Every method in this course, from the very first numerical algorithm onward, is chasing the same target: a point where a smooth objective has nowhere better to go. In Calculus III that target had a name, a critical point, and a fixed test for telling three fates apart once you found one. Both come back completely unchanged here -- this course adds how to search for one when no algebra shortcut exists, never a new definition of what counts as a candidate in the first place.
Picture standing on a hilly landscape and asking one question: is the ground level in every direction I could step, not just the two I happen to be facing? A critical point is exactly a spot where the answer is yes -- both partial derivatives, the tilt north-south and the tilt east-west, read zero at once. That still leaves three very different kinds of level spots. A bowl's bottom is level and every nearby step goes up. A dome's top is level and every nearby step goes down. A mountain pass is level too, but only along the trail -- step off it either side and the ground drops away, while stepping along the ridge it climbs. The number D is what tells those three apart without having to walk around and check by hand: a positive D means the surface curves the same way in every direction (so whichever way f_{xx} curves settles it, bowl or dome), and a negative D means it curves opposite ways depending on which direction you look, which is exactly the mountain-pass shape.
A point (a,b) in the interior of f's domain is a critical point if \nabla f(a,b)=\mathbf 0, meaning f_x(a,b)=0 and f_y(a,b)=0 at once (or if either partial fails to exist there, though every function in this review is smooth everywhere). Finding candidates is step one; classifying each one is step two, and that is where the second derivative test earns its keep. Compute the three second partials at a critical point and form D=f_{xx}f_{yy}-f_{xy}^2, the determinant of the Hessian matrix \begin{pmatrix}f_{xx}&f_{xy}\\f_{xy}&f_{yy}\end{pmatrix}. If D>0 and f_{xx}>0, the point is a local minimum; if D>0 and f_{xx}<0, a local maximum; if D<0, a saddle point, curving up in one direction and down in another at the same spot; if D=0, the test is silent and a direct comparison of nearby values is the fallback.
A bowl-shaped critical point
Drag the point across the domain floor and watch the height above it. The surface only goes flat directly above the marked critical point -- everywhere else still has an uphill or downhill direction.
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Where the gradient arrows vanish
This is f(x,y)=x^2-y^2. Watch the indigo gradient arrows shrink to nothing exactly at the marked point (0,0), even though the surface keeps rising in some directions and falling in others there -- a saddle, not a flat region.
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Classify each critical point from D and f_xx alone
Each card gives the second-derivative test's two numbers, D and f_{xx}, at some critical point -- no picture of the surface itself. Click a card, then click the bin naming what that point must be.
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Worked example
Classifying the critical points of f(x,y)=x^2y-3x^2-y^2
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1
Find and classify every critical point of f(x,y)=x^2y-3x^2-y^2.
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2factoring f_x makes the case split explicit -- either x=0 or y=3
Set both partials to zero: f_x=2xy-6x=2x(y-3)=0 and f_y=x^2-2y=0.
factoring f_x makes the case split explicit -- either x=0 or y=3
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3
From f_x=0: either x=0 or y=3. If x=0, then f_y=0 gives y=0, so (0,0) is critical. If y=3, then f_y=0 gives x^2=6, so x=\pm\sqrt6, giving (\sqrt6,3) and (-\sqrt6,3).
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4D positive and f_xx negative together mean a local maximum
Second partials: f_{xx}=2y-6, f_{yy}=-2, f_{xy}=2x. At (0,0): D=(-6)(-2)-0^2=12>0 and f_{xx}=-6<0.
D positive and f_xx negative together mean a local maximum
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5
At (\pm\sqrt6,3): D=(0)(-2)-(2(\pm\sqrt6))^2=-24<0 in both cases.
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6
\boxed{(0,0)\text{ is a local maximum; }(\sqrt6,3)\text{ and } (-\sqrt6,3)\text{ are both saddle points.}} Nothing here needed a picture -- the sign pattern alone settled every case.
Where this goes next
Every later chapter's optimality conditions
This test is the finite, hand-computable special case of a much larger idea this course spends its next several chapters building: certifying a candidate point as a genuine optimum rather than guessing from a picture. The sign pattern on D and f_{xx} is really a statement about the Hessian matrix's eigenvalues -- both positive, both negative, or mixed signs -- restated in a form that needs no eigenvalue computation at all in two variables. The moment a third variable or an inequality constraint enters, that shortcut runs out and the eigenvalue-based version from section 3 takes over directly.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
How many critical points does f(x,y) = x^4 + y^4 - 4xy have?
