When the Test Does Not Decide
A second derivative test can classify a stationary point when the second derivative has a definite, nonzero sign there. But two common outcomes do not provide that conclusion: the second derivative is zero, or it does not exist. In either case, the test is inconclusive. This means the test supplies no classification; it does not mean that the function has no local extremum.
The next step is to examine the function more closely. A direct comparison of \(f(x)\) with \(f(c)\) near the point \(c\) can establish a local maximum or minimum. Alternatively, a first-derivative sign analysis can show whether the function changes from decreasing to increasing, or vice versa. The examples below show why one cannot infer the answer from the inconclusive second-derivative outcome alone.
For an interior point where \(f'(c)=0\), the second derivative test uses the sign of \(f''(c)\): when \(f''(c)>0\), it gives a local minimum; when \(f''(c)<0\), it gives a local maximum. When \(f''(c)=0\), it does not decide between these possibilities or the possibility of neither. If \(f''(c)\) does not exist, this test cannot be applied at \(c\). In both situations, another analysis is needed.
Zero Second Derivative: Three Different Outcomes
Worked Example: A Stationary Point That Is a Local Minimum
Consider \(f(x)=x^4\) at \(c=0\). Differentiate:
Thus \(0\) is stationary. The second derivative is
The second derivative test is inconclusive. To classify the point, compare function values directly. For every \(x\),
For \(x\ne 0\), \(x^4>0\). Therefore \(f(x)>f(0)\) for every \(x\ne 0\), so \(0\) is a strict local minimum (in fact, a global minimum). As a check, substituting \(x=0\) gives \(f(0)=0\), consistent with the comparison.
Worked Example: A Stationary Point That Is a Local Maximum
Now consider \(g(x)=-x^4\) at \(c=0\). Its derivatives are
Again, the second derivative test is inconclusive. For every \(x\), however,
For \(x\ne 0\), \(-x^4<0\). Hence \(g(x)<g(0)\) for every \(x\ne 0\), and \(0\) is a strict local maximum (also a global maximum). Substitution confirms \(g(0)=0\), as used in the comparison.
Worked Example: A Stationary Point That Is Neither
Consider \(h(x)=x^3\) at \(c=0\). Then
The same test outcome occurs, but the classification differs. If \(x>0\), then \(h(x)=x^3>0=h(0)\). If \(x<0\), then \(h(x)=x^3<0=h(0)\). Every neighborhood of zero contains points on both sides of zero, so there are nearby function values both above and below \(h(0)\). Thus \(0\) is neither a local minimum nor a local maximum. Substituting \(x=0\) verifies \(h(0)=0\), the value used in both comparisons.
Why Zero Gives No Classification
The three examples give a complete reason not to treat \(f''(c)=0\) as a hidden conclusion. The functions \(x^4\), \(-x^4\), and \(x^3\) all have the same value, first derivative, and second derivative at zero:
Nevertheless, the first has a strict local minimum, the second has a strict local maximum, and the third has neither. Therefore those data alone cannot determine which behavior occurs. This is not a limitation that can be resolved by interpreting zero as weakly positive or weakly negative: the examples have genuinely different outcomes.
When the Second Derivative Does Not Exist
A stationary point does not have to have a second derivative. It is enough for \(f'(c)\) to exist and equal zero for \(c\) to be stationary. If \(f''(c)\) is unavailable, the second derivative test cannot be used there. Direct comparison still works if the function values can be compared near the point.
Worked Example: A Local Minimum with No Second Derivative
Define \(p(x)=x^2+x|x|\). Since \(|x|=-x\) for \(x<0\), and \(|x|=x\) for \(x\geq 0\), the function can be written as
At zero, \(p(0)=0\). For \(h<0\), the difference quotient for the first derivative is
For \(h>0\), it is
Both one-sided limits are zero, so \(p'(0)=0\). More generally, \(p'(x)=0\) for \(x<0\), while \(p'(x)=4x\) for \(x>0\). To test whether \(p''(0)\) exists, differentiate \(p'\) at zero using its difference quotient. For \(h<0\), \(p'(h)=0\), and for \(h>0\), \(p'(h)=4h\). Thus
The one-sided limits disagree, so \(p''(0)\) does not exist. Yet \(p(x)=0\) for \(x<0\), and \(p(x)=2x^2>0\) for \(x>0\). Therefore \(p(x)\geq p(0)\) for all \(x\), and zero is a local minimum, though not a strict one. The piecewise formula also verifies \(p(0)=0\).
