When the Test Gives No Answer
The second derivative test is useful when a function has a stationary point \(c\), so that \(f'(c)=0\), and the second derivative there is nonzero. A positive second derivative indicates a local minimum, while a negative second derivative indicates a local maximum.
But sometimes \(f'(c)=0\) and \(f''(c)=0\). In that case, the test is inconclusive: its conditions do not determine whether \(f\) has a local maximum, a local minimum, or neither at \(c\). This does not mean that there is no extremum. It means that we need other evidence.
The important evidence is how the function moves as \(x\) passes through \(c\). The first derivative describes that movement: positive values mean the function is increasing, and negative values mean it is decreasing. We will use these signs to settle cases that the second derivative test cannot decide.
What “Inconclusive” Means
A function \(f\) has a local maximum at \(c\) if there is an open interval around \(c\) on which \(f(x)\leq f(c)\). It has a local minimum at \(c\) if, on some open interval around \(c\), \(f(x)\geq f(c)\). A local extremum can be non-strict: equality at other nearby points is allowed.
For the situation considered here, \(c\) is a stationary point: \(f'(c)=0\). If \(f''(c)=0\) as well, the second derivative test does not classify it. The point may still be an extremum, as with \(f(x)=x^4\) at \(0\), or not, as with \(f(x)=x^3\) at \(0\). In both examples the two derivatives at \(0\) vanish.
A sign analysis of \(f'\) resolves these examples. The sign must be checked on intervals immediately to the left and right of \(c\), not just at \(c\): since \(f'(c)=0\), its value at the point itself cannot show which way the function is moving nearby.
Why the Sign Change Works
The connection between derivative signs and function values follows from the Mean Value Theorem. If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then some \(d\in(a,b)\) satisfies
Since \(b-a>0\), a positive derivative throughout an interval makes \(f(b)-f(a)>0\) whenever \(a<b\) there. Thus \(f\) is increasing on that interval. A negative derivative throughout makes \(f(b)-f(a)<0\), so \(f\) is decreasing. Applying this reasoning on both sides of \(c\) gives the classification.
Check that \(f'(c)=0\) and \(f''(c)=0\).
Determine whether \(f'\) is positive or negative on each side of \(c\), close enough to \(c\).
Increasing then decreasing gives a maximum; decreasing then increasing gives a minimum; increasing on both sides or decreasing on both sides gives neither.
Worked Examples
Worked Example: A Local Minimum Despite a Zero Second Derivative
Classify \(f(x)=x^4\) at \(c=0\).
First calculate the derivatives:
At \(c=0\),
The second derivative test is inconclusive. To classify the point, inspect the first derivative. If \(x<0\), then \(4x^3<0\), so \(f'(x)<0\). If \(x>0\), then \(4x^3>0\), so \(f'(x)>0\). The function decreases to \(0\) and increases after \(0\); therefore \(0\) is a local minimum.
Check the conclusion directly: \(f(0)=0\), and \(x^4>0\) for every \(x\ne 0\). Thus \(f(x)>f(0)\) for every nearby \(x\ne0\), confirming a strict local minimum. Substitution also verifies both derivative values used above are zero.
Worked Example: A Stationary Point That Is Not an Extremum
Classify \(g(x)=x^3\) at \(c=0\).
Differentiate and evaluate:
Again the second derivative test is inconclusive. For every \(x\ne0\), \(3x^2>0\), so \(g'(x)>0\) on both sides of \(0\). The function is increasing through the point rather than changing direction. Therefore \(0\) is not a local extremum.
Check directly: if \(x<0\), then \(g(x)=x^3<0=g(0)\); if \(x>0\), then \(g(x)=x^3>0=g(0)\). Values occur both below and above \(g(0)\) in every neighborhood of \(0\). Thus \(0\) is neither a local maximum nor a local minimum.
Worked Example: A Local Maximum Despite a Zero Second Derivative
Classify \(h(x)=-x^4+5\) at \(c=0\).