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Solving the system gives more than one solution.
There's one more solution than that -- check whether x=0,y=0 also solves the system.
Recheck the system: f_x=4x^3-4y=0 and f_y=4y^3-4x=0 combine to x^9=x, which has 3 real roots, not 4.
Why
f_x=4x^3-4y=0 \implies y=x^3. Substituting into f_y=4y^3-4x=0: 4x^9-4x=0 \implies x(x^8-1)=0, giving x=0,1,-1 -- three real solutions, hence three critical points: (0,0), (1,1), (-1,-1).
Solve f_x = 0 for y in terms of x, then substitute into f_y = 0.
From Calculus III — Multivariable Calculus
Find the x-coordinate of the critical point of f(x,y) = e^{x^2+y^2-4y} (there is exactly one, and it's where the exponent's own critical point sits).
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Why
The exponent g(x,y)=x^2+y^2-4y has g_x=2x=0 \implies x=0; since e^{(\cdot)} is strictly increasing, f's critical point matches g's.
Correct answer
Also accepted:
Why
e to any exponent is always positive and increasing, so f and its exponent share the same critical point.
From Calculus III — Multivariable Calculus
For f(x,y) = x^3 - 12x + y^2 - 6y, find the y-coordinate of its (unique) critical point.
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Why
f_y=2y-6=0 \implies y=3 (the x-coordinates come from f_x=3x^2-12=0, giving x = 2 or x = -2, so there are actually two critical points sharing this y-value).
Correct answer
Also accepted:
Why
Set f_y = 0 and solve for y -- the x-coordinate doesn't affect this equation at all.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
For f(x,y) = xy - x^2 - y^2 - 2x - 2y + 4, find f_{xx} (a constant).
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Why
f_x = y-2x-2, so f_{xx}=-2.
Correct answer
Also accepted:
Why
Find f_x first, then differentiate that result with respect to x again.
From Calculus III — Multivariable Calculus
f(x,y) = x^3-3x+y^2 has critical points at (1,0) and (-1,0). Classify (-1,0).
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Check f_xx at x=-1: f_xx = 6x = -6, and check the sign of D before deciding.
D is negative at this point, which rules out a local minimum.
D is nonzero here.
Why
f_xx=6x, f_yy=2, f_xy=0. At (-1,0): D=(-6)(2)-0^2=-12<0, a saddle point.
Compute D = f_xx f_yy - f_xy^2 at x=-1, y=0.
From Calculus III — Multivariable Calculus
For f(x,y) = x^3 - 3x + y^2, find D at the critical point (1, 0).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
f_{xx}=6x, f_{yy}=2, f_{xy}=0. At (1,0): D=(6)(2)-0^2=12.
Correct answer
Also accepted:
Why
First confirm (1,0) is a critical point: f_x = 3x^2-3 = 0 at x=1, f_y=2y=0 at y=0.
Practice: Finding and classifying critical points, start to finish (3 problems · doesn’t affect your progress)
Find the y-value of the critical point of f(x,y)=x^2+y^2-4x+2y. (Hint: set both partials to zero first.)
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
f_x=2x-4=0\Rightarrow x=2; f_y=2y+2=0\Rightarrow y=-1.
Correct answer
Also accepted:
Why
For f(x,y)=x^2+y^2-4x+2y, compute D=f_{xx}f_{yy}-f_{xy}^2 at its critical point.
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Why
f_{xx}=2, f_{yy}=2, f_{xy}=0, so D=(2)(2)-0^2=4.
Correct answer
Also accepted:
Why
Given D=4>0 and f_{xx}=2>0 from the previous problem, what kind of critical point is (2,-1)?
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A local maximum needs f_{xx}<0 when D>0 -- here f_{xx}=2 is positive.
A saddle needs D<0 -- here D=4 is positive, ruling out a saddle entirely.
D\\ne0 here, so the test is fully conclusive -- no need to fall back on comparing nearby values.
Why
D>0 and f_{xx}>0 together are exactly the local-minimum case.
Local minimum