Exercises — Second derivative test — Hessian determinant

WHAT we do: plug into $D=f_{xx}{f}_{yy}-(f_{xy}{)}^2$. $D=(4)(9)-0^2=36>0.$ WHY check $f_{xx}$: $D>0$ says "extremum", but not which. The sign of $f_{xx}=4>0$ says the surface smiles upward. Verdict: local minimum.

$D=(-2)(-5)-(1{)}^2=10-1=9>0.$ Both diagonal entries are negative, and indeed $D>0$ forces them to share a sign. $f_{xx}=-2<0$ → the surface frowns. Verdict: local maximum.

$D=(3)(1)-(4{)}^2=3-16=-13<0.$ WHY stop: $D<0$ settles it alone — no need to look at $f_{xx}$. Geometrically the cross-term $f_{xy}=4$ is so large it tilts the bowl into a Pringle. Verdict: saddle point.

Step 1 — find where the ground is flat. WHY: the test only applies at $\nabla f=0$. $f_x=2x-2=0\Rightarrow x=1,f_y=8y+8=0\Rightarrow y=-1.$ Critical point: $(1,-1)$. Step 2 — second derivatives (all constants here): $f_{xx}=2,f_{yy}=8,f_{xy}=0.$ Step 3 — discriminant: $D=(2)(8)-0^2=16>0,f_{xx}=2>0.$ Verdict: local minimum at $(1,-1)$.

Step 1: $f_x=2x-y=0,f_y=-x+2y-3=0.$ From the first, $y=2x$. Substitute into the second: $-x+2(2x)-3=0\Rightarrow 3x=3\Rightarrow x=1$, so $y=2$. Critical point: $(1,2)$. Step 2: $f_{xx}=2,f_{yy}=2,f_{xy}=-1$. Step 3: $D=(2)(2)-(-1{)}^2=4-1=3>0,f_{xx}=2>0.$ Verdict: local minimum at $(1,2)$.

Step 1: $f_x=y=0$ and $f_y=x=0$ → only critical point $(0,0)$. Step 2: $f_{xx}=0,f_{yy}=0,f_{xy}=1$. Step 3: $D=(0)(0)-(1{)}^2=-1<0.$ Verdict: saddle point at $(0,0)$. Look at the figure: along $y=x$ the product $xy$ climbs, along $y=-x$ it falls — the signature of a saddle.

Step 1: $f_x=3x^2-12=0\Rightarrow x^2=4\Rightarrow x=\pm 2$. And $f_y=2y=0\Rightarrow y=0$. Two critical points: $(2,0)$ and $(-2,0)$. WHY two: $x^3$ makes $f_x$ quadratic, so two roots. Step 2: $f_{xx}=6x,f_{yy}=2,f_{xy}=0$. Note $f_{xx}$ depends on the point. Step 3, at $(2,0)$: $f_{xx}=12$, so $D=(12)(2)-0=24>0$, $f_{xx}>0$ → local minimum. Step 3, at $(-2,0)$: $f_{xx}=-12$, so $D=(-12)(2)-0=-24<0$ → saddle point. Lesson: the same formula gives different verdicts at different points because $f_{xx}=6x$ changes sign.

Step 1: $f_x=3x^2-3y=0\Rightarrow y=x^2$; $f_y=3y^2-3x=0\Rightarrow x=y^2$. Substitute $y=x^2$: $x=(x^2{)}^2=x^4\Rightarrow x^4-x=0\Rightarrow x(x^3-1)=0\Rightarrow x=0\text{ or }x=1$. Critical points: $(0,0)$ and $(1,1)$. Step 2: $f_{xx}=6x,f_{yy}=6y,f_{xy}=-3$. At $(0,0)$: $D=(0)(0)-(-3{)}^2=-9<0$ → saddle point. At $(1,1)$: $D=(6)(6)-(-3{)}^2=36-9=27>0$, $f_{xx}=6>0$ → local minimum.

Step 1: $f_x=2x+ky$, $f_y=kx+2y$; both vanish at $(0,0)$, so it is critical for every $k$. Step 2: $f_{xx}=2,f_{yy}=2,f_{xy}=k$. Step 3: $D=(2)(2)-k^2=4-k^2$. A saddle needs $D<0$: $4-k^2<0\Rightarrow k^2>4\Rightarrow |k|>2$. Verdict: saddle for $k>2$ or $k<-2$. (At $k=\pm 2$, $D=0$: inconclusive. For $|k|<2$: minimum.)

