Second Derivative Test: Solved Examples

Assumption is the mother of all screw-ups.

If you choose not to decide, you still have made a choice, Neil Peart

Steps to Perform the Second Derivative Test

1. Find the critical points of the function f(x, y) by solving the system of equations: $f_x(x, y) = 0$ and $f_y(x, y) = 0$.
2. Calculate the second partial derivatives. Let A = $f_{xx}(x_0,y_0)= \frac{\partial^2 f}{\partial x^2}(x_0, y_0); B = f_{xy}(x_0, y_0) = \frac{\partial^2 f}{\partial x\partial y}(x_0, y_0) = \frac{\partial^2 f}{\partial y\partial x}(x_0, y_0); C = f_{yy}(x_0, y_0) = \frac{\partial^2 f}{\partial y^2}(x_0, y_0).$
3. Compute the Hessian matrix H at each critical point, H = $(\begin{smallmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y}\\\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{smallmatrix})$.
4. Calculate the determinant of the Hessian matrix at each critical point $(x_0, y_0)$: $D = f_{xx}f_{yy} - (f_{xy})^2$.
5. Analyze the sign of D and $f_{xx}$ to classify the point:

• If $D > 0$ and $f_{xx} > 0$, then the critical point is a local minimum.
• If $D > 0$ and $f_{xx} < 0$, then the critical point is a local maximum.
• If $D < 0$, then the critical point is a saddle point.
• If $D = 0$, the test is inconclusive. It could be a minimum, maximum, or saddle.

General Case: The Quadratic Form

Consider the function $f(x, y) = ax^2 + bxy + cy^2$.

1. First partial derivatives: $f_x = \frac{\partial f}{\partial x} = 2ax + by, f_y = \frac{\partial f}{\partial y} = bx + 2cy$.
2. Critical points: Setting both partial derivatives to zero yields a system where $(0, 0)$ is a solution.
3. Second partial derivatives: $f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2a, f_{xy}= \frac{\partial^2 f}{\partial x \partial y} = b, \frac{\partial^2 f}{\partial y^2} =2c$.
4. Hessian matrix ($H$) = $(\begin{smallmatrix}2a & b\\\ b & 2c\end{smallmatrix})$
5. **Determinant of the Hessian matrix $D = f_{xx} \cdot f_{yy} - (f_{xy})^2 = (2a)(2c) - b^2 = 4ac - b^2$
6. Analyze the nature of the critical points:

• If $4ac - b^2 > 0$ and $2a > 0$ (meaning $a > 0$), the critical point is a local minimum.
• If $4ac - b^2 > 0$ and $2a < 0$ (meaning $a < 0$), the critical point is a local maximum.
• If $4ac - b^2 < 0$, the critical point is a saddle point.
• If $4ac - b^2 = 0$, the test is inconclusive (a degenerate critical point). Notice that the previous results verify this theorem, and our last case corresponds to degenerate critical points.

Solved examples

• Find the local maximum, minimum values, and saddle points for $f(x, y) = x^2 + y^2$.

1. First, we find the partial derivatives of f with respect to x and y: $\frac{\partial f}{\partial x} = 2x, \frac{\partial f}{\partial y} = 2y.$
2. To find the critical points, we set these derivatives equal to zero: $2x = 0 \implies x = 0, 2y = 0 \implies y = 0.$ Therefore, the only critical point is $(0, 0)$.
3. Analysis: Because the function $f(x,y) = x^2 + y^2 \geq 0$ for all real numbers $x$ and $y$, and the function evaluates to exactly $f(0, 0) = 0$, the origin is not just a local minimum, but a global minimum.

• Find the local maximum, minimum values, and saddle points for $f(x, y) = x^3 - 3xy^2$.

