What the Second Derivative Test Tells You
At an interior point where a differentiable function has a horizontal tangent, the second derivative can help classify the point. It measures how the first derivative is changing nearby: a positive second derivative corresponds to upward bending, while a negative second derivative corresponds to downward bending.
The test is useful because it can classify a local extremum with a calculation at one point. But that conclusion is justified only when the required conditions hold. In particular, the point must satisfy \(f'(c)=0\), and the second derivative must have the appropriate nonzero sign.
- If \(f''(c)>0\), \(f\) has a strict local minimum at \(c\).
- If \(f''(c)<0\), \(f\) has a strict local maximum at \(c\).
A local minimum means that there is an interval around \(c\) on which \(f(x)\geq f(c)\); a strict local minimum has \(f(x)>f(c)\) for nearby \(x\ne c\). A local maximum is defined with the inequalities reversed. The theorem concerns an interior point and a local extremum; it does not by itself establish an absolute maximum or minimum over a larger domain.
Why the Sign Gives the Classification
Here is the reasoning for \(f''(c)>0\). Continuity of \(f''\) at \(c\) means that if \(f''(c)\) is positive, then \(f''\) remains positive sufficiently close to \(c\). On that neighborhood, the Mean Value Theorem applied to \(f'\) shows that \(f'\) is increasing. Since \(f'(c)=0\), \(f'\) is negative just to the left of \(c\) and positive just to the right. Thus \(f\) decreases toward \(c\) and increases away from it.
If \(f''(c)>0\), continuity ensures \(f''(x)>0\) for all \(x\) sufficiently close to \(c\).
The Mean Value Theorem shows that \(f'\) increases through its value \(f'(c)=0\), so \(f'(x)<0\) to the left and \(f'(x)>0\) to the right.
The function decreases before \(c\) and increases after \(c\), so nearby function values are greater than \(f(c)\).
The sign argument can be written precisely. Choose a neighborhood of \(c\) where \(f''(x)>0\). For any \(x\ne c\) in that neighborhood, apply the Mean Value Theorem to \(f'\) between \(c\) and \(x\). For some \(u\) between those points,
Because \(f'(c)=0\) and \(f''(u)>0\), this gives \(f'(x)>0\) when \(x>c\), and \(f'(x)<0\) when \(x<c\). Now apply the Mean Value Theorem to \(f\) between \(c\) and \(x\). For some \(v\) between them,
If \(x>c\), both factors on the right are positive. If \(x<c\), both are negative. In either case \(f(x)-f(c)>0\), proving that \(c\) is a strict local minimum. If \(f''(c)<0\), continuity instead gives \(f''<0\) nearby. The same Mean Value Theorem reasoning makes \(f'\) positive to the left and negative to the right, so \(f(x)-f(c)<0\) on both sides. Therefore \(c\) is a strict local maximum.
Worked Examples
Worked Example: A Positive Second Derivative
Classify the stationary point of \(f(x)=x^2+4x+7\) using the second derivative test.
First find the first derivative and set it equal to zero:
The candidate is \(c=-2\). Differentiate again and evaluate:
The function is a polynomial, so it is twice differentiable and its second derivative is continuous. Since \(f'(-2)=0\) and \(f''(-2)>0\), the test proves that \(f\) has a strict local minimum at \(x=-2\). Its value there is
Thus the local minimum point is \((-2,3)\). As a check, completing the square gives \(f(x)=(x+2)^2+3\). Since \((x+2)^2\geq 0\), the function is at least \(3\), with equality at \(x=-2\), confirming the classification.
Worked Example: Classifying Two Stationary Points
For \(g(x)=x^3-3x\), find and classify all stationary points using the second derivative test.
