Editing Maximum and minimum (section)

==Examples==
[[Image:xth root of x.svg|thumb|right|The global maximum of {{math|{{sqrt|''x''|''x''}}}} occurs at {{math|''x'' {{=}} ''[[e (mathematical constant)|e]]''}}.]]
{|class="wikitable"
!Function!!Maxima and minima
|-
| ''x''<sup>2</sup>||Unique global minimum at ''x'' = 0.
|-
| ''x''<sup>3</sup> ||No global minima or maxima. Although the first derivative (3''x''<sup>2</sup>) is 0 at ''x'' = 0, this is an [[inflection point]]. (2nd derivative is 0 at that point.)
|-
| <big><math>\sqrt[x]{x}</math></big> ||Unique global maximum at ''x'' = ''[[e (mathematical constant)|e]]''. (See figure at right)
|-
| ''x''<sup>−''x''</sup> ||Unique global maximum over the positive real numbers at ''x'' = 1/''e''.
|-
| ''x''<sup>3</sup>/3 − ''x'' ||First derivative ''x''<sup>2</sup> − 1 and [[second derivative]] 2''x''.  Setting the first derivative to 0 and solving for ''x'' gives [[stationary point]]s at −1 and +1. From the sign of the second derivative, we can see that −1 is a local maximum and +1 is a local minimum. This function has no global maximum or minimum.
|-
| <nowiki> |</nowiki>''x''<nowiki>|</nowiki> ||Global minimum at ''x'' = 0 that cannot be found by taking derivatives, because the derivative does not exist at ''x'' = 0.
|-
| cos(''x'') ||Infinitely many global maxima at 0, ±2{{pi}}, ±4{{pi}}, ..., and infinitely many global minima at ±{{pi}}, ±3{{pi}}, ±5{{pi}}, ....
|-
| 2 cos(''x'') − ''x'' ||Infinitely many local maxima and minima, but no global maximum or minimum.
|-
| cos(3{{pi}}''x'')/''x'' with {{nowrap|0.1 ≤ ''x'' ≤ 1.1}} ||Global maximum at ''x''&nbsp;= 0.1 (a boundary), a global minimum near ''x''&nbsp;= 0.3, a local maximum near ''x''&nbsp;= 0.6, and a local minimum near ''x''&nbsp;= 1.0. (See figure at top of page.)
|-
|''x''<sup>3</sup> + 3''x''<sup>2</sup> − 2''x'' + 1 defined over the closed interval (segment) [−4,2] || Local maximum at ''x''&nbsp;= −1−{{radic|15}}/3, local minimum at ''x''&nbsp;= −1+{{radic|15}}/3, global maximum at ''x''&nbsp;= 2 and global minimum at ''x''&nbsp;= −4.
|}

For a practical example,<ref name="minimization_maximization_refresher">{{cite web|author=Garrett, Paul|title=Minimization and maximization refresher|url=https://mathinsight.org/minimization_maximization_refresher}}</ref> assume a situation where someone has <math>200</math> feet of fencing and is trying to maximize the square footage of a rectangular enclosure, where <math>x</math> is the length, <math>y</math> is the width, and <math>xy</math> is the area:

:<math> 2x+2y = 200 </math>
:<math> 2y = 200-2x </math>
:<math> \frac{2y}{2} = \frac{200-2x}{2} </math>
:<math> y = 100 - x</math>
:<math> xy=x(100-x) </math>

The derivative with respect to <math>x</math> is:
:<math>\begin{align}
\frac{d}{dx}xy&=\frac{d}{dx}x(100-x) \\
&=\frac{d}{dx} \left(100x-x^2 \right) \\
&=100-2x
\end{align}</math>
Setting this equal to <math>0</math>
:<math>0=100-2x</math>
:<math>2x=100</math>
:<math>x=50</math>
reveals that <math>x=50</math> is our only [[Critical_point_(mathematics)|critical point]].
Now retrieve the [[Interval_(mathematics)|endpoints]] by determining the interval to which <math>x</math> is restricted. Since width is positive, then <math>x>0</math>, and since {{nowrap|<math>x=100-y</math>,}} that implies that {{nowrap|<math>x < 100</math>.}}
Plug in critical point {{nowrap|<math>50</math>,}} as well as endpoints <math>0</math> and {{nowrap|<math>100</math>,}} into {{nowrap|<math>xy=x(100-x)</math>,}} and the results are <math>2500, 0,</math> and <math>0</math> respectively.

Therefore, the greatest area attainable with a rectangle of <math>200</math> feet of fencing is {{nowrap|<math>50 \times 50 = 2500</math>.}}<ref name="minimization_maximization_refresher"></ref>