Critical points — finding, classifying: Exercises

Before we begin, the two tools we lean on the whole way:

Here $f_{x}$ means "the slope of $f$ if you only walk in the $x$ -direction, freezing $y$ ", and $f_{xx}$ means "how that $x$ -slope itself is changing" — the curvature along $x$ . $f_{x y}$ is the cross term: how the $x$ -slope changes as you shift $y$ . If those words feel new, re-read the parent Critical points — finding, classifying first; the derivation lives in Taylor Series in Several Variables and Quadratic Forms and Definiteness.

Figure — Critical points — finding, classifying

The picture above is your answer key for step 3 of the recipe. Read it left to right:

Left (bowl): both curvatures point up, they agree → $D = f_{xx} f_{y y} - (f_{x y})^{2} > 0$ with $f_{xx} > 0$ . This is Exercise 3's $(4, - 2)$ and Exercise 8's $(0, 0)$ .
Middle (dome): both curvatures point down, still agreeing → $D > 0$ but $f_{xx} < 0$ . This is Exercise 7(b)'s $g = - x^{4} - y^{4}$ at the origin — a local maximum.
Right (saddle): the two curvatures fight — up along one axis, down along the other → $D < 0$ . This is Exercise 5's $(0, 0)$ and every saddle you meet below.

So when you compute $D$ , you are literally asking the picture: "do the curvatures agree (extremum) or fight (saddle)?" The amber dot marks the critical point in each panel — the one flat spot where the recipe applies.

L1 — Recognition

Recall Solution 1

What we do: just plug into $D = f_{xx} f_{y y} - (f_{x y})^{2}$ . No gradient, no Taylor — the second derivatives are handed to us. $D = (4) (9) - (2)^{2} = 36 - 4 = 32 > 0.$ Since $D > 0$ , it's an extremum, not a saddle. Now $f_{xx} = 4 > 0$ ("cup, smiley"), so it curves upward. Match this to the left panel of the figure above. Conclusion: local minimum.

Recall Solution 2

$D = - 7 < 0$ . The instant $D < 0$ , the two curvatures have opposite signs — up one way, down the other — so it is a saddle point, full stop (the right panel of the figure). The value of $f_{xx}$ is irrelevant here; checking it is wasted effort. The student's error: they tried to use $f_{xx}$ to pick max vs min, but that branch only exists when $D > 0$ . With $D < 0$ there is no max/min to describe. Conclusion: saddle.

L2 — Application

Recall Solution 3

Step 1 — gradient (why: extrema need zero slope in every direction). $f_{x} = 2 x + y - 6, f_{y} = x + 2 y .$ Step 2 — solve the system. From $f_{y} = 0$ : $x = - 2 y$ . Substitute into $f_{x} = 0$ : $2 (- 2 y) + y - 6 = 0 \Rightarrow - 3 y - 6 = 0 \Rightarrow y = - 2, x = - 2 (- 2) = 4.$ One critical point: $(4, - 2)$ . Step 3 — discriminant. $f_{xx} = 2, f_{y y} = 2, f_{x y} = 1$ (constants — this $f$ is a pure quadratic, so curvature is the same everywhere). $D = (2) (2) - (1)^{2} = 3 > 0, f_{xx} = 2 > 0.$ Conclusion: local minimum at $(4, - 2)$ — the bowl (left panel of the figure).

Recall Solution 4

Step 1. $f_{x} = 3 x^{2} - 12, f_{y} = 2 y - 4$ . Step 2. $3 x^{2} - 12 = 0 \Rightarrow x = \pm 2$ ; $2 y - 4 = 0 \Rightarrow y = 2$ . Two candidates: $(2, 2)$ and $(- 2, 2)$ . Step 3. $f_{xx} = 6 x, f_{y y} = 2, f_{x y} = 0$ , so $D = 12 x$ .

At $(2, 2)$ : $D = 24 > 0$ , $f_{xx} = 12 > 0$ → local minimum.
At $(- 2, 2)$ : $D = - 24 < 0$ → saddle. Why they differ: the $x$ -curvature $f_{xx} = 6 x$ flips sign with $x$ . Where it curves up (positive $x$ ) it agrees with the $y$ -bowl → min; where it curves down (negative $x$ ) they fight → saddle.

L3 — Analysis

Recall Solution 5

Step 1. $f_{x} = 2 x - 3 y, f_{y} = 2 y - 3 x$ . Step 2. $2 x = 3 y$ and $2 y = 3 x$ . Substitute $x = \frac{3}{2} y$ into $2 y = 3 x$ : $2 y = \frac{9}{2} y \Rightarrow y (2 - \frac{9}{2}) = 0 \Rightarrow y = 0 \Rightarrow x = 0$ . Only $(0, 0)$ . Step 3. $f_{xx} = 2, f_{y y} = 2, f_{x y} = - 3$ . $D = (2) (2) - (- 3)^{2} = 4 - 9 = - 5 < 0.$ Conclusion: saddle at $(0, 0)$ — even though $f_{xx} = 2 > 0$ and $f_{y y} = 2 > 0$ . The big cross term $f_{x y} = - 3$ overpowers both, so along the line $x = y$ the surface actually dips (the right panel of the figure, despite both diagonal curvatures reading positive).

