{
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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $f \\colon U \\to \\mathbb{R}$ be of [[class C^2|0TCBZMEYGN]] on $U$, and let $a \\in U$ with $\\nabla f(a) = 0$.\nLet $H = H_f(a)$ be the [[Hessian matrix|0SCSQWXRA4]] of $f$ at $a$.\n\n1. If $H$ is [[positive definite|01RBSSGMY5]], then $f$ has a strict [[local minimum|04CPG9K1C4]] at $a$.\n2. If $H$ is negative definite, then $f$ has a strict local maximum at $a$.\n3. If $H$ has a positive and a negative eigenvalue, then $a$ is a [[saddle point|00N3T18QNA]] of $f$.\n\nIn the other cases the test gives no conclusion.\n\n## Remarks\n\nFor $n = 2$, let $D = f_{xx}(a) f_{yy}(a) - f_{xy}(a)^2$ be the determinant of $H$. If $D \u003e 0$ and $f_{xx}(a) \u003e 0$, then $a$ is a local minimum. If $D \u003e 0$ and $f_{xx}(a) \u003c 0$, then $a$ is a local maximum. If $D \u003c 0$, then $a$ is a saddle point.\nThe proof uses the [[second-order Taylor expansion|0P30YYZN85]]. If $H$ is positive definite, then $h^{\\mathsf{T}} H h \\ge m \\|h\\|^2$ for some $m \u003e 0$, and this term dominates the remainder.\nThe functions $x^4 + y^4$, $-(x^4 + y^4)$ and $x^4 - y^4$ all have zero Hessian at the origin. The origin is a minimum, a maximum and a saddle point, respectively, so the test cannot decide when $H$ is singular and semidefinite.\nBy Sylvester's criterion, $H$ is positive definite if and only if all its leading principal minors are positive.\n\n## Explanations\n\n### L1 human\n\nAt a critical point the tangent plane is flat, so the second derivatives decide the shape of the graph.\nIf the graph bends upward in every direction, the point is a minimum. If it bends down in every direction, the point is a maximum.\nIf it bends up in some directions and down in others, the point is a saddle.\n",
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