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In math is the local max and local min just any peak ... point where slope of the function changes from positive to negative or vice-versa... Or are the LOCAL max and min just the highest point of the function and lowest point of a function in a given range? If that were the case wouldn't there be infinite local max and min since you can have infinite ranges? Or inorder for something to be considered a local maxima or minima does slope have to change? (Rather than slope, is it that differentiability has to become non-existant at that point?)

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When one is talking about a local or global maximum of some real-valued function $f$ the domain of reference has to be specified precisely. Many functions are defined by an expression in terms of some input variable $x$ which may take integer, positive, real, complex, or vector values. Only after the exact domain $\Omega$ of admissible $x$ for the problem at hand has been specified (say, the closed unit ball with center $0\in{\mathbb R}^n$) can we start thinking about maxima or minima of $f$.

The function $f:\ \Omega\to{\mathbb R}$ takes a global maximum at the point $\xi\in\Omega$, if $f(\xi)\geq f(x)$ for all $x\in\Omega$. This definition should cause no problems.

The function $f:\ \Omega\to{\mathbb R}$ with $\Omega\subset{\mathbb R}^n$ takes a local maximum at the point $\xi\in\Omega$, if there is a neighborhood $U(\xi)\subset{\mathbb R}^n$ such that $f(\xi)\geq f(x)$ for all $x\in\Omega\cap U(\xi)$. This means that far away from $\xi$ there may still be points $x\in\Omega$ with $f(x)>f(\xi)$.

The difficulty in this definition consists in the word "local". When the domain $\Omega$ of $f$ is a subset of ${\mathbb R}^n$, for example, then "local" refers to the so-called relative topology of $\Omega$. In this respect consider the following example: $$f(x):=(x-1)^2\ ,\qquad \Omega:=[0,3]\ .$$ Then there is a unique point $\xi\in\Omega$ where $f$ takes a global maximum, namely the point $\xi=3$. At the point $\xi'=0$ the function $f$ takes a local maximum with respect to $\Omega$, since $$f(0)\geq f(x)\qquad\forall x\in\ ]{-1},1[\ \cap\Omega\ .$$

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If you think of a function $f$ of one variable, then a local maximum of $f$ is a point $x^*$ in the domain of $f$ such that there is an open interval $I$ which contains $x^*$ and that $f(x^*) \ge f(x)$ for all $x \in I$ which is in the domain of $f$.

For domains lying in more general space, we replace the word “interval” in the above with “neighbourhood”.

So a local maximum is indeed a peak, except that it does not have to be “peaked”. If $f(x)=2$ for all $x$ then every $x$ is a local maximum of the function. For the function $f(x)=2x$ defined on $[0,1]$, $x^*=1$ is a local maximum.

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    $\begingroup$ (I think that needs to be an open interval though) $\endgroup$ – Eric Stucky Oct 16 '13 at 15:45
  • $\begingroup$ @EricStucky. Thanks, corrected. $\endgroup$ – Jyotirmoy Bhattacharya Oct 16 '13 at 15:46
  • $\begingroup$ In the case of $f(x)=2x$ on $[0,1]$, some books (e.g. Stewart.'s Calculus) does not consider it as a local maximum. Since there is no open interval $I$ of $\mathbb R$ containing $1$ and contained in $I$. So, according to these books, some absolute maximums are not local :/ $\endgroup$ – Taladris Oct 16 '13 at 15:57

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