Recall Solution 6

Step 1. $f_{x} = 3 x^{2} - 3 y, f_{y} = 3 y^{2} - 3 x$ . Step 2. From $f_{x} = 0$ : $y = x^{2}$ . Sub into $f_{y} = 0$ : $3 (x^{2})^{2} - 3 x = 0 \Rightarrow 3 x^{4} - 3 x = 0 \Rightarrow 3 x (x^{3} - 1) = 0$ . So $x = 0$ (→ $y = 0$ ) or $x = 1$ (→ $y = 1$ ). Critical points $(0, 0)$ and $(1, 1)$ . Step 3. $f_{xx} = 6 x, f_{y y} = 6 y, f_{x y} = - 3$ , so $D = 36 x y - 9$ .

At $(0, 0)$ : $D = 0 - 9 = - 9 < 0$ → saddle.
At $(1, 1)$ : $D = 36 - 9 = 27 > 0$ , $f_{xx} = 6 > 0$ → local minimum.

L4 — Synthesis

Recall Solution 7

For all three, $f_{x}, f_{y}, f_{xx}, f_{y y}, f_{x y}$ all vanish at $(0, 0)$ (every term is degree 4, so first and second derivatives are zero there). Hence $D = 0$ for all three — the test is inconclusive. Why $D = 0$ means "inconclusive" (the key idea, not just a rule): $D$ is built from the second-order (quadratic) piece of the Taylor expansion — the bowl/dome/saddle shape that approximates the surface near the point. When every second derivative is zero, that quadratic approximation is completely flat: to second order these three surfaces look identical (a flat table). The shape only reveals itself in the quartic ( $x^{4}, y^{4}$ ) terms, which the Hessian never sees. So we must inspect those higher-order terms by hand. We do so directly:

(a) $x^{4} + y^{4} \geq 0$ , zero only at origin → local (global) minimum (the bowl, but built from $4$ th powers).
(b) $- x^{4} - y^{4} \leq 0$ , zero only at origin → local (global) maximum — this is the concrete dome case, matching the middle panel of the figure above.
(c) $x^{4} - y^{4}$ : along $x$ -axis it's $+ x^{4} > 0$ ; along $y$ -axis it's $- y^{4} < 0$ → up one way, down another → saddle. Lesson: when $D = 0$ the quadratic (2nd-order) picture is flat; the quartic terms decide, and you must look at them by hand.

Recall Solution 8

Step 1. Use the product rule. With $u = x^{2} + y^{2}$ , $v = e^{- x}$ : $f_{x} = (2 x) e^{- x} + (x^{2} + y^{2}) (- e^{- x}) = e^{- x} (2 x - x^{2} - y^{2}),$ $f_{y} = (2 y) e^{- x} .$ Step 2. $e^{- x}$ is never zero, so $f_{y} = 0 \Rightarrow y = 0$ ; then $f_{x} = 0 \Rightarrow 2 x - x^{2} = 0 \Rightarrow x (2 - x) = 0 \Rightarrow x = 0$ or $x = 2$ . Critical points $(0, 0)$ and $(2, 0)$ . Step 3 — second derivatives. Differentiating again (carefully, since both factors depend on $x$ ): $f_{xx} = e^{- x} (x^{2} - 4 x + 2 + y^{2}), f_{y y} = 2 e^{- x}, f_{x y} = - 2 y e^{- x} .$

At $(0, 0)$ : $f_{xx} = 1 \cdot (0 - 0 + 2 + 0) = 2$ , $f_{y y} = 2$ , $f_{x y} = 0$ . $D = (2) (2) - 0 = 4 > 0$ , $f_{xx} = 2 > 0$ → local minimum (value $0$ ).
At $(2, 0)$ : $f_{xx} = e^{- 2} (4 - 8 + 2 + 0) = - 2 e^{- 2}$ , $f_{y y} = 2 e^{- 2}$ , $f_{x y} = 0$ . $D = (- 2 e^{- 2}) (2 e^{- 2}) = - 4 e^{- 4} < 0$ → saddle.

L5 — Mastery

Recall Solution 9

Method — Lagrange multipliers (constraint present; see Lagrange Multipliers). Minimise $F = x^{2} + y^{2} + z^{2}$ subject to $g = x + 2 y + 3 z - 6 = 0$ . Set $\nabla F = λ \nabla g$ : $2 x = λ (1), 2 y = λ (2), 2 z = λ (3) .$ So $x = \frac{λ}{2}, y = λ, z = \frac{3 λ}{2}$ . Plug into the constraint: $\frac{λ}{2} + 2 λ + 3 \cdot \frac{3 λ}{2} = 6 \Rightarrow \frac{λ}{2} + 2 λ + \frac{9 λ}{2} = 6 \Rightarrow 7 λ = 6 \Rightarrow λ = \frac{6}{7} .$ Then $x = \frac{3}{7}, y = \frac{6}{7}, z = \frac{9}{7}$ . Minimum distance: $x^{2} + y^{2} + z^{2} = \frac{9 + 36 + 81}{49} = \frac{126}{49} = \frac{126}{7} = \frac{3 14}{7}$ . Why it is the minimum: squared distance grows without bound as you move far away on the plane, so the single stationary point Lagrange returns must be the global minimum. (Sanity check with the point–plane formula $\frac{∣ - 6 ∣}{1 + 4 + 9} = \frac{6}{14} = \frac{3 14}{7}$ ✓.)

