When the Second Derivative Test Does Not Decide
The second derivative test is a quick way to classify some critical points. But it does not classify every one. In particular, the test gives no conclusion when the second derivative is zero at a critical point, and it cannot be applied when one of its required derivatives is unavailable.
In either situation, the point still may be a local minimum, a local maximum, or neither. To decide, examine how the function behaves near the point: nearby signs of the first derivative can show whether the function decreases or increases, and direct comparisons of function values can settle the question as well.
Critical Points and the Test’s Limits
For a function \(f\), a critical number is a number \(c\) in the domain of \(f\) where \(f'(c)=0\) or \(f'(c)\) does not exist. A local minimum at \(c\) means that \(f(c)\leq f(x)\) for all domain values \(x\) sufficiently close to \(c\); a local maximum is defined with the inequality reversed.
The second derivative test is used at a stationary point, where \(f'(c)=0\), provided \(f''(c)\) exists. Under the usual hypotheses of the test, \(f''(c)>0\) indicates a local minimum, and \(f''(c)<0\) indicates a local maximum. If \(f''(c)=0\), the test is inconclusive. If \(f''(c)\) does not exist, the test is unavailable. Neither case, by itself, tells us whether a local extremum exists.
Example: A Zero Second Derivative with a Minimum
Worked Example: Classifying \(f(x)=x^4\) at \(x=0\)
Differentiate:
At \(c=0\), \(f'(0)=0\), so \(0\) is a critical number. Also,
The second derivative test therefore gives no classification. Check the first-derivative signs instead. If \(x<0\), then \(4x^3<0\); if \(x>0\), then \(4x^3>0\). Thus \(f'\) changes from negative to positive at \(0\): the function decreases to the point and increases away from it.
The function values confirm the conclusion. For every \(x\ne 0\),
Therefore, \(x=0\) is a local minimum (in fact, a global minimum). The answer comes from the nearby behavior, not from the inconclusive second derivative test.
Use Derivative Signs with the Right Conditions
A sign change in \(f'\) can classify an extremum, but continuity at the point matters. A useful version of the first-derivative test assumes that \(f\) is continuous at \(c\), differentiable on each side sufficiently close to \(c\), and has the stated derivative signs on those sides. If \(f'(c)\) exists, continuity at \(c\) is guaranteed; when \(f'(c)\) does not exist, check continuity rather than assuming it.
Here is why the negative-to-positive pattern gives a minimum. Choose \(x<c\) close enough that \(f'<0\) on \((x,c)\). Continuity on \([x,c]\) and differentiability on \((x,c)\) allow the Mean Value Theorem: for some \(u\) with \(x<u<c\),
Because \(f'(u)<0\) and \(c-x>0\), this difference is negative, so \(f(x)>f(c)\). For \(y>c\) close enough, apply the same theorem on \([c,y]\). For some \(v\) with \(c<v<y\),
Thus nearby function values on both sides exceed \(f(c)\), which proves that \(c\) is a local minimum. Reversing both derivative signs proves the local-maximum case. The continuity condition is what permits the theorem to compare the function value at \(c\) with nearby values.
Example: Zero Second Derivative but No Extremum
Worked Example: Classifying \(g(x)=x^3\) at \(x=0\)
Differentiate:
At \(0\), \(g'(0)=0\) and \(g''(0)=0\). The second derivative test is inconclusive. For every \(x\ne0\) on either side of zero, \(3x^2>0\), so \(g'\) is positive on both sides. The function is continuous at \(0\), so the first-derivative sign test applies: there is no local extremum.
Direct function values verify this. If \(x<0\), then \(g(x)=x^3<0=g(0)\). If \(x>0\), then \(g(x)=x^3>0=g(0)\). Arbitrarily close to zero there are values both below and above \(g(0)\), so zero is neither a local minimum nor a local maximum.