1. First, we find the partial derivatives of f with respect to x and y: $\frac{\partial f}{\partial x} = 3x^2-3y^2, \frac{\partial f}{\partial y} = -6xy.$
2. To find the critical points, we set these derivatives equal to zero: $3x^2-3y^2 = 0 \implies x^2 = y^2 \implies x = \pm y, -6xy = 0 \implies x = 0~\text{or}~y = 0$. Combining these equations, we get that the only critical point is $(0, 0)$.
3. Evaluate the second partial derivatives and use the second derivative test: $f_{xx} = 6x, \quad f_{yy} = -6x, \quad f_{xy} = -6y$
   Evaluating the determinant $D$ at $(0, 0)$: $D = f_{xx}(0, 0)f_{yy}(0, 0) - [f_{xy}(0, 0)]^2 = (0)(0) - (0)^2 = 0$
   Analysis: Because $D = 0$, the second derivative test is inconclusive. You cannot determine the nature of $(0, 0)$ using this test alone.
   Graphically, this forms a “monkey saddle” where the surface has three alternating dips and rises.
   $\star$ Along the line (y = 0): $f(x, 0) = x^3$. For small $x > 0, f > 0$; for small $x < 0, f < 0$.
   $\star$ Along the line (x = 0): $f(0, y) = 0$. So along this line, the function is identically zero.
   $\star$ Along the line (y = x): $f(x, x) = x^3 - 3x(x^2) = x^3 - 3x^3 = -2x^3$. For small $x > 0, f < 0$; for small $x < 0, f > 0$.
4. Conclusion: So in any neighborhood of $(0, 0)$, you can find points where $f > 0$, points where $f < 0$, and points where $f = 0$. That means $(0, 0)$ is not a local maximum or minimum. Instead, the function changes sign around $(0, 0)$ in multiple directions — this is exactly the hallmark of a saddle point (“a monkey saddle”).

• Verifying via Completing the Square. Find the local extrema for $f(x, y) = x^2 - 2xy + 3y^2 + 2x - 2y$.

1. First, we find the partial derivatives of f with respect to x and y: $\frac{\partial f}{\partial x} = 2x -2y +2, \frac{\partial f}{\partial y} = -2x +6y -2$
2. To find the critical points, we set these partial derivatives equal to zero:
   $\begin{cases} 2x -2y +2 = 0 ~(i) \\ -2x +6y -2 = 0 ~(ii) \end{cases}$
   We now solve this system of equations to find the critical points.
   First, let’s add (i) and (ii) equations together to eliminate x: $(2x−2y+2)+(−2x+6y−2)=0 \implies[\text{Simplifying this, we get:}] 4y = 0 \implies y = 0.$
   Now, substitute y = 0 into the first equation, $2x -2y +2 = 0: 2x -2 \cdot 0 + 2 = 0 \implies 2x + 2 = 0 \implies x = -1$. Therefore, the critical point is $(-1, 0)$.
3. To determine whether (−1, 0) is a local minimum, maximum, or saddle point, we are going to complete the square for the function $f(x, y)$.
   Group the perfect square for $(x-y)^2$: $f(x, y) = (x-y)^2 + 2(x-y) + 2y^2$
   Add and subtract $1$ to complete the square again, this time for the $(x-y)$ term: $f(x, y) = (x-y)^2 + 2(x-y) + 1 - 1 + 2y^2$
   $f(x, y) = ((x-y) + 1)^2 + 2y^2 - 1$.
   Since obviously $((x-y) + 1)^2 \geq 0$ and $2y^2 \geq 0$ for all real numbers, the lowest possible value of this function occurs when those squared terms are zero, leaving $-1$. Since our critical point $(-1, 0)$ satisfies $f(-1, 0) = -1$, the point $(-1, 0)$ is indeed a minimum.

• Find the local maximum, minimum values, and saddle points for $f(x, y) = -2x^2 + 2xy - y^2 + 12x - 4y - 7$.

1. Find the first partial derivatives: $f_x = -4x + 2y + 12$, $f_y = 2x - 2y - 4$.
2. Critical Points: Solve the system:
   $\begin{cases} -4x + 2y + 12 = 0 ~(i)\\ 2x -2y -4 = 0 ~(ii)\end{cases}$
   Adding these two equations (i) + (ii): $-2x + 8 = 0 \implies x = \frac{-8}{-2} = 4$.
   Substitute $x = 4$ into (i): -4·4 + 2y + 12 = 0 ↭ -4 + 2y = 0 ↭ y = $\frac{4}{2}= 2.$ Therefore, the critical point is $(4, 2)$.
3. Evaluate the second partial derivatives and use the second derivative test: $f_{xx} = -4, \quad f_{yy} = -2, \quad f_{xy} = 2$
   Calculate the determinant $D$ of the Hessian matrix: $D = (-4)(-2) - (2)^2 = 8 - 4 = 4$
   Analysis: Because $D = 4 > 0$ and $f_{xx} = -4 < 0$, the critical point evaluates to a local maximum. The value at this maximum is $f(4, 2) = 13$.

• Polynomial Extrema. Find the local maximum, minimum values, and saddle points for $f(x, y) = 4 + x^3 + y^3 - 3xy$.

