Local Extrema and the Second Derivative Test

0. Warmup: one-variable review

Before tackling two variables, recall the single-variable case. The function $y = x^2$ has a minimum at $x = 0$ because the tangent line there is horizontal — the derivative vanishes.

y = x² — minimum at origin

The tangent is horizontal at the minimum: $y'(0) = 0$ . The same idea extends to two variables using the gradient.

1. Local maxima and minima

Definition. A function $f(x,y)$ has a local maximum at $(a,b)$ if $f(a,b) \geq f(x,y)$ for all $(x,y)$ in some open disk around $(a,b)$ . It has a local minimum if $f(a,b) \leq f(x,y)$ in that disk.

Informally: a local maximum is a hilltop, a local minimum is a valley bottom — both from the local perspective of a neighborhood.

2. Critical points

Theorem. If $f$ has a local maximum or minimum at $(a,b)$ and the partial derivatives exist there, then

$\boxed{\nabla f(a,b) = \mathbf{0}, \quad \text{i.e.,} \quad f_x(a,b) = 0 \text{ and } f_y(a,b) = 0.}$

A point where $\nabla f = \mathbf{0}$ (or $\nabla f$ does not exist) is called a critical point.

Warning. Not every critical point is a local extremum. Just as $x = 0$ is a critical point of $y = x^3$ but not an extremum, functions of two variables can have critical points that are saddle points — neither a max nor a min.

3. Three-dimensional illustration: the paraboloid

The surface $z = x^2 + y^2$ has $f_x = 2x$ and $f_y = 2y$ . Both vanish only at $(0,0)$ . The tangent plane there is $z = 0$ — horizontal, confirming the minimum.

Blue = surface z = x² + y². Orange = tangent plane at critical point (0, 0, 0) where ∇f = 0.

The surface curves upward in every direction from the origin, so $(0,0)$ is indeed a local (and global) minimum.

4. Example: $f(x,y) = x^2 + y^2 - 2x - 6y + 14$

Find and classify the critical point.

Set the partial derivatives to zero:

$f_x = 2x - 2 = 0 \implies x = 1,$ $f_y = 2y - 6 = 0 \implies y = 3.$

So $(1,3)$ is the only critical point. Completing the square confirms it is a minimum:

$f(x,y) = (x-1)^2 + (y-3)^2 + 4 \geq 4,$

with equality at $(1,3)$ . The minimum value is $f(1,3) = 4$ .

Blue = surface z = (x−1)² + (y−3)² + 4. Orange = tangent plane at critical point (1, 3, 4) where ∇f = 0.

5. Example: $f(x,y) = y^2 - x^2$

Find and classify the critical point.

$f_x = -2x = 0, \qquad f_y = 2y = 0 \implies (0,0) \text{ is the only critical point.}$

But $f(0,0) = 0$ , while $f(1,0) = -1 < 0$ and $f(0,1) = 1 > 0$ . So $(0,0)$ is neither a local max nor a local min — it is a saddle point.

Blue = surface z = y² − x². Orange = tangent plane at critical point (0, 0, 0) where ∇f = 0.

6. The second derivative test

To classify a critical point without completing the square, use the discriminant

$D(a,b) = f_{xx}(a,b)\, f_{yy}(a,b) - \bigl[f_{xy}(a,b)\bigr]^2.$

Second derivative test. Let $(a,b)$ be a critical point of $f$ .

Condition	Conclusion
$D > 0$ and $f_{xx}(a,b) > 0$	local minimum
$D > 0$ and $f_{xx}(a,b) < 0$	local maximum
$D < 0$	saddle point
$D = 0$	test is inconclusive

7. Hessian matrix and concavity

The Hessian matrix of $f$ at a point is

$H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{pmatrix}.$

For a unit vector $\mathbf{u} = \langle a, b \rangle$ , the second directional derivative is

$D_{\mathbf{u}}^2 f = \mathbf{u}^T H \mathbf{u} = a^2 f_{xx} + 2ab\, f_{xy} + b^2 f_{yy}.$

This is the Hessian quadratic form. The sign of $D_{\mathbf{u}}^2 f$ for all directions $\mathbf{u}$ determines concavity:

$H$ is positive definite $\iff$ $D_{\mathbf{u}}^2 f > 0$ for every $\mathbf{u} \neq \mathbf{0}$ $\iff$ the surface is concave up in every direction $\iff$ local minimum.
$H$ is negative definite $\iff$ $D_{\mathbf{u}}^2 f < 0$ for every $\mathbf{u} \neq \mathbf{0}$ $\iff$ the surface is concave down in every direction $\iff$ local maximum.
$H$ is indefinite (positive in some directions, negative in others) $\iff$ saddle point.

The discriminant $D = \det(H) = f_{xx}f_{yy} - f_{xy}^2$ is the determinant of the Hessian. When $D > 0$ the Hessian is definite (the sign is determined by $f_{xx}$ ); when $D < 0$ it is indefinite.

The paraboloid $z = x^2 + y^2$ has $H = \begin{pmatrix}2&0\\0&2\end{pmatrix}$ — positive definite, concave up in every direction:

Blue = surface z = x² + y². Orange = tangent plane at critical point (0, 0, 0) where ∇f = 0.

