Why Use the Second Derivative Test?
A stationary point is a point where the derivative is zero. At such a point, the graph has a horizontal tangent, but that fact alone does not tell us whether the point is a local maximum, a local minimum, or neither. The Second Derivative Test uses the sign of the second derivative to classify some stationary points efficiently.
The key idea is that the second derivative describes how the first derivative is changing. If \(f''(c)>0\), the first derivative is increasing through zero near \(c\); if \(f''(c)<0\), it is decreasing through zero. The resulting behavior of \(f\) distinguishes a local minimum from a local maximum.
The Second Derivative Test
To apply the test at \(x=c\), first locate a stationary point by solving \(f'(c)=0\). Then evaluate \(f''(c)\). The test applies when \(f\) is differentiable in an interval around \(c\) and \(f'\) is differentiable at \(c\), so that \(f''(c)\) exists. These conditions ensure that the derivative information needed for the conclusion is available.
- If \(f''(c)>0\), then \(f\) has a strict local minimum at \(c\).
- If \(f''(c)<0\), then \(f\) has a strict local maximum at \(c\).
- If \(f''(c)=0\), the test gives no conclusion. The point may be a local minimum, a local maximum, or neither.
The test gives a local classification. It does not establish that the point is the greatest or least value of \(f\) over its entire domain. It also does not apply just because \(f''(c)\) has a particular sign: the stationary-point condition \(f'(c)=0\) is essential.
Why the Test Works
Here is a proof for the case \(f''(c)>0\). The negative case follows by reversing the inequalities. Since \(f''(c)\) is the derivative of \(f'\) at \(c\), the definition of derivative gives the limit below. Because the limit is positive, the quotient is positive whenever \(x\) is sufficiently close to, but different from, \(c\).
Using \(f'(c)=0\), the numerator in this quotient is \(f'(x)\). Thus \(f'(x)/(x-c)>0\) for nearby \(x\ne c\). For \(x<c\), the denominator is negative, so \(f'(x)<0\). For \(x>c\), the denominator is positive, so \(f'(x)>0\).
Now take a nearby \(x<c\). By the Mean Value Theorem, there is a number \(\xi\) between \(x\) and \(c\) such that \(f(c)-f(x)=f'(\xi)(c-x)\). Here \(\xi<c\), so \(f'(\xi)<0\), while \(c-x>0\). Therefore \(f(c)-f(x)<0\), or \(f(x)>f(c)\). For nearby \(x>c\), the Mean Value Theorem gives \(f(x)-f(c)=f'(\xi)(x-c)\) for some \(\xi\) between \(c\) and \(x\). Both factors are positive, so \(f(x)>f(c)\) again. Consequently, \(c\) is a strict local minimum.
If \(f''(c)<0\), the quotient above is negative near \(c\). It follows that \(f'(x)>0\) just to the left and \(f'(x)<0\) just to the right. Applying the Mean Value Theorem on either side shows \(f(x)<f(c)\) for nearby \(x\ne c\), so \(c\) is a strict local maximum.
A Reliable Application Process
Differentiate \(f\) and solve \(f'(x)=0\) within the domain.
Confirm that \(f\) is differentiable around each point and that the second derivative exists at it.
Substitute each stationary value \(c\) into \(f''\).
A positive value gives a strict local minimum; a negative value gives a strict local maximum; zero is inconclusive.
Worked Examples
Worked Example: Classify Two Stationary Points
Let \(f(x)=x^3-3x\). Find and classify its stationary points, and give their coordinates.
Differentiate and solve for stationary values:
Differentiate again and evaluate at both values:
The test classifies \(x=-1\) as a strict local maximum and \(x=1\) as a strict local minimum. Evaluate the original function to obtain the coordinates:
Thus the local maximum is \((-1,2)\), and the local minimum is \((1,-2)\). As a check, \(f'(x)=3(x-1)(x+1)\) is positive for \(x<-1\), negative for \(-1<x<1\), and positive for \(x>1\). The function therefore changes from increasing to decreasing at \(-1\), and from decreasing to increasing at \(1\), in agreement with the classifications.
