# Calculus/Extreme Value Theorem

 ← Euler's Method Calculus Rolle's Theorem → Extreme Value Theorem
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  and $f'(x)>0\$  for $x>c\$ , then $f(c)\$  is a local minimum.
• If $f'(x)>0\$  for $x  and $f'(x)<0\$  $x>c\$ , then $f(c)\$  is a local maximum.

Example 1:

Let $f(x)=3x^{2}+4x-5\$ . Find all local extrema.

• 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 $x=-{\frac {2}{3}}$ .

Answer: local minimum: $x=-{\frac {2}{3}}$ .


## 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:

Let $f(x)=x^{3}+2x+7\$ .  Find any points of inflection on the graph of $f(x)\$ .

• 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\$ .

Answer: point of inflection: $x=0\$ .