Calculate the derivatives:
At \(0\),
For \(x<0\), \(x^3<0\), so \(-4x^3>0\). For \(x>0\), \(x^3>0\), so \(-4x^3<0\). Therefore \(h'\) changes from positive to negative: the function increases toward \(0\) and decreases after \(0\). It has a local maximum at \(0\).
Check by substitution: \(h(0)=5\), while \(h(x)=5-x^4<5\) for every \(x\ne0\). This confirms a strict local maximum. The derivative calculations also give \(h'(0)=h''(0)=0\), as required for the inconclusive case.
A Justification Using the Mean Value Theorem
Here is why the sign patterns above establish the claimed extrema. Suppose \(f'(x)>0\) immediately to the left of \(c\), and \(f'(x)<0\) immediately to the right. Choose any \(x<c\) sufficiently close to \(c\). The Mean Value Theorem on \([x,c]\) gives a number \(d\in(x,c)\) such that
Here \(f'(d)>0\) and \(c-x>0\), so \(f(c)-f(x)>0\), or \(f(x)<f(c)\). For any \(y>c\) sufficiently close to \(c\), the theorem on \([c,y]\) gives a number \(e\in(c,y)\) such that
Here \(f'(e)<0\) and \(y-c>0\), so \(f(y)-f(c)<0\), or \(f(y)<f(c)\). Thus nearby function values on both sides are below \(f(c)\), which proves that \(c\) is a local maximum. Reversing the signs proves the local-minimum case.
If \(f'\) is positive on both sides, the same theorem gives \(f(x)<f(c)\) for nearby \(x<c\) and \(f(y)>f(c)\) for nearby \(y>c\). Those values lie on opposite sides of \(f(c)\), so \(c\) is neither type of extremum. If \(f'\) is negative on both sides, the inequalities reverse and again values lie on opposite sides of \(f(c)\).
Common Pitfall: Treating “Inconclusive” as “No Extremum”
A frequent error is to reason that because \(f''(c)=0\), the point cannot be an extremum. For instance, one might write:
The conclusion does not follow. A zero second derivative makes the second derivative test inconclusive; it does not rule out an extremum. For \(x^4\), the first derivative is negative to the left of \(0\) and positive to the right. The function therefore has a local minimum there, as the direct check \(x^4>0=f(0)\) for \(x\ne0\) confirms.
A related mistake is to inspect only \(f''(c)\) and skip the signs of \(f'\) nearby. The whole purpose of the follow-up analysis is to find out whether the function changes direction as it passes through \(c\). Evaluate or factor \(f'(x)\), then determine its sign on each side.
Limits of This Sign Pattern
The conclusions here use a strict sign for \(f'\) on each side of \(c\), in an interval close enough to the point. A derivative equal to zero at \(c\) is expected; what matters is its behavior nearby. If \(f'\) has additional zeros or does not keep a consistent sign on one side, do not assume that a simple left-right pattern applies. Use a sufficiently detailed sign analysis to determine what happens on intervals around \(c\).
The derivative signs classify the behavior of the function, not the reason the second derivative test failed. In each worked example, both derivatives vanish at the stationary point, yet the conclusions differ. That is why the sign change, rather than the pair of zero values, supplies the decisive evidence.
Check Your Understanding
For each question, use the derivatives and their signs near the stated point to justify the classification.
- For \(p(x)=x^6\), verify that \(p'(0)=p''(0)=0\), then classify \(0\).
- For \(q(x)=-x^6\), verify that \(q'(0)=q''(0)=0\), then classify \(0\).
- For \(r(x)=x^5\), verify that \(r'(0)=r''(0)=0\), then decide whether \(0\) is a local extremum.
- What does \(f'(c)=0\) and \(f''(c)=0\) establish, and what does it fail to establish?
- Explain how the Mean Value Theorem connects a positive derivative on an interval to increasing function values.