1. Find the first partial derivatives: $f_x = 3x^2 - 3y$, $f_y = 3y^2 - 3x$
2. Critical points: Set both derivatives to zero.
   $\begin{cases} 3x^2-3y = 0 ~(i)\\ 3y^2-3x = 0 ~(ii)\end{cases}$
   From (i), we get $3x^2 - 3y = 0 \implies y = x^2$.
   Substitute $y = x^2$ into (ii): $3(x^2)^2 - 3x = 0 \implies 3x(x^3 - 1) = 0$
   This yields $x = 0$ or $x = 1$. Since $y = x^2$, the critical points are $(0, 0)$ and $(1, 1)$.
3. Second partial derivatives: $f_{xx} = 6x, \quad f_{yy} = 6y, \quad f_{xy} = -3$
4. Second Derivative Test: The determinant $D$ of the Hessian matrix is $D = f_{xx}f_{yy} - (f_{xy})^2 = (6x)(6y) - (-3)^2 = 36xy - 9$.
5. At the critical point $(0, 0): D(0, 0) = 36\cdot 0\cdot 0 -9 = -9 < 0 \implies$ $(0, 0)$ is a saddle point.
   At the critical point $(1, 1): D(1, 1) = 36\cdot 1\cdot 1 -9 = 27 > 0$ and $f_{x}(1, 1) = 6 > 0 \implies $ $(1, 1)$ is a relative (local) minimum with $f(1, 1) = 3$.

• Rational Function Extrema. f(x, y) = $2xy + \frac{64}{y} +\frac{32}{x}$. Show that $(2, 4)$ is a critical point and a local minimum for $f(x, y) = 2xy + \frac{64}{y} + \frac{32}{x}$.

1. First, we need to compute the partial derivatives of f with respect to x and y: $f_x = 2y -\frac{32}{x^2}, f_y = 2x -\frac{64}{y^2}$.
2. To find the critical points, we set these partial derivatives equal to zero and solve for x and y: $2y -\frac{32}{x^2} = 0, 2x -\frac{64}{y^2} = 0$
   From the first equation: $2y = \frac{32}{x^2} \implies y = \frac{16}{x^2}$.
   Substitute $y = \frac{16}{x^2}$ into the second equation: $2x = \frac{64}{\frac{16^2}{x^4}} = \frac{64}{\frac{256}{x^4}} = \frac{64x^4}{256} = \frac{x^4}{4}.$
   Simplify and solve for x: $2x = \frac{x^4}{4} \implies 8x = x^4 \implies x(x^3-8) = 0$. This gives us $x = 0$ or $x^3 = 8,~ (x = 2)$.
   Since x = 0 is not valid (division by zero), we have x = 2. Substitute x = 2 back into $y = \frac{16}{x^2}, y = \frac{16}{4} = 4$. Thus, the only critical point is $(2, 4)$.
3. Let’s verify that $(2, 4)$ is indeed a critical point by plugging it back into the partial derivatives: $f_x(2, 4) = 8-\frac{32}{4} = 8-8 = 0; f_y(2, 4) =4-\frac{64}{16} = 4 -4 = 0.$
4. Now, we compute the second partial derivatives: $f_{xx} = \frac{64}{x^3}, f_{xy} = f_{yx} =2, f_{yy} = \frac{128}{y^3}$.
5. Then, evaluating these at (2, 4), we get $f_{xx} = \frac{64}{x^3}|_{(2, 4)} = 8$. $f_{xy} = f_{yx} =2.$ $f_{yy} = \frac{128}{y^3}|_{(2, 4)} =2$.
   The second derivative test involves computing the determinant D of the Hessian matrix H: $AC - B^2 = \text{det}(\begin{smallmatrix}8 & 2\\\ 2 & 2\end{smallmatrix}) = 8 \cdot 2-2^2 = 12 > 0$ and $f_{xx}(2, 4) = 8 > 0 \implies$ $(2, 4)$ is a local minimum.

• Find the extrema for $f(x, y) = x^3 - 3x + y^3 - 3y^2$.

