View source for Extreme values on closed and bounded domains

<blockquote style="background: white; border: 1px solid black; padding: 1em;"> 
'''Extreme Value Theorem:'''
If ''f'' is a continuous function and closed on the interval [<math>a,b</math>], then ''f'' has both a minimum and a maximum.</blockquote>

This introduces us to the aspect of global extrema and local extrema. (Also known as absolute extrema or relative extrema respectively.)

How is this so? Let us use an example.

<math> f(x) = x^2 </math> and is closed on the interval [-1,2]. Find all extrema.

:<math> \frac{dy}{dx} = 2x</math>

A critical point (a point where the derivative is zero) exists at (0,0). Just for practice, let us use the second derivative test to evaluate whether or not it is a minimum or maximum. (You should know it is a minimum from looking at the graph.)

:<math> \frac{d^2y}{dx^2} = 2</math>

<math> f''(c) > 0</math>, thus it must be a minimum.

As mentioned before, one can find global extrema on a closed interval. How? Evaluate the ''y'' coordinate at the endpoints of the interval and compare it to the ''y'' coordinates of the critical point.  When you are finding extrema on a closed interval it is called a local extremum and when it's for the whole graph it's called a global extremum.

1: Critical Point: (0,0) This is the lowest value in the interval. Therefore, it is a local minimum which also happens to be the global minimum.

2: Left Endpoint (-1, 1) This point is not a critical point nor is it the highest/lowest value, therefore it qualifies as nothing.

3: Right Endpoint (2, 4) This is the highest value in the interval, and thus it is a local maximum.

This example was to show you the '''extreme value theorem'''. The quintessential point is this: on a closed interval, the function will have both minima and maxima. However, if that interval was an open interval of all real numbers, (0,0) would have been a local minimum. On a closed interval, always remember to evaluate endpoints to obtain global extrema.

==First Derivative Test==

Recall that the first derivative of a function describes the slope of the graph of the function at every point along the graph for which the function is defined and differentiable.

<blockquote style="background: white; border: 1px solid black; padding: 1em;"> 
'''Increasing/Decreasing:'''

* If <math> f'(x) < 0 \ </math>, then <math> f(x) \ </math> is decreasing.

* If <math> f'(x) > 0 \ </math>, then <math> f(x) \ </math> is increasing.

'''Local Extrema:'''

* If <math> \frac{dy}{dx}|_{x=c} = f'(c) = 0 </math> and <math> f'(x) \ </math> changes signs at <math> x = c \ </math>, then there exists a local extremum at <math> x = c \ </math>.

* If <math> f'(x) < 0 \ </math> for <math> x < c \ </math> and <math> f'(x) > 0 \ </math> for <math> x > c \ </math>, then <math> f(c) \ </math> is a local minimum.

* If <math> f'(x) > 0 \ </math> for <math> x < c \ </math> and <math> f'(x) < 0 \ </math> <math> x > c \ </math>, then <math> f(c) \ </math> is a local maximum.
</blockquote>

'''Example 1:'''

 Let <math> f(x) = 3x^2 + 4x - 5 \ </math>. Find all local extrema.

* Find <math> \frac{dy}{dx} </math>

: <math> f(x) = 3x^2 + 4x - 5 \ </math>
: <math> f'(x) = 6x + 4 \ </math>

* Set <math> \frac{dy}{dx} = 0 </math> to find local extrema.

: <math> 6x + 4 = 0 \ </math>
: <math> 6x = -4 \ </math>
: <math> x = - \frac{2}{3} </math>

* Determine whether there is a local minimum or maximum at <math> x = - \frac{2}{3} </math>.

: Choose an ''x'' value smaller than <math> - \frac{2}{3} </math>:

: <math> f'(-1) = 6(-1) + 4 = -2 < 0 \ </math>

: Choose an ''x'' value larger than <math> - \frac{2}{3} </math>:

: <math> f'(1) = 6(1) + 4 = 10 > 0 \ </math>

Therefore, there is a local minimum at <math> x = - \frac{2}{3} </math> because <math> f'(- \frac{2}{3}) = 0 </math> and <math> f'(x) \ </math> changes signs at <math> x = - \frac{2}{3} </math>.

 Answer: local minimum: <math> x = - \frac{2}{3} </math>.

==Second Derivative Test== 

Recall that the second derivative of a function describes the concavity of the graph of that function.

<blockquote style="background: white; border: 1px solid black; padding: 1em;"> 
* If <math> \frac{d^2y}{dx^2}|_{x=c} = f''(c) = 0 </math> and <math> f''(c) \ </math> changes signs at <math> x = c \ </math>, then there is a point of inflection (change in concavity) at <math> x = c \ </math>.

* If <math> f''(x) < 0 \ </math>, then the graph of <math> f(x) \ </math> is concave down.

* If <math> f''(x) > 0 \ </math>, then the graph of <math> f(x) \ </math> is concave up.
</blockquote>

'''Example 2:'''

 Let <math> f(x) = x^3 + 2x + 7 \ </math>.  Find any points of inflection on the graph of <math> f(x) \ </math>.

* Find <math> \frac{d^2y}{dx^2} </math>.

: <math> f(x) = x^3 + 2x + 7 \ </math>
: <math> f'(x) = 3x^2 + 2 \ </math>
: <math> f''(x) = 6x \ </math>

* Set <math> \frac{d^2y}{dx^2} = 0 </math>.

: <math> 6x = 0 \ </math>
: <math> x = 0 \ </math>

* Determine whether <math> f''(x) \ </math> changes signs at <math> x = 0 \ </math>.

: Choose an ''x'' value that is smaller than 0:

: <math> f''(-1) = 6(-1) = -6 < 0 \ </math>

: Choose an ''x'' value that is larger than 0:

: <math> f''(1) = 6(1) = 6 > 0 \ </math>

Therefore, there exists a point of inflection at <math> x = 0 \ </math> because <math> f''(0) = 0 \ </math> and <math> f''(x) \ </math> changes signs at <math> x = 0 \ </math>.

 Answer: point of inflection: <math> x = 0 \ </math>.

==Resources==
* [https://en.wikibooks.org/wiki/Calculus/Extreme_Value_Theorem Extreme Value Theorem], Wikibooks: Calculus