From a Stationary Point to a Classification
A stationary point is a point \(x=c\) where \(f'(c)=0\). The tangent line there is horizontal, but that fact alone does not tell us whether the graph has a local maximum, a local minimum, or neither. The second derivative test uses the value of \(f''(c)\) to classify certain stationary points efficiently.
The key is to check the test's conditions before using its conclusion. The point must be in the interior of the domain, the function must be twice differentiable on an interval around it, and its first derivative must be zero there. When those conditions hold, a negative second derivative indicates a local maximum; a positive second derivative indicates a local minimum.
What the Test Concludes
A local maximum at \(c\) means that, for every \(x\) sufficiently close to \(c\), \(f(x)\leq f(c)\). A local minimum means that, for every \(x\) sufficiently close to \(c\), \(f(x)\geq f(c)\). These are local comparisons: the function need not be largest or smallest on its entire domain.
The test classifies the point \(x=c\). If a problem asks for the corresponding point on the graph, report \((c,f(c))\), not just the \(x\)-coordinate. It is also useful to keep the two derivative roles separate: \(f'(c)=0\) identifies a stationary point, while the sign of \(f''(c)\) classifies it under the theorem's hypotheses.
A Justification Using Derivative Signs
The reason the test works is that the second derivative describes how the first derivative changes near \(c\). At a stationary point, \(f'(c)=0\). The definition of \(f''(c)\) gives the limit
Since \(f'(c)=0\), the quotient in this limit is \(f'(x)/(x-c)\). If its limit is positive, then sufficiently near \(c\), for \(x\ne c\), the quotient is positive. If its limit is negative, the quotient is negative. Because \(x-c\) is negative to the left of \(c\) and positive to the right, these quotient signs determine the signs of \(f'(x)\) on either side.
Near \(c\), \(f'(x)/(x-c)>0\). Thus \(f'(x)<0\) just left of \(c\), and \(f'(x)>0\) just right of \(c\).
The function decreases as it approaches \(c\) from the left and increases after \(c\). Therefore \(f(c)\) is less than nearby function values: \(c\) is a local minimum.
Near \(c\), \(f'(x)/(x-c)<0\). Thus \(f'(x)>0\) just left of \(c\), and \(f'(x)<0\) just right of \(c\). The function increases and then decreases, so \(c\) is a local maximum.
This argument uses the relationship between derivative signs and monotonicity. The quotient's sign remains the same as its nonzero limit sufficiently close to \(c\); this is why a strict inequality in the theorem matters.
A Reliable Procedure
For a candidate point \(c\), use the following sequence. In particular, do not begin by looking only at the second derivative: first establish that the candidate is in a setting where the theorem applies.
- Check that \(c\) is in the interior of the domain.
- Check that \(f\) is twice differentiable on an open interval containing \(c\).
- Verify that \(f'(c)=0\).
- Compute \(f''(c)\), keeping the value exact where possible.
- If \(f''(c)<0\), conclude that \(c\) is a local maximum. If \(f''(c)>0\), conclude that \(c\) is a local minimum.
- Find \(f(c)\) if the problem asks for the point on the graph or the local extreme value.
Worked Examples
Worked Example: Classify Two Stationary Points
Classify the stationary points of \(f(x)=x^3-3x^2-9x+1\).
This is a polynomial, so it is twice differentiable for every real \(x\). Its domain is \(\mathbb{R}\), and every real point is interior to the domain. Differentiate:
Setting \(f'(x)=0\) gives \(x=3\) or \(x=-1\). The second derivative is
At \(c=-1\), \(f''(-1)=6(-1)-6=-12<0\), so \(x=-1\) is a local maximum. Its function value is
At \(c=3\), \(f''(3)=6(3)-6=12>0\), so \(x=3\) is a local minimum. Its function value is
Check the stationary-point calculations: \(f'(-1)=3(-1-3)(-1+1)=0\), and \(f'(3)=3(3-3)(3+1)=0\). Thus the classifications apply, and the graph points are \((-1,6)\) for the local maximum and \((3,-26)\) for the local minimum.
Worked Example: A Local Maximum of a Trigonometric Function
Classify the stationary point of \(f(x)=\sin x+\cos x\) at \(c=\frac{\pi}{4}\).
Sine and cosine are twice differentiable for all real \(x\), and \(\frac{\pi}{4}\) is an interior point of the domain. Differentiate:
Since \(\sin(\frac{\pi}{4})=\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\),
and
The hypotheses hold, so \(x=\frac{\pi}{4}\) is a local maximum. The function value is
Check: substituting into the first derivative gives \(f'(\frac{\pi}{4})=0\), and the second derivative is negative there. The local maximum point is \((\frac{\pi}{4},\sqrt{2})\).
Worked Example: A Local Minimum Involving a Logarithm
Classify the stationary point of \(f(x)=e^x-2x\).
The exponential function and the linear function are twice differentiable for all real \(x\), so the domain and differentiability conditions hold at any real candidate. First find the stationary point:
The second derivative is \(f''(x)=e^x\), so
Therefore \(x=\ln 2\) is a local minimum. Its function value is
Check: \(f'(\ln 2)=e^{\ln 2}-2=2-2=0\), and \(f''(\ln 2)=2>0\). The local minimum point is \((\ln 2,2-2\ln 2)\).
Common Pitfall: Skipping the Hypothesis Check
A positive second derivative does not, by itself, classify every point as a local minimum. The test requires a stationary point and the stated differentiability and domain conditions. For instance, consider \(g(x)=x^2\) at \(c=1\). Although \(g''(1)=2>0\), the first derivative is \(g'(x)=2x\), so \(g'(1)=2\ne0\). The second derivative test in the theorem cannot classify \(x=1\), because it is not a stationary point.
Another common error is reversing the conclusions. For example, let \(h(x)=-x^2+4x\) and consider \(c=2\). Then \(h'(x)=-2x+4\), so \(h'(2)=-4+4=0\). Also, \(h''(x)=-2\), giving \(h''(2)=-2<0\). The correct conclusion is a local maximum, not a local minimum. As a check, \(h(2)=-4+8=4\), while \(h(2+t)=4-t^2<4\) for every nonzero \(t\), confirming the maximum.
Always pair the second-derivative sign with the verified stationary-point condition: negative means local maximum; positive means local minimum.
Why the Test Is Useful
Once stationary points have been found, the second derivative test can classify them by evaluating one additional derivative at each point. This is especially convenient when the second derivative is simple, as in a polynomial or exponential example. It also supplies a concise mathematical justification: the sign of \(f''(c)\) determines the nearby signs of \(f'\), which determine whether the function decreases then increases or increases then decreases.
The conclusions are local and conditional. The test does not state that a local maximum is an absolute maximum, and it does not apply merely because a second derivative has a particular sign somewhere. Its value comes from combining a checked stationary point with a strict second-derivative sign under the required hypotheses.
Check Your Understanding
For each question, state any hypothesis that must be checked before applying the test.
- If \(f'(c)=0\) and \(f''(c)=7\), what classification follows when the test's hypotheses hold?
- If \(f'(c)=0\) and \(f''(c)=-3\), what classification follows when the test's hypotheses hold?
- For \(p(x)=x^2+6x+5\), find its stationary point and classify it with the second derivative test.
- Why does the condition \(f'(c)=0\) matter when using the second derivative test?
- What does the test tell you about the function value at the classified point, and what does it not establish about the entire domain?