Differentiate and solve \(g'(x)=0\):
The second derivative is
At \(x=-1\), \(g'(-1)=3(1)-3=0\) and \(g''(-1)=6(-1)=-6<0\). The second derivative test proves a strict local maximum there. Its function value is
At \(x=1\), \(g'(1)=3(1)-3=0\) and \(g''(1)=6(1)=6>0\). The test proves a strict local minimum there, with value
The classifications can also be checked directly. Around \(x=-1\), write \(x=-1+h\). Then
For sufficiently small nonzero \(h\), \(-3h^2+h^3<0\), so the nearby values are below \(2\). Around \(x=1\), write \(x=1+h\):
For sufficiently small nonzero \(h\), \(3h^2+h^3>0\), so nearby values are above \(-2\). Both checks agree with the test.
Worked Example: A Function Defined on a Restricted Domain
Classify the stationary point of \(h(x)=\ln x-\frac{x}{2}\), whose domain is \(x>0\).
Differentiate on the domain and solve for a stationary point:
The candidate \(2\) is in the domain. The second derivative and its value at the candidate are
The function is twice differentiable with continuous second derivative for \(x>0\). Because \(h'(2)=0\) and \(h''(2)<0\), the second derivative test proves that \(h\) has a strict local maximum at \(x=2\). The function value is
The result is the local maximum point \((2,\ln 2-1)\). As a check on the direction of change, \(h'(x)>0\) for \(0<x<2\), since \(1/x>1/2\), and \(h'(x)<0\) for \(x>2\). The function increases before \(2\) and decreases after it, consistent with a local maximum.
Writing a Complete Justification
In an AP response, a conclusion such as “there is a minimum” should be connected to the theorem’s hypotheses and evidence. A concise justification identifies the candidate, verifies the first-derivative condition, evaluates the second derivative there, and states the classification.
If a problem asks for the function value as well, evaluate \(f(c)\) and report the point \((c,f(c))\). Keep the candidate coordinate and the function value distinct: the test classifies the behavior at \(x=c\), while \(f(c)\) gives the corresponding vertical coordinate.
A Common Sign Error
The sign of \(f''(c)\) is easy to reverse. For example, for \(q(x)=x^2\), a mistaken calculation might say, “\(q''(0)=2>0\), so \(x=0\) is a local maximum.” The numerical derivative is correct, but the conclusion is not: a positive second derivative gives a local minimum, not a local maximum.
The correction is \(q'(x)=2x\), so \(q'(0)=0\), and \(q''(x)=2\), so \(q''(0)=2>0\). The second derivative test therefore gives a strict local minimum at \(x=0\). Directly, \(q(x)=x^2\geq 0=q(0)\), which checks the conclusion.
Another important safeguard is to evaluate \(f''\) at the stationary point, not at an unrelated input. The test uses the pair of values \(f'(c)\) and \(f''(c)\) at the same \(c\). If \(f'(c)\ne 0\), the test’s hypothesis is not satisfied, so the sign of \(f''(c)\) alone does not classify a local extremum there.
When the Test Gives a Classification
The decisive cases are \(f'(c)=0\) together with a continuous, nonzero second derivative at \(c\). A positive value proves a strict local minimum; a negative value proves a strict local maximum. These are sufficient conditions: once they hold, no further sign chart is needed to establish the local classification.
The requirement that \(f'(c)=0\) matters because the test concerns a stationary point. The requirement that the second derivative be nonzero matters because the conclusion depends on its sign. If a required condition is absent, do not claim that this test has classified the point. In particular, a zero second derivative does not supply either of the positive- or negative-sign conclusions in the theorem.
Finally, a local classification answers a question about nearby inputs only. To determine an absolute extremum on a closed interval, one must compare the relevant values across the interval, including its endpoints. The second derivative test alone makes no such global comparison.
Check Your Understanding
For each question, identify the hypotheses you would check before stating a conclusion from the second derivative test.
- If \(f'(3)=0\), \(f''(3)=5\), and \(f''\) is continuous near \(3\), what local classification follows?
- If \(g'(a)=0\) and \(g''(a)=-7\), what classification follows when the other hypotheses of the test hold?
- For \(p(x)=x^2-6x+10\), find the stationary point and use the second derivative test to classify it.
- Why is the sign of \(f''(c)\) alone insufficient when \(f'(c)\ne 0\)?
- Does a local maximum established by this test automatically have to be an absolute maximum on the function’s entire domain? Explain.
