Second Derivative Test

Curvature at a Stationary Point

The previous tutorial considered global extrema: comparisons with every value on a domain. The second derivative test addresses a more local question. When the first derivative vanishes at an interior point, the second derivative can indicate whether nearby function values lie above or below the value at that point. This test classifies local behavior; by itself, it does not establish a global extremum.

Recall Fermat’s Theorem: if a differentiable function has a local extremum at an interior point, its derivative there is zero. Such a point is called stationary when its derivative is zero. The first derivative condition identifies candidates, but it does not distinguish a local minimum from a local maximum, or from a point that is neither. The Second-Derivative Test for a Strict Local Extremum, established in “Local Extrema,” supplies that distinction when the second derivative is nonzero.

Second-Derivative Test (recall): Suppose $f$ is twice differentiable on a neighborhood of an interior point $a$, and $f'(a)=0$. If $f''(a)>0$, then $f$ has a strict local minimum at $a$. If $f''(a)<0$, then $f$ has a strict local maximum at $a$. If $f''(a)=0$, the test gives no conclusion.

The sign has a geometric interpretation: positive second derivative corresponds to upward curvature at the stationary point, while negative second derivative corresponds to downward curvature. The hypotheses matter. The point must be interior, the function must be twice differentiable as required, and its first derivative must vanish. A value of $f''(a)$ alone does not classify an arbitrary point where $f'(a)\ne0$.

Applying the Test

To use the test, first solve $f'(x)=0$ to find stationary points. Then evaluate $f''$ at each one. A positive value identifies a strict local minimum; a negative value identifies a strict local maximum. If the second derivative is zero, stop short of a classification and inspect the function by another method, such as comparing its values directly or studying its derivative signs.

Worked Example: A Strict Local Minimum

Let $f(x)=2x^2-8x+1$. Its derivatives are

f'(x)=4x-8,\qquad f''(x)=4.

The stationary-point equation $4x-8=0$ gives $a=2$. Since $f''(2)=4>0$, the Second-Derivative Test gives a strict local minimum at $2$. Its value is

$$ f(2)=2(2)^2-8(2)+1=8-16+1=-7. $$

In this example, direct algebra also confirms the classification:

$$ f(x)-f(2)=2x^2-8x+8=2(x-2)^2. $$

For $x\ne2$, the square $(x-2)^2$ is positive, so $f(x)-f(2)>0$. This verifies that nearby values, and indeed all other values, exceed $f(2)$. The test itself only asserts a local minimum; the algebra reveals the stronger global comparison.

Worked Example: A Strict Local Maximum

Consider $g(x)=-x^2+6x-4$. Differentiating gives

g'(x)=-2x+6,\qquad g''(x)=-2.

The only stationary point is $a=3$, because $-2(3)+6=0$. At that point $g''(3)=-2<0$, so the Second-Derivative Test gives a strict local maximum. The function value is

$$ g(3)=-(3)^2+6(3)-4=-9+18-4=5. $$

Completing the square checks the comparison explicitly:

g(x)=-(x-3)^2+5, \qquad g(x)-g(3)=-(x-3)^2.

For every $x\ne3$, this difference is negative. Thus $g(3)$ exceeds the value at every other real input, although the second derivative test alone only provides a local conclusion.

When the Second Derivative Is Zero

The case $f''(a)=0$ is not a weaker version of a minimum or maximum conclusion. It is genuinely inconclusive: different kinds of behavior are possible under the same first- and second-derivative conditions. The value zero does not mean that the point is flat in a way that rules out an extremum; nor does it indicate which kind of behavior occurs.

Worked Example: Three Outcomes When the Test Is Inconclusive

Compare the functions $u(x)=x^4$, $v(x)=-x^4$, and $w(x)=x^3$ at $a=0$. For each function, the first derivative and second derivative vanish at zero:

\begin{aligned} u'(x)&=4x^3, & u''(x)&=12x^2,\\ v'(x)&=-4x^3, & v''(x)&=-12x^2,\\ w'(x)&=3x^2, & w''(x)&=6x, \end{aligned} \qquad u'(0)=v'(0)=w'(0)=0,\quad u''(0)=v''(0)=w''(0)=0.