Worked Example: No Second Derivative and No Extremum
Let \(q(x)=x|x|\). Its piecewise form is
Since \(q(0)=0\), the difference quotient for \(q'(0)\) is \(-h\) when \(h<0\) and \(h\) when \(h>0\). Both approach zero, so \(q'(0)=0\). For nonzero \(x\), differentiation gives
The derivative of \(q'\) at zero would require the limit of \(q'(h)/h\). It equals \(-2\) for \(h<0\) and \(2\) for \(h>0\), so the limit does not exist and \(q''(0)\) is unavailable.
For \(x<0\), \(q(x)=-x^2<0=q(0)\); for \(x>0\), \(q(x)=x^2>0=q(0)\). Both kinds of values occur arbitrarily close to zero. Hence zero is neither a local minimum nor a local maximum. The formula gives \(q(0)=0\), confirming the reference value in the comparison.
A Reliable Approach to an Inconclusive Point
Once the second derivative test does not decide, avoid trying to extract a classification from it. Instead, identify what information is available and use a method that directly addresses the local behavior.
Confirm that \(c\) is in the domain and that \(f'(c)=0\). Then determine whether \(f''(c)\) exists and, if it does, whether it is zero.
Try to establish whether \(f(x)-f(c)\) is nonnegative or nonpositive near \(c\). If it has both signs arbitrarily close to \(c\), neither local-extremum inequality holds.
A first-derivative sign analysis or an algebraic comparison may classify the point. State the reasoning that supports the conclusion rather than reporting only that the second derivative test failed.
There is an important distinction between “the test is inconclusive” and “the function is neither.” In the \(x^3\) and \(x|x|\) examples, the points are indeed neither, but that conclusion follows from comparing function values on both sides. In the \(x^4\) and \(-x^4\) examples, the same second-derivative outcome accompanies extrema instead.
Common Mistake: Treating Zero as a Minimum
A frequent error is to assume that \(f''(c)=0\) means the graph is sufficiently flat to have a local minimum. For example, someone might reason, incorrectly, that for \(f(x)=-x^4\), the conditions \(f'(0)=0\) and \(f''(0)=0\) imply a minimum because the second derivative is “not negative.” But the direct calculation gives
Thus zero is a strict maximum, not a minimum. The correction is to treat a zero second derivative as no classification from this test and then check the function near the point. Likewise, the absence of \(f''(c)\) is not evidence for any particular outcome: the examples \(p\) and \(q\) have no second derivative at zero, but one has a minimum and the other has neither.
Why This Distinction Matters
The second derivative test is useful because, under its applicable conditions, it can classify a stationary point efficiently. Its conclusion depends on a nonzero second derivative. When that condition is missing, careful analysis prevents false certainty. It also makes clear what a complete solution must show: either a comparison with \(f(c)\), a suitable derivative sign argument, or another justified way to establish the local behavior.
In practice, check the conclusion against the function, not just the derivative values recorded at a single point. The sign of \(f(x)-f(c)\) near \(c\) tells directly whether nearby values lie above, below, or on both sides of the value at the point. That comparison is the basis of the classifications in these examples.
Check Your Understanding
For each question, explain what the available information does and does not establish.
- For \(r(x)=x^6\) at zero, find \(r'(0)\) and \(r''(0)\), then classify the point by comparing \(r(x)\) with \(r(0)\).
- Why do \(x^4\), \(-x^4\), and \(x^3\) show that \(f'(0)=f''(0)=0\) is not enough to classify a stationary point?
- For \(s(x)=-x^2+x|x|\), determine whether \(s'(0)\) exists, whether \(s''(0)\) exists, and whether zero is a local maximum, a local minimum, or neither.
- State the difference between an inconclusive second derivative test and a conclusion that a point is neither a local maximum nor a local minimum.
- If \(f'(c)=0\) but \(f''(c)\) does not exist, what kind of comparison could still establish that \(c\) is a local extremum?
