$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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Local Extremum and Critical Points

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, and $ \vec{a} $ $f:A\to\mathbb{R}$. If $\vec{a}$ is a local minimum or maximum and the gradient of the function exists at $\vec{a}$ (ie. $\nabla f(\vec{a})$ exists), then $\vec{a}$ is a critical point of $f$.

Concepts

In multivariable calculus, a critical point is either a local minimum, a local maximum, or a saddle point. This theorem says that if a point inside the interior of the preimage is a local minimum or maximum, then it must be a critical point.
Only on the boundry points might a point be a local maximum or minimum and not be a critical point.

Hypothesis

Coming soon

Results

Coming soon

Proof

The idea for the proof is that we use directional derivatives, and a scalar $h$ to approach $0$.
We show that if $\vec{a}$ is a local maximum, then $\vec{a}$ is a critical point of $f$.
If $\vec{a}$ is a local maximum, then that means there exists some $\delta > 0$ such that for all $\vec{x}\in B_{\delta}(\vec{a})$, we have that $f(\vec{x})\leq f(\vec{a})$. Let $\vec{u}$ be a unit vector. That means that $f(\vec{a}+h\vec{u})-f(\vec{a})\leq 0$ for all $h<\delta$. Then we get $$\begin{align*} \lim_{h\to 0^+}\frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}\leq 0 \end{align*}$$ and $$\begin{align*} \lim_{h\to 0^-}\frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}\geq 0, \end{align*}$$ as coming from the left $h$ is negative. Since the gradient exists, that means the limit exists for all unit vectors $u$, and hence we require that the limits be equal for all $u$, which means for all unit vectors $u$ the derivative is $$\begin{align*} \lim_{h\to 0 } \frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}=0 \end{align*}$$.
A similar argument follows for showing that the local minimum is a critical point.

Local Extremum and Critical Points

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, and $ \vec{a} $ $f:A\to\mathbb{R}$. If $\vec{a}$ is a local minimum or maximum and the gradient of the function exists at $\vec{a}$ (ie. $\nabla f(\vec{a})$ exists), then $\vec{a}$ is a critical point of $f$.

Concepts

In multivariable calculus, a critical point is either a local minimum, a local maximum, or a saddle point. This theorem says that if a point inside the interior of the preimage is a local minimum or maximum, then it must be a critical point.
Only on the boundry points might a point be a local maximum or minimum and not be a critical point.

Hypothesis

Coming soon

Results

Coming soon

Proof

The idea for the proof is that we use directional derivatives, and a scalar $h$ to approach $0$.
We show that if $\vec{a}$ is a local maximum, then $\vec{a}$ is a critical point of $f$.
If $\vec{a}$ is a local maximum, then that means there exists some $\delta > 0$ such that for all $\vec{x}\in B_{\delta}(\vec{a})$, we have that $f(\vec{x})\leq f(\vec{a})$. Let $\vec{u}$ be a unit vector. That means that $f(\vec{a}+h\vec{u})-f(\vec{a})\leq 0$ for all $h<\delta$. Then we get $$\begin{align*} \lim_{h\to 0^+}\frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}\leq 0 \end{align*}$$ and $$\begin{align*} \lim_{h\to 0^-}\frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}\geq 0, \end{align*}$$ as coming from the left $h$ is negative. Since the gradient exists, that means the limit exists for all unit vectors $u$, and hence we require that the limits be equal for all $u$, which means for all unit vectors $u$ the derivative is $$\begin{align*} \lim_{h\to 0 } \frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}=0 \end{align*}$$.
A similar argument follows for showing that the local minimum is a critical point.
FullPage
result
concepts
hypothesis
implications
proof
FullPage
result
concepts
hypothesis
implications
proof

Local Extremum and Critical Points

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, and $ \vec{a} $ $f:A\to\mathbb{R}$. If $\vec{a}$ is a local minimum or maximum and the gradient of the function exists at $\vec{a}$ (ie. $\nabla f(\vec{a})$ exists), then $\vec{a}$ is a critical point of $f$.

Concepts

In multivariable calculus, a critical point is either a local minimum, a local maximum, or a saddle point. This theorem says that if a point inside the interior of the preimage is a local minimum or maximum, then it must be a critical point.
Only on the boundry points might a point be a local maximum or minimum and not be a critical point.

Hypothesis

Coming soon

Results

Coming soon

Proof

The idea for the proof is that we use directional derivatives, and a scalar $h$ to approach $0$.
We show that if $\vec{a}$ is a local maximum, then $\vec{a}$ is a critical point of $f$.
If $\vec{a}$ is a local maximum, then that means there exists some $\delta > 0$ such that for all $\vec{x}\in B_{\delta}(\vec{a})$, we have that $f(\vec{x})\leq f(\vec{a})$. Let $\vec{u}$ be a unit vector. That means that $f(\vec{a}+h\vec{u})-f(\vec{a})\leq 0$ for all $h<\delta$. Then we get $$\begin{align*} \lim_{h\to 0^+}\frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}\leq 0 \end{align*}$$ and $$\begin{align*} \lim_{h\to 0^-}\frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}\geq 0, \end{align*}$$ as coming from the left $h$ is negative. Since the gradient exists, that means the limit exists for all unit vectors $u$, and hence we require that the limits be equal for all $u$, which means for all unit vectors $u$ the derivative is $$\begin{align*} \lim_{h\to 0 } \frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}=0 \end{align*}$$.
A similar argument follows for showing that the local minimum is a critical point.

Local Extremum and Critical Points

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, and $ \vec{a} $ $f:A\to\mathbb{R}$. If $\vec{a}$ is a local minimum or maximum and the gradient of the function exists at $\vec{a}$ (ie. $\nabla f(\vec{a})$ exists), then $\vec{a}$ is a critical point of $f$.

Concepts

In multivariable calculus, a critical point is either a local minimum, a local maximum, or a saddle point. This theorem says that if a point inside the interior of the preimage is a local minimum or maximum, then it must be a critical point.
Only on the boundry points might a point be a local maximum or minimum and not be a critical point.

Hypothesis

Coming soon

Results

Coming soon

Proof

The idea for the proof is that we use directional derivatives, and a scalar $h$ to approach $0$.
We show that if $\vec{a}$ is a local maximum, then $\vec{a}$ is a critical point of $f$.
If $\vec{a}$ is a local maximum, then that means there exists some $\delta > 0$ such that for all $\vec{x}\in B_{\delta}(\vec{a})$, we have that $f(\vec{x})\leq f(\vec{a})$. Let $\vec{u}$ be a unit vector. That means that $f(\vec{a}+h\vec{u})-f(\vec{a})\leq 0$ for all $h<\delta$. Then we get $$\begin{align*} \lim_{h\to 0^+}\frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}\leq 0 \end{align*}$$ and $$\begin{align*} \lim_{h\to 0^-}\frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}\geq 0, \end{align*}$$ as coming from the left $h$ is negative. Since the gradient exists, that means the limit exists for all unit vectors $u$, and hence we require that the limits be equal for all $u$, which means for all unit vectors $u$ the derivative is $$\begin{align*} \lim_{h\to 0 } \frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}=0 \end{align*}$$.
A similar argument follows for showing that the local minimum is a critical point.
FullPage
result
concepts
hypothesis
implications
proof