Multivariable-Calculus · Unit 9 · Video 2 · Interactive Practice

The Second Derivative Test for f(x,y): AC − B² and Why It Works

IKey Formulas

Formula	Name	What it says
$A = f_{xx}(x_0, y_0),\ \ B = f_{xy}(x_0, y_0),\ \ C = f_{yy}(x_0, y_0)$	Second derivatives at a critical point	Three numbers, taken where $f_x = f_y = 0$
$AC - B^2 > 0$ : min if $A > 0$ , max if $A < 0$ ; $\ AC - B^2 < 0$ : saddle; $\ AC - B^2 = 0$ : no conclusion	Second derivative test	The sign of one number sorts the point
$w = ax^2 + bxy + cy^2 = y^2\left[at^2 + bt + c\right],\ \ t = \frac{x}{y}$	Quadratic by direction	Discriminant $b^2 - 4ac > 0$ : saddle; $< 0$ : min or max
$\Delta f \approx \frac{A}{2}(x - x_0)^2 + B(x - x_0)(y - y_0) + \frac{C}{2}(y - y_0)^2$	Quadratic approximation	$4\cdot\frac{A}{2}\cdot\frac{C}{2} - B^2 = AC - B^2$

Key Insight: $AC - B^2$ is exactly the number $4ac - b^2$ of the quadratic approximation, so wherever the quadratic terms decide the shape ( $AC - B^2 \neq 0$ ) the test inherits the completing-the-square classification.

IIVisualization 1 — The Sign of $w$ Direction by Direction

Along the direction $t = x/y$ , $w$ has the sign of $at^2 + bt + c$ , so two real roots mean a saddle.

a

b

c

IIIVisualization 2 — Near the Point, $f$ Behaves Like Its Quadratic Part

For $f = x^3 + y^3 - 3xy$ , shrinking the window makes $\Delta f$ and its quadratic approximation agree in sign.

Critical point Window h = 2.5

IVVisualization 3 — When $AC - B^2 = 0$

Four functions share $A = 2$ , $B = 0$ , $C = 0$ at the origin, yet the higher-order terms give them different shapes.

Function

VQuiz Questions

Problem 1 · Reading the Test

Given: at a critical point of $f$ , $\ f_{xx} = -4$ , $\ f_{xy} = 2$ , $\ f_{yy} = -3$ . Classify the point.

✅ Correct!

AC - B^2 = 12 - 4 = 8 > 0

, and

A = -4 < 0

makes it a maximum.

❌ Close, but…

AC - B^2 > 0

only says "min or max". The sign of

A

decides which, and

A = -4 < 0

❌ Check the product.

AC = (-4)(-3) = +12

, so

AC - B^2 = 12 - 4 = 8

, which is positive.

❌ Not quite. The test is inconclusive only when

AC - B^2 = 0

; here

AC - B^2 = (-4)(-3) - 2^2 = 8

Show solution

Name the second derivatives: $A = -4$ , $B = 2$ , $C = -3$ . Then

AC - B^2 = (-4)(-3) - 2^2 = 12 - 4 = 8 > 0.

A positive $AC - B^2$ means a local minimum or maximum, and the sign of $A$ chooses: $A = -4 < 0$ , so the point is a local maximum.

Problem 2 · A Negative $f_{xx}$ Is Not Enough

Given: $f(x, y) = x^3 - 3x + y^2 - 2y$ has a critical point at $(-1, 1)$ . Classify it.

✅ Correct!

AC - B^2 = (-6)(2) - 0^2 = -12 < 0

: a saddle, whatever the sign of

A

❌ Close, but… the sign of

A

matters only after

AC - B^2 > 0

. Here

AC - B^2 = (-6)(2) - 0 = -12 < 0

❌ Not quite.

B = 0

does not make the test inconclusive; only

AC - B^2 = 0

does, and here

AC - B^2 = -12

❌ Not quite. Evaluate the second derivatives at

(-1, 1)

f_{xx} = 6x = -6

f_{xy} = 0

f_{yy} = 2

Show solution

First derivatives: $f_x = 3x^2 - 3$ and $f_y = 2y - 2$ , both zero at $(-1, 1)$ , so it is a critical point.

Second derivatives: $f_{xx} = 6x$ , $f_{xy} = 0$ , $f_{yy} = 2$ . At $(-1, 1)$ : $A = -6$ , $B = 0$ , $C = 2$ .

AC - B^2 = (-6)(2) - 0^2 = -12 < 0.

So $(-1, 1)$ is a saddle point: $f$ curves down in the $x$ -direction and up in the $y$ -direction. (The other critical point, $(1, 1)$ , has $AC - B^2 = 12 > 0$ with $A = 6 > 0$ : a local minimum.)