This example and \(x^4\) have the same second-derivative-test outcome, but different classifications. That is why \(f''(c)=0\) cannot be treated as evidence for either an extremum or its absence.
When the Second Derivative Does Not Exist
A second derivative can be unavailable even when the first derivative exists and equals zero. In that case, the second derivative test still gives no classification, but the first-derivative signs or function values may work.
Worked Example: Classifying \(p(x)=x^{4/3}\) at \(x=0\)
For real \(x\), write \(p(x)=(\sqrt[3]{x})^4\). To find the derivative at zero from its definition, use \(p(0)=0\):
For \(x\ne0\), differentiation gives
The formula for \(p''(x)\) is undefined at \(x=0\). The second derivative there also fails to exist as a finite derivative: from the definition of the derivative of \(p'\), for \(h\ne0\),
This quotient grows without bound as \(h\to0\), so it has no finite limit. Now use the first-derivative signs: \(p'(x)<0\) for \(x<0\), while \(p'(x)>0\) for \(x>0\). Since \(p\) is continuous at zero, the sign test applies and gives a local minimum. Directly, \(p(x)=x^{4/3}\geq0=p(0)\), with strict inequality for \(x\ne0\), confirming the result.
When the First Derivative Is Also Unavailable
A critical number need not be a point where the first derivative is zero. If the derivative does not exist, nearby derivative signs can still be informative, provided the function is continuous at the point and differentiable on each side. For instance, consider \(q(x)=|x|\) at zero. The function is continuous, but \(q'(0)\) does not exist: the difference quotient is \(-1\) for \(h<0\) and \(1\) for \(h>0\). Away from zero, \(q'(x)=-1\) on the left and \(q'(x)=1\) on the right. The sign change, with continuity verified, establishes a local minimum. The function values also show it directly: \(|x|\geq0=|0|\).
This example illustrates the distinction between the test and the conclusion. The second derivative test is unavailable here, but the critical point can still be classified.
A Common Mistake: Treating “Inconclusive” as a Classification
A common error is to calculate \(f''(c)=0\) and conclude that \(c\) is not an extremum. For \(f(x)=x^4\), that incorrect reasoning would say:
The implication is false. The second derivative test stops at “inconclusive”; it does not supply the conclusion shown. In fact, \(x^4>0=f(0)\) whenever \(x\ne0\), so zero is a local minimum. The correction is to identify the test’s outcome accurately, then use a valid method—such as the first-derivative sign test under its hypotheses or a direct comparison of function values.
A Reliable Classification Process
Confirm that \(c\) is in the domain and that \(f'(c)=0\) or \(f'(c)\) does not exist.
At a stationary point, evaluate \(f''(c)\) if it exists. A positive or negative value gives the usual test conclusion; zero is inconclusive, and nonexistence makes the test unavailable.
Use signs of \(f'\) on either side when the continuity and differentiability conditions are satisfied. Otherwise, compare \(f(x)\) directly with \(f(c)\) for nearby \(x\).
Conclude local minimum, local maximum, or neither, and explain the sign pattern or function-value comparison that justifies the classification.
This process matters in AP free-response solutions because naming a test is not enough: the conclusion must follow from the test’s hypotheses and the evidence near the point. When a test is inconclusive, say so, then provide the additional reasoning needed to classify the point.
Check Your Understanding
For each question, distinguish what the second derivative test says from what further evidence is needed.
- If \(f'(c)=0\) and \(f''(c)=0\), what does the second derivative test conclude?
- What continuity condition is needed to use a nearby first-derivative sign change when \(f'(c)\) does not exist?
- If \(f'\) changes from positive to negative around a continuous point \(c\), what type of local extremum occurs?
- For \(r(x)=x^6\) at zero, compute \(r'(0)\) and \(r''(0)\). What does the second derivative test say, and how can function values classify the point?
- Why does the fact that \(f''(c)\) does not exist fail to prove that \(c\) is not a local extremum?