The Taylor picture. Near a point $(a,b)$, expanding to second order with $u=x-a,v=y-b$: $f(a+u,b+v)\approx f(a,b)+\underset{\text{linear}}{\underbrace{f_{x}u+f_{y}v}}+\frac{1}{2}(f_{xx}{u}^2+2f_{xy}uv+f_{yy}{v}^2).$ The test reads shape from the quadratic part only. That is legitimate only when the linear part is zero — i.e. $f_x=f_y=0$. Otherwise the linear term $f_{x}u+f_{y}v$ dominates for small $u,v$ (it beats the quadratic near the point), so the surface is tilted, and "min/max/saddle" is not even the right question — it's just a slope. At $(1,0)$ for $f=x^2+y^2$: $f_x=2x=2\ne 0$. The gradient is nonzero; the surface simply slopes uphill in $x$. Blindly computing $D=f_{xx}{f}_{yy}-f_{xy}^2=4>0$ and declaring "minimum" is meaningless — $(1,0)$ is on the side of the bowl, not the bottom. Always solve $\nabla f=0$ first.

(a) Direct: $D=f_{xx}{f}_{yy}-f_{xy}^2=(3)(3)-(1{)}^2=9-1=8$. (b) Eigenvalues solve $\det (H-\lambda I)=0$: $\det \left(\begin{array}{cc}3-\lambda & 1\\ 1& 3-\lambda \end{array}\right)=(3-\lambda {)}^2-1=0\Rightarrow 3-\lambda =\pm 1\Rightarrow \lambda =2\text{ or }4.$ Product: ${\lambda }_1{\lambda }_2=2\times 4=8=D$. ✓ Geometry: both eigenvalues are positive → the surface curves up along both principal axes → positive definite → local minimum. The eigenvalues $2$ and $4$ are the curvatures along the two principal directions of the bowl.

WHAT we need: critical point at $(2,-1)$, downward curvature, no cross-term. A shifted downward paraboloid works: $f(x,y)=-3(x-2{)}^2-2(y+1{)}^2.$ Check critical point: $f_x=-6(x-2)=0\Rightarrow x=2$; $f_y=-4(y+1)=0\Rightarrow y=-1$. ✓ Check derivatives: $f_{xx}=-6$ ✓, $f_{yy}=-4$, $f_{xy}=0$ ✓. Test: $D=(-6)(-4)-0^2=24>0$ and $f_{xx}=-6<0$ → local maximum. ✓ Construction succeeds.

Step 1: $f_x=4x^3=0$, $f_y=4y^3=0$ → critical at $(0,0)$. Step 2: $f_{xx}=12x^2$, $f_{yy}=12y^2$, $f_{xy}=0$. At $(0,0)$: $f_{xx}=0$, $f_{yy}=0$, $f_{xy}=0$. Step 3: $D=(0)(0)-0^2=0$ → inconclusive. The quadratic Taylor part is entirely flat, so second-order info tells us nothing. By hand: $f(x,y)=x^4+y^4\ge 0$ for all $(x,y)$, with equality only at $(0,0)$ (both fourth powers are $\ge 0$). So $f(0,0)=0$ is the smallest value $f$ takes anywhere → global (hence local) minimum. The curvature is quartic, not quadratic, which is exactly why the second-derivative test can't see it.

Step 1: $f_x=3x^2=0\Rightarrow x=0$; $f_y=0$ for all $y$. So the whole line $x=0$ is critical! Focus on $(0,0)$. Step 2: $f_{xx}=6x$, $f_{yy}=0$, $f_{xy}=0$. At $(0,0)$: all zero. Step 3: $D=0$ → inconclusive. By hand: moving in $+x$, $f=x^3>0$; moving in $-x$, $f=x^3<0$. So $f$ increases on one side and decreases on the other along the same axis — like the point of inflection of $y=x^3$ lifted into a surface. It is not a strict min (values below it exist), not a strict max (values above it exist). Moving in $y$, $f$ stays flat. This is a degenerate inflection ridge, not a genuine saddle (a saddle needs curvature of opposite signs; here the $x$-behaviour is cubic, not quadratic). This is exactly the "$D=0$ = investigate by hand" case.

Step 1 — critical points: $f_x=4x^3-4x=4x(x^2-1)=0\Rightarrow x=0,1,-1$. And $f_y=2y=0\Rightarrow y=0$. Three critical points: $(-1,0),(0,0),(1,0)$. Step 2 — derivatives: $f_{xx}=12x^2-4$, $f_{yy}=2$, $f_{xy}=0$. So $D=(12x^2-4)(2)$. At $(0,0)$: $f_{xx}=-4$, $D=(-4)(2)=-8<0$ → saddle point. At $(1,0)$: $f_{xx}=12-4=8$, $D=(8)(2)=16>0$, $f_{xx}>0$ → local minimum. At $(-1,0)$: $f_{xx}=8$, $D=16>0$, $f_{xx}>0$ → local minimum. The shape: a "double well" — two valleys at $(\pm 1,0)$ separated by a saddle ridge over the origin (the classic $x^4-2x^2$ double-well profile, lifted by the bowl $y^2$).