Worked Example: Classify Three Stationary Points
Let \(g(x)=x^4-4x^2+1\). Find and classify all its stationary points.
First differentiate and factor:
Solving \(g'(x)=0\) gives \(x=0\), \(x=\sqrt{2}\), and \(x=-\sqrt{2}\). The second derivative is
At the three stationary values,
The point with \(x=0\) is a strict local maximum; the points with \(x=\pm\sqrt{2}\) are strict local minima. To find their coordinates, substitute into \(g\):
The local maximum is \((0,1)\), and the local minima are \((\sqrt{2},-3)\) and \((-\sqrt{2},-3)\). As a check on the derivative calculations, differentiating \(g'(x)=4x^3-8x\) gives \(g''(x)=12x^2-8\), as used above. The classifications are local; this calculation by itself does not claim that either minimum is the global minimum over the domain.
Worked Example: A Zero Second Derivative Is Inconclusive
Consider \(h(x)=x^4\) at \(x=0\). Its first and second derivatives are
The point is stationary, but the Second Derivative Test gives no classification because the second derivative is zero. We can determine what happens directly: \(x^4\geq 0\) for every real \(x\), and \(h(0)=0\). Thus \(x=0\) is a strict local minimum (in fact, a global minimum). This confirms that a zero second derivative does not mean that no extremum exists.
For contrast, \(q(x)=x^3\) also has \(q'(0)=0\) and \(q''(0)=0\), but \(q(x)<0\) for \(x<0\) and \(q(x)>0\) for \(x>0\). Hence \(x=0\) is neither a local maximum nor a local minimum. The same test outcome, \(f''(0)=0\), is consistent with different behaviors.
Common Mistake: Treating Zero as a Verdict
A frequent error is to interpret \(f''(c)=0\) as proof that \(c\) is not an extremum. For \(h(x)=x^4\), that reasoning would be: \(h'(0)=0\), \(h''(0)=0\), therefore \(x=0\) is not an extremum. The conclusion is wrong, because \(x^4\geq 0=h(0)\) for every \(x\).
The correction is to keep the test's third case separate: when \(f''(c)=0\), the test is inconclusive. Use other information to decide what happens, such as a direct comparison of function values or a separate analysis of the derivative. Do not replace “the test gives no conclusion” with either “there is an extremum” or “there is no extremum.”
What the Test Can and Cannot Tell You
The Second Derivative Test is useful because one evaluation can classify a stationary point without requiring a full analysis of the function. But it answers a specific question under specific conditions: at a stationary point where the second derivative exists, does its sign certify a local maximum or minimum?
- If \(f'(c)\ne 0\), the Second Derivative Test does not apply at \(c\), even if \(f''(c)\) exists.
- If \(f'(c)=0\) but \(f''(c)=0\), the test is inconclusive; additional reasoning is needed.
- If the test identifies a local extremum, that conclusion concerns values near \(c\), not necessarily values across the entire domain.
- A point at the endpoint of a domain needs separate consideration: the two-sided neighborhood argument in the test is not available there.
Keep the sequence clear: find a stationary point, verify the required derivatives exist, evaluate the second derivative, and state only the conclusion its sign supports. That discipline makes the test both efficient and reliable.
Check Your Understanding
For each question, distinguish what the test establishes from what would require further analysis.
- State the Second Derivative Test at a stationary point when \(f''(c)>0\) and when \(f''(c)<0\).
- Why is the condition \(f'(c)=0\) required before using the test?
- If \(f'(c)=0\) and \(f''(c)=0\), what can you conclude from the test alone?
- For \(p(x)=x^3-6x\), find the stationary values and classify each using the Second Derivative Test.
- Does identifying a local minimum establish that it is the global minimum on the domain? Explain.