For $u$, every $x\ne0$ satisfies $u(x)=x^4>0=u(0)$, so zero is a strict local minimum. For $v$, every $x\ne0$ satisfies $v(x)=-x^4<0=v(0)$, so zero is a strict local maximum. For $w$, if $x>0$ then $w(x)>w(0)$, while if $x<0$ then $w(x)<w(0)$. Every neighborhood of zero therefore contains both larger and smaller function values, and zero is neither a local maximum nor a local minimum.

All three examples satisfy $f'(0)=f''(0)=0$, yet their classifications differ. When the second derivative test is inconclusive, additional information is necessary. The derivative sign-change criterion from “Local Extrema” is one possible tool: it can classify a point by tracking the sign of $f'$ on either side.

A common error is to treat $f''(a)=0$ as evidence for a minimum because the graph may appear to flatten there. The examples show why that is unsafe. Another error is to use $f''(a)>0$ to claim that $a$ is a global minimum. The second-derivative test is local; establishing a global conclusion requires a comparison over the whole domain, as discussed in “Global Extrema.”

Curvature Bounds Give More Than Classification

A pointwise sign test tells us which kind of strict local extremum occurs, but not how much the function changes as we move away from the point. A lower bound on the second derivative throughout an interval gives a quantitative estimate. The next theorem turns a uniform positive curvature bound into a lower bound on the increase in function values. The proof uses the Second-Derivative Criterion for Convexity and the Supporting-Line Inequality established earlier in the course.

Theorem (Quadratic Separation from a Positive Curvature Bound): Let $I$ be an open interval, let $a\in I$, and suppose $f:I\to\mathbb{R}$ is twice differentiable with $f'(a)=0$. If there is a constant $m>0$ such that $f''(x)\geq m$ for every $x\in I$, then, for every $x\in I$, $$ f(x)-f(a)\geq \frac{m}{2}(x-a)^2. $$ In particular, $a$ is the unique global-minimum point of $f$ on $I$.

Proof. Define $g:I\to\mathbb{R}$ by

g(x)=f(x)-\frac{m}{2}(x-a)^2.

Its second derivative is $g''(x)=f''(x)-m\geq0$ throughout $I$. By the Second-Derivative Criterion for Convexity, $g$ is convex on $I$. Also, $g(a)=f(a)$, and differentiation gives

$$ g'(a)=f'(a)-m(a-a)=0. $$

Apply the Supporting-Line Inequality to $g$ at $a$. For every $x\in I$,

g(x)\geq g(a)+g'(a)(x-a)=g(a).

Substituting the definition of $g$, and using $g(a)=f(a)$, yields

f(x)-\frac{m}{2}(x-a)^2\geq f(a),

which is the claimed inequality. If $x\ne a$, then $(x-a)^2>0$ and $m>0$, so $f(x)-f(a)\geq \frac{m}{2}(x-a)^2>0$. Thus $f(a)$ is strictly less than $f(x)$ at every other point of $I$, proving that $a$ is the unique global-minimum point on $I$. $\square$

This theorem is stronger than the pointwise test in two ways. It gives a numerical lower bound on the increase, and it compares $f(a)$ with every point in the interval $I$, not merely with points sufficiently close to $a$. The interval and the uniform bound matter: knowing only that $f''(a)>0$ does not provide a fixed positive lower bound for $f''$ across an entire interval.

Worked Example: Estimating the Increase from Curvature Bounds

Let $p(x)=x^2+\frac14x^4$ on $I=(-1,1)$. Its derivatives are

p'(x)=2x+x^3,\qquad p''(x)=2+3x^2.