1. To find the critical points, we first compute the partial derivatives: $f_x = 3x^2 -3 = 3(x^2-1) = 3(x-1)(x+1), f_y = 3y^2-6y = 3(y^2-2y)=3y(y-2).$
2. Set the partial derivatives equal to zero: $3(x-1)(x+1) = 0 \implies x = \pm 1$.
   $3(y^2 -2y) = 3y(y-2) = 0 \implies y = 0$ or $y = 2$. Thus, there are four critical points, namely $(1, 0), (1, 2), (-1, 0), \text{ and } (-1, 2)$.
3. Now, compute the second partial derivatives, $f_{xx} = 6x, f_{xy} = 0, f_{yy} = 6y -6$.
4. The second derivative test involves computing the determinant D of the Hessian matrix H: $D = AC -B^2 = (6x)(6y-6) -0^2 = 36xy -36x = 36x(y-1)$.
   $D(1, 2) = 36x(y-1)|_{(1, 2)} = 36 > 0, A = 6 > 0 \implies$ $(1, 2)$ is a local minimum.
   $D(-1, 2) = 36x(y-1)|_{(-1, 2)} = -36 < 0 \implies $ $(-1, 2)$ is a saddle point.
   $D(1, 0) = 36x(y-1)|_{(1, 0)} = -36 < 0 \implies$ $(1, 0)$ is a saddle point.
   $D(-1, 0) = 36x(y-1)|_{(-1, 0)} = 36 > 0, A = 6 \cdot (-1) < 0 \implies $ $(-1, 0)$ is a local maximum.

A saddle point occurs when a surface curves upward in one direction and downward in another, creating a shape mathematically identical to a horse's saddle.

• Classify the critical points for $f(x, y) = x + y + \frac{1}{xy}$ where $x, y > 0$.

1. To find the critical points, we first compute the partial derivatives: $f_x =1-\frac{1}{x^2y}, f_y = 1-\frac{1}{xy^2}$.
2. Set the partial derivatives equal to zero: $1-\frac{1}{x^2y} = 0, 1-\frac{1}{xy^2} = 0$.
   $1-\frac{1}{x^2y} = 0 \implies \frac{1}{x^2y} = 1 \implies x^2y = 1$ (i)
   $1-\frac{1}{xy^2} = 0 \implies \frac{1}{xy^2} = 1 \implies xy^2 = 1$ (ii)
   Divide (i) by (ii): $xy^{-1} = 1 \implies x = y$.
   Substitute x = y into (ii): $x^3 = 1$ [By assumption, $x, y > 0$] $x = 1 \implies[x = y] y = 1$. Thus, there is only one critical point $(1, 1)$.
3. Let’s compute the second partial derivatives, $f_{xx}=\frac{2}{x^3y}, f_{xy}=\frac{1}{x^2y^2}, f_{yy}=\frac{2}{xy^3}$.
4. Evaluate the Second Partial Derivatives at the Critical Point $(1, 1)$: A = 2, B = 1, and C = 2.
5. The second derivative test involves computing the determinant D of the Hessian matrix H: $D = AC -B^2 = 2\cdot 2 -1^2 = 3 > 0$, and $A = f_{xx}(1, 1) > 0$.
   Therefore, the critical point $(1, 1)$ is a local minimum.
6. Consider the Behavior at Infinity:
   $\lim_{x \to \infty} x + y + \frac{1}{xy} = \lim_{y \to \infty} x + y + \frac{1}{xy} =[\text{with y > 0 or x > 0 fixed respectively}] \infty.$
   Similarly, consider the limits as $x \to 0^+$ or $y \to 0^+$: $\lim_{x \to 0^+} x + y + \frac{1}{xy} = \lim_{y \to 0^+} x + y + \frac{1}{xy} =[\text{with y > 0 or x > 0 fixed respectively}] \infty$.
7. Arithmetic Mean–Geometric Mean. For any three positive numbers: $\frac{a+b+c}{3} \ge \sqrt[3]{abc}$.
8. Use the AM–GM inequality on the three positive numbers x, y, and $\frac{1}{xy}$: $\frac{x + y + \frac{1}{xy}}{3} \ge \sqrt[3]{x \cdot y \cdot \frac{1}{xy}} = \sqrt[3]{1} = 1.$
   Multiplying both sides by 3 yields: $x + y + \frac{1}{xy} \ge 3$ for all $x > 0, y > 0$.
   Because the AM–GM inequality states that equality holds if and only if all terms are equal, we set them equal to solve for the boundary condition: $x = y = \frac{1}{xy} \quad \Rightarrow \quad x = y = 1.$
   Therefore, global minimum value $f_{min} = 3$. It occurs uniquely at $(1, 1)$.
9. Because the function increases without bound as either variable approaches zero or infinity, the local minimum at $(1,1)$ is actually the global minimum for the domain $x, y > 0$.
   There’s no way to “escape” to a smaller value at the boundary of the domain; everything blows up, while the interior point $(1,1)$ sits at the bottom with value 3.
   Therefore, in general we must always check both the interior critical points and the boundary behaviors/limits to understand if a local extremum is also an absolute extremum.

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