# Calculus/Extreme Value Theorem

If f is a continuous function and closed on the interval [$a,b$], then f has both a minimum and a maximum.
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.
$f(x)=x^{2}$ and is closed on the interval [-1,2]. Find all extrema.
    ${\frac {dy}{dx}}=2x$
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.)
    ${\frac {d^{2}y}{dx^{2}}}=2$
$f''(c)>0$, 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.
Increasing/Decreasing:
* If $f'(x)<0\$, then $f(x)\$ is decreasing.
* If $f'(x)>0\$, then $f(x)\$ is increasing.
Local Extrema:
* If ${\frac {dy}{dx}}|_{x=c}=f'(c)=0$ and $f'(x)\$ changes signs at $x=c\$, then there exists a local extremum at $x=c\$.
* If $f'(x)<0\$ for $x<c\$ and $f'(x)>0\$ for $x>c\$, then $f(c)\$ is a local minimum.
* If $f'(x)>0\$ for $x<c\$ and $f'(x)<0\$ $x>c\$, then $f(c)\$ is a local maximum.
Example 1:
* Find ${\frac {dy}{dx}}$
    $f(x)=3x^{2}+4x-5\$
    $f'(x)=6x+4\$
* Set ${\frac {dy}{dx}}=0$ to find local extrema.
    $6x+4=0\$
    $6x=-4\$
    $x=-{\frac {2}{3}}$
* Determine whether there is a local minimum or maximum at $x=-{\frac {2}{3}}$.
    Choose an x value smaller than $-{\frac {2}{3}}$:
    $f'(-1)=6(-1)+4=-2<0\$
    Choose an x value larger than $-{\frac {2}{3}}$:
    $f'(1)=6(1)+4=10>0\$
Therefore, there is a local minimum at $x=-{\frac {2}{3}}$ because $f'(-{\frac {2}{3}})=0$ and $f'(x)\$ changes signs at $x=-{\frac {2}{3}}$.

## Second Derivative Test

Recall that the second derivative of a function describes the concavity of the graph of that function.
* If ${\frac {d^{2}y}{dx^{2}}}|_{x=c}=f''(c)=0$ and $f''(c)\$ changes signs at $x=c\$, then there is a point of inflection (change in concavity) at $x=c\$.
* If $f''(x)<0\$, then the graph of $f(x)\$ is concave down.
* If $f''(x)>0\$, then the graph of $f(x)\$ is concave up.
Example 2:
* Find ${\frac {d^{2}y}{dx^{2}}}$.
    $f(x)=x^{3}+2x+7\$
    $f'(x)=3x^{2}+2\$
    $f''(x)=6x\$
* Set ${\frac {d^{2}y}{dx^{2}}}=0$.
    $6x=0\$
    $x=0\$
* Determine whether $f''(x)\$ changes signs at $x=0\$.
    Choose an x value that is smaller than 0:
    $f''(-1)=6(-1)=-6<0\$
    Choose an x value that is larger than 0:
    $f''(1)=6(1)=6>0\$
Therefore, there exists a point of inflection at $x=0\$ because $f''(0)=0\$ and $f''(x)\$ changes signs at $x=0\$.
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