Problem 3 · Building the Quadratic Approximation

Given: at a critical point $(x_0, y_0)$ , $\ A = f_{xx} = 4$ , $\ B = f_{xy} = 3$ , $\ C = f_{yy} = 2$ . Write $u = x - x_0$ , $v = y - y_0$ . Find the quadratic approximation of $\Delta f$ and the type of the point.

Quadratic approximation of Δf

Type of the critical point

✅ Correct!

\Delta f \approx 2u^2 + 3uv + v^2

has

4ac - b^2 = 8 - 9 = -1

, which equals

AC - B^2 = 8 - 9 = -1 < 0

: a saddle.

❌ Missing the halves. The square terms carry

\tfrac{1}{2}

\tfrac{A}{2}u^2 = 2u^2

and

\tfrac{C}{2}v^2 = v^2

❌ The middle term has no ½. It collects two equal contributions,

\tfrac{1}{2}f_{xy} + \tfrac{1}{2}f_{yx} = B

, so it is

3uv

❌ Counted twice. The two halves

\tfrac{1}{2}f_{xy}

and

\tfrac{1}{2}f_{yx}

add to

B = 3

, not

2B

❌ Check the approximation.

\Delta f \approx \tfrac{A}{2}u^2 + Buv + \tfrac{C}{2}v^2

❌ Check the coefficients. With the correct

a = 2

b = 3

c = 1

4ac - b^2 = 8 - 9 = -1 < 0

. A positive value comes from dropping the halves (

4 \cdot 4 \cdot 2 - 9 = 23

) or halving

B

❌ Check the type.

AC - B^2 = 4 \cdot 2 - 3^2 = -1

, which is negative.

Show solution

At a critical point the linear terms vanish, and the quadratic terms are

\Delta f \approx \tfrac{1}{2}f_{xx}u^2 + f_{xy}uv + \tfrac{1}{2}f_{yy}v^2 = \tfrac{4}{2}u^2 + 3uv + \tfrac{2}{2}v^2 = 2u^2 + 3uv + v^2.

As a quadratic $au^2 + buv + cv^2$ : $a = 2$ , $b = 3$ , $c = 1$ , so

4ac - b^2 = 4(2)(1) - 3^2 = -1, \qquad AC - B^2 = 4 \cdot 2 - 3^2 = -1.

The two numbers agree, as they must, and both are negative: the point is a saddle.

Problem 4 · A Valley Floor Along $y = x$

Given: $f(x, y) = (x - y)^2 - x^4$ has a critical point at the origin. Find $AC - B^2$ there, then decide the type of the point.

Value of AC − B² at (0, 0)

Type of the critical point

✅ Correct!

AC - B^2 = 4 - 4 = 0

, and along the floor

y = x

the quadratic part vanishes, leaving

f(x, x) = -x^4 < 0

, while

f(x, 0) = x^2 - x^4 > 0

for small

x \neq 0

❌ Watch the sign.

B = -2

but

B^2 = 4

, so

AC - B^2 = 4 - 4

, not

4 + 4

❌ Missing $B$ . The cross term

-2xy

(x - y)^2

gives

f_{xy} = -2

❌ Recompute.

f_{xx} = 2 - 12x^2

f_{xy} = -2

f_{yy} = 2

at the origin.

❌ Close, but…

(x - y)^2 \geq 0

is zero along the whole line

y = x

, and there

-x^4

is all that is left.

❌ Not quite. The test cannot decide, but inspecting

f

along lines through the origin can.

❌ Not quite. Compare

f

along the

x

-axis with

f

along the line

y = x

Show solution

Derivatives: $f_x = 2(x - y) - 4x^3$ , $f_y = -2(x - y)$ , both zero at the origin. Then $f_{xx} = 2 - 12x^2$ , $f_{xy} = -2$ , $f_{yy} = 2$ , so at $(0, 0)$ :

A = 2,\quad B = -2,\quad C = 2,\qquad AC - B^2 = 4 - (-2)^2 = 0.

The test cannot conclude. The quadratic part $(x - y)^2$ is a flat-bottomed valley whose floor is the line $y = x$ , so the higher-order term decides:

f(x, x) = -x^4 < 0 \ \ (x \neq 0), \qquad f(x, 0) = x^2 - x^4 = x^2(1 - x^2) > 0 \ \ (0 < |x| < 1).

$f$ takes both signs arbitrarily close to the origin, where $f = 0$ : a degenerate saddle.

Solved: 0 / 4

The Second Derivative Test for f(x,y): AC − B² and Why It Works

IKey Formulas

IIVisualization 1 — The Sign of www Direction by Direction

IIIVisualization 2 — Near the Point, fff Behaves Like Its Quadratic Part

IVVisualization 3 — When AC−B2=0AC - B^2 = 0AC−B2=0

VQuiz Questions

IIVisualization 1 — The Sign of $w$ Direction by Direction

IIIVisualization 2 — Near the Point, $f$ Behaves Like Its Quadratic Part

IVVisualization 3 — When $AC - B^2 = 0$