The saddle $z = y^2 - x^2$ has $H = \begin{pmatrix}-2&0\\0&2\end{pmatrix}$ — indefinite, concave up in the $y$ -direction but down in the $x$ -direction:

Blue = surface z = y² − x². Orange = tangent plane at critical point (0, 0, 0) where ∇f = 0.

8. Applying the second derivative test to $f = y^2 - x^2$

At the critical point $(0,0)$ :

$f_{xx} = -2, \qquad f_{yy} = 2, \qquad f_{xy} = 0.$

$D = f_{xx}\,f_{yy} - f_{xy}^2 = (-2)(2) - 0 = -4 < 0.$

Since $D < 0$ , the origin is a saddle point — confirming the observation from Section 5.

9. Absolute extrema on a closed region

Local extrema are found at critical points inside a region. But to find absolute (global) extrema on a closed, bounded region $D$ , we must also check the boundary of $D$ .

Procedure:

Find all critical points of $f$ inside $D$ and evaluate $f$ there.
Find the maximum and minimum of $f$ on the boundary of $D$ .
The overall maximum (or minimum) is the largest (or smallest) of all these values.

Analogy in one variable. Consider $y = x^2$ on the interval $[-1, 1]$ . The interior critical point is $x = 0$ (minimum, $y = 0$ ), but the absolute maximum occurs at the boundary points $x = \pm 1$ (where $y = 1$ ).

y = x² on [−1, 1]: absolute max at the boundary endpoints

The marked points at $x = \pm 1$ are the boundary; the absolute maximum value $1$ is attained there, not at the interior minimum.

10. Exercises

Solution.

$f_x = 3x^2 - 3 = 0 \implies x = \pm 1.$ $f_y = 2y - 2 = 0 \implies y = 1.$

Critical points: $(1, 1)$ and $(-1, 1)$ .

Second derivatives: $f_{xx} = 6x$ , $f_{yy} = 2$ , $f_{xy} = 0$ . So $D = 12x$ .

At $(1,1)$ : $D = 12 > 0$ and $f_{xx} = 6 > 0$ → local minimum. $f(1,1) = 1 - 3 + 1 - 2 = -3$ .
At $(-1,1)$ : $D = -12 < 0$ → saddle point.

Solution.

$f_x = 2x + y - 3 = 0,$ $f_y = x + 2y = 0 \implies x = -2y.$

Substituting: $2(-2y) + y - 3 = 0 \implies -3y = 3 \implies y = -1$ , $x = 2$ .

Critical point: $(2, -1)$ .

$f_{xx} = 2,\quad f_{yy} = 2,\quad f_{xy} = 1.$ $D = (2)(2) - 1^2 = 3 > 0, \quad f_{xx} = 2 > 0.$

Local minimum at $(2,-1)$ . $f(2,-1) = 4 - 2 + 1 - 6 = -3$ .

Solution.

$f_x = 4x^3 = 0$ and $f_y = 4y^3 = 0$ give the only critical point $(0,0)$ .

The second derivative test gives $D = (12 \cdot 0)(12 \cdot 0) - 0 = 0$ , so it is inconclusive.

However, $f(x,y) = x^4 + y^4 \geq 0$ for all $(x,y)$ , and $f(0,0) = 0$ . Therefore $(0,0)$ is a global minimum by direct argument.

Solution.

Interior critical points. $f_x = 2x - 2 = 0 \implies x = 1$ ; $f_y = 2y = 0 \implies y = 0$ . Critical point $(1,0)$ is inside the disk. $f(1,0) = 1 - 2 = -1$ .

Boundary. On $x^2 + y^2 = 4$ use the parametrization $x = 2\cos t$ , $y = 2\sin t$ :

$g(t) = 4 - 4\cos t.$

Maximum when $\cos t = -1$ : point $(-2, 0)$ , $f(-2,0) = 4 + 4 = 8$ .

Minimum when $\cos t = 1$ : point $(2, 0)$ , $f(2,0) = 4 - 4 = 0$ .

Conclusion. Absolute minimum $-1$ at $(1,0)$ ; absolute maximum $8$ at $(-2,0)$ .

Local Extrema and the Second Derivative Test

Local Extrema and the Second Derivative Test

0. Warmup: one-variable review

y = x² — minimum at origin

1. Local maxima and minima

2. Critical points

3. Three-dimensional illustration: the paraboloid

4. Example: f(x,y)=x2+y2−2x−6y+14f(x,y) = x^2 + y^2 - 2x - 6y + 14f(x,y)=x2+y2−2x−6y+14

5. Example: f(x,y)=y2−x2f(x,y) = y^2 - x^2f(x,y)=y2−x2

6. The second derivative test

7. Hessian matrix and concavity

8. Applying the second derivative test to f=y2−x2f = y^2 - x^2f=y2−x2

9. Absolute extrema on a closed region

y = x² on [−1, 1]: absolute max at the boundary endpoints

10. Exercises

4. Example: $f(x,y) = x^2 + y^2 - 2x - 6y + 14$

5. Example: $f(x,y) = y^2 - x^2$

8. Applying the second derivative test to $f = y^2 - x^2$