> [!theorem] > > Suppose that the [[Derivative#Higher Derivatives|second derivative]] $\frac{d^2 f}{dx^2}$ of some [[Real Numbers|real]]-valued [[Science/Math/Functions/Function|function]] $f$ is [[Continuity|continuous]] near some point $c$: > - If $f^\prime(c) = 0$ and $f^{\prime\prime}(c) > 0$, then $f$ has a local minimum at $c$. > - If $f^\prime(c) = 0$ and $f^{\prime\prime}(c) < 0$, then $f$ has a local maximum at $c$. > [!theorem] > > Let $f: \real^2 \to \real$ be a function of $2$ variables. Suppose that the second derivatives of $f$ are continuous over a disk with centre $\vec{p}$ and suppose that the [[Gradient|gradient]] $\nabla f (\vec{p}) = \vec{0}$. Let > $ > D(\vec{p}) = \frac{\partial^2 f}{\partial x^2}(\vec{p})\frac{\partial^2 f}{\partial y^2}(\vec{p}) - \parens{\frac{\partial^2 f}{\partial x \partial y}(\vec{p})}^2 > $ > If the signs of $\frac{\partial^2 f}{\partial x^2}$ and $\frac{\partial^2 f}{\partial y^2}$ are different, then the point is a saddle point and $\vec{p}$ would not be a local extremum. > > If they have the same sign, then the diagonal concavity should be somewhere between them, and thus $\vec{p}$ is a local extremum based on the sign of $\frac{\partial^2 f}{\partial x^2}$. > > However, if $D(\vec{p}) = 0$, then the test is inconclusive.