Recall Solution 10

Step 1 — solve the gradient system honestly. $f_{x} = 2 x + cy, f_{y} = c x + 2 y$ . Setting both to zero gives the linear system ${2 x + cy = 0 c x + 2 y = 0 ⟺ (2 c c 2) (x y) = (00) .$ A homogeneous linear system has only the trivial solution $(0, 0)$ when its matrix is invertible, i.e. when $det (2 c c 2) = 4 - c^{2} \neq = 0 ⟺ c \neq = \pm 2.$ When $det = 0$ (i.e. $c = \pm 2$ ) the two equations become multiples of one another, so they collapse to a single line of solutions — infinitely many critical points. (Notice this determinant is exactly $D$ ; that is not a coincidence for a pure quadratic, whose Hessian is this matrix.) Step 2 — discriminant. $f_{xx} = 2, f_{y y} = 2, f_{x y} = c$ , so $D = (2) (2) - c^{2} = 4 - c^{2} .$ Step 3 — cover every case in $c$ :

$D > 0 ⟺ - 2 < c < 2$ : unique critical point $(0, 0)$ , and $f_{xx} = 2 > 0$ → local minimum.
$D < 0 ⟺ c < - 2$ or $c > 2$ : unique critical point $(0, 0)$ → saddle.
$c = + 2$ (so $D = 0$ ): the gradient vanishes along the whole line $x + y = 0$ . Directly, $f = x^{2} + 2 x y + y^{2} = (x + y)^{2} \geq 0$ , equal to $0$ exactly on that line → a non-strict (degenerate) minimum valley; every point of $x + y = 0$ is a (non-isolated) critical point.
$c = - 2$ (so $D = 0$ ): the gradient vanishes along $x - y = 0$ . Directly, $f = (x - y)^{2} \geq 0$ → a valley along $x = y$ . Answer: minimum for $∣ c ∣ < 2$ (unique point); saddle for $∣ c ∣ > 2$ (unique point); at $c = \pm 2$ the system degenerates into an entire line of critical points forming a flat, non-strict valley (a positive-semidefinite form).

Common mistake L5 trap: forgetting the constrained point may not be a minimum, and skipping the degenerate boundary.

Why it feels right: Lagrange gives one clean point, so it "must" be the answer; and for the $c$ -family, $∣ c ∣ < 2$ vs $∣ c ∣ > 2$ seems to cover everything. Why it's wrong: Lagrange finds stationary points — you still argue it's a minimum (Ex 9: squared distance $\to \infty$ far away, so the lone stationary point is the min). And the strict inequalities skip $c = \pm 2$ , where the Hessian's determinant is zero, the linear system degenerates, and you get a whole line of critical points — a real, testable case. Fix: always state why a Lagrange stationary point is the extremum you want (boundedness/geometry), and when $D = det H = 0$ solve the gradient system explicitly — it may have infinitely many solutions.

Active Recall

Recall Order of operations in the second-derivative test?

Compute $D$ first; only if $D > 0$ do you read $f_{xx}$ (positive → min, negative → max). $D < 0$ → saddle immediately; $D = 0$ → inspect directly.

Recall Why does a large

$∣ f_{x y} ∣$ create saddles even when $f_{xx}, f_{y y} > 0$ ? The mixed term measures cross-curvature; in $D = f_{xx} f_{y y} - f_{x y}^{2}$ a big $f_{x y}^{2}$ can push $D$ negative, meaning some diagonal direction curves downward — an indefinite quadratic form.

Recall What do you do when

$D = 0$ ? The Hessian is silent. Bound $f$ directly, test along axes and diagonals, or expand to higher order (Taylor). $D = 0$ can hide a min, max, saddle, or (for a pure quadratic) a whole line of critical points.

Connections

Critical points — finding, classifying — the parent recipe these exercises drill.
Gradient and Directional Derivatives — Step 1 of every problem: $\nabla f = 0$ .
Hessian Matrix — where $D = det H$ lives; a zero determinant signals degeneracy (Ex 10).
Quadratic Forms and Definiteness — why positive diagonal need not mean positive definite (Ex 5, 10).
Taylor Series in Several Variables — the $D = 0$ escape hatch (Ex 7).
Lagrange Multipliers — constrained extrema (Ex 9).
Single-variable Second Derivative Test — the 1-D reflex that misleads at L1.