where \(f''\) is undefined. To test the concavity, examine the second derivative away from zero. For \(x<0\), \(x^{1/3}<0\), so \(x^{-1/3}<0\), giving \(f''(x)<0\). For \(x>0\), \(x^{-1/3}>0\), giving \(f''(x)>0\). Therefore
The graph is concave down on the left and concave up on the right. Thus \((0,0)\) is an inflection point. In this example the first derivative exists at the point and equals zero, even though the second derivative does not exist there.
Check: Substituting \(x=-1\) gives \(f''(-1)=-10/9\), and substituting \(x=1\) gives \(f''(1)=10/9\). These signs confirm the change in concavity; \(f(0)=0\) confirms the proposed point lies on the graph.
A Reliable Test for Undefined-Second-Derivative Points
When a calculation reveals that \(f''(c)\) is undefined, use a sign-chart argument rather than treating nonexistence as a conclusion. The essential steps are:
Check that \(f(c)\) is defined and that \(f\) is continuous at \(c\).
Compute \(f''(x)\) on intervals just to the left and right, where the derivative exists.
Establish whether \(f''\) is positive or negative throughout each side interval; a test value can help check the sign.
Opposite signs establish opposite concavities and an inflection point. The same signs do not establish a change in concavity.
If the sign of \(f''\) varies repeatedly on one side, a single test value is not enough to establish concavity on that whole side. The sign must be justified on the relevant interval. The second-derivative criterion applies only where the function is twice differentiable, so first verify that the intervals used for the test avoid the point where the derivative is undefined.
Common Mistake: Treating Undefined as a Sign Change
A common error is to argue, “\(f''(c)\) is undefined, so the graph changes concavity at \(c\).” This confuses a failure of the derivative to exist at one point with a change in the graph’s behavior on intervals. In the cusp example \(f(x)=x^{2/3}\), the incorrect shortcut would label \((0,0)\) an inflection point because \(f''(0)\) is undefined. But the actual calculation gives
so the graph is concave down on both sides and does not change concavity there.
The correction is to keep two questions separate: where does \(f''\) exist, and what is the concavity on each side of the candidate? Only the second question determines whether there is a concavity change.
Why This Test Matters
Testing only points where \(f''(x)=0\) would miss examples such as \(x^{1/3}\) and \(x^{5/3}\), whose inflection points occur where the second derivative is undefined. Checking undefined points as well makes the search for possible inflection points more complete, while the sign comparison prevents every cusp or derivative failure from being mislabeled.
For fractional powers, take particular care with signs on negative inputs. If the denominator of the exponent is odd, the real-valued power is defined for negative \(x\), but its sign depends on the exponent. For example, \(x^{5/3}\) is negative for negative \(x\), so \(x^{-1/3}\) is negative there. Replacing a negative input by its absolute value without justification can reverse the sign chart and lead to an incorrect concavity conclusion.
Check Your Understanding
For each question, explain how the behavior on both sides of the point supports your conclusion.
- For \(f(x)=x^{2/3}\), what is the sign of \(f''\) on each side of zero, and is \((0,0)\) an inflection point?
- For \(f(x)=x^{1/3}\), which side is concave up and which side is concave down near zero?
- For \(f(x)=x^{5/3}\), does \(f'(0)\) exist? Does \(f''(0)\) exist? Explain using the difference quotients.
- Why is the fact that \(f''(c)\) is undefined insufficient to establish an inflection point?
- What must be checked about \(f\) at \(c\) before calling \((c,f(c))\) an inflection point?