We have $p'(0)=0$. For every $x\in(-1,1)$, $0\leq x^2<1$, so $2\leq p''(x)<5$. In particular, the theorem applies with $a=0$ and $m=2$, giving

p(x)-p(0)\geq \frac{2}{2}(x-0)^2=x^2.

Here $p(0)=0$, and direct calculation shows $p(x)=x^2+\frac14x^4\geq x^2$, as the theorem predicts. For every nonzero $x\in(-1,1)$, this bound is strictly positive, so zero is the unique global-minimum point on that interval. The conclusion is quantitative: the function rises by at least $x^2$ above its minimum.

A Positive Curvature Bound Limits Stationary Points

Uniform positive curvature also prevents the derivative from returning to zero after it has crossed zero. This gives a useful uniqueness result on an interval. It differs from the pointwise second-derivative test: it concerns all stationary points in the interval, not just the behavior near one candidate.

Theorem (At Most One Stationary Point Under a Positive Curvature Bound): Let $I$ be an open interval, and suppose $f:I\to\mathbb{R}$ is twice differentiable. If $f''(x)\geq m>0$ for every $x\in I$, then $f$ has at most one stationary point in $I$.

Proof. Let $u,v\in I$ with $u<v$. The function $f'$ is continuous on $[u,v]$, because it is differentiable there, and differentiable on $(u,v)$. By the Mean Value Theorem, there is a point $c\in(u,v)$ such that

$$ f'(v)-f'(u)=f''(c)(v-u). $$

Since $f''(c)\geq m>0$ and $v-u>0$, this gives $f'(v)-f'(u)>0$. Thus $f'(v)>f'(u)$ whenever $u<v$, so $f'$ is strictly increasing on $I$. A strictly increasing function cannot equal zero at two distinct points: if $f'(u)=f'(v)=0$ with $u<v$, strict increase would require $f'(u)<f'(v)$, a contradiction. Therefore $f$ has at most one stationary point in $I$. $\square$

The theorem does not assert that a stationary point exists. A strictly increasing derivative may remain positive throughout the interval, for example. If a stationary point does exist, however, the quadratic-separation theorem shows that it is the unique global minimum on the interval. Existence and classification are separate questions: the curvature bound supplies uniqueness and behavior, but not necessarily a point where the derivative vanishes.

Worked Example: A Unique Stationary Point on an Interval

For $p(x)=x^2+\frac14x^4$ on $(-1,1)$, the curvature estimate in the preceding example gives $p''(x)=2+3x^2\geq2$. The positive-curvature-bound theorem therefore guarantees at most one stationary point. We can locate it by solving

$$ p'(x)=2x+x^3=x(2+x^2)=0. $$

For every real $x$, $2+x^2\geq2>0$, so the equation holds exactly when $x=0$. This verifies that a stationary point exists and that it is the only one in $(-1,1)$. Since $p(0)=0$, the quadratic-separation result further gives $p(x)\geq x^2>0$ for every nonzero $x$ in the interval. The stationary point is thus the unique global minimum there.

The second derivative test is most direct when $f''(a)$ is nonzero: its sign classifies a stationary point as a strict local minimum or maximum. If $f''(a)=0$, no classification follows without further analysis. When the second derivative has a uniform positive or negative bound on an interval, stronger conclusions become available, including quantitative separation and uniqueness of a stationary point. These conclusions depend on interval-wide information and should not be confused with what a single value $f''(a)$ can establish.

Check Your Understanding

Use the second derivative test and the curvature results to answer the following questions.

What classification follows if $f'(a)=0$ and $f''(a)<0$, and why is the conclusion local?
Give two different behaviors that can occur when $f'(a)=f''(a)=0$.
In the quadratic-separation theorem, why must the lower bound $m$ be strictly positive?
What inequality does the quadratic-separation theorem give when $f'(a)=0$ and $f''(x)\geq m>0$ throughout an interval?
Why does a positive lower bound on $f''$ imply at most one stationary point, but not guarantee that one exists?

What You'll Learn