# 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 [${\displaystyle a,b}$], 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.

${\displaystyle f(x)=x^{2}}$ and is closed on the interval [-1,2]. Find all extrema.

${\displaystyle {\frac {dy}{dx}}=2x}$

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.)

${\displaystyle {\frac {d^{2}y}{dx^{2}}}=2}$

${\displaystyle f''(c)>0}$, 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 ${\displaystyle f'(x)<0\ }$ , then ${\displaystyle f(x)\ }$  is decreasing.
• If ${\displaystyle f'(x)>0\ }$ , then ${\displaystyle f(x)\ }$  is increasing.

Local Extrema:

• If ${\displaystyle {\frac {dy}{dx}}|_{x=c}=f'(c)=0}$  and ${\displaystyle f'(x)\ }$  changes signs at ${\displaystyle x=c\ }$ , then there exists a local extremum at ${\displaystyle x=c\ }$ .
• If ${\displaystyle f'(x)<0\ }$  for ${\displaystyle x  and ${\displaystyle f'(x)>0\ }$  for ${\displaystyle x>c\ }$ , then ${\displaystyle f(c)\ }$  is a local minimum.
• If ${\displaystyle f'(x)>0\ }$  for ${\displaystyle x  and ${\displaystyle f'(x)<0\ }$  ${\displaystyle x>c\ }$ , then ${\displaystyle f(c)\ }$  is a local maximum.

Example 1:

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

• Find ${\displaystyle {\frac {dy}{dx}}}$
${\displaystyle f(x)=3x^{2}+4x-5\ }$
${\displaystyle f'(x)=6x+4\ }$
• Set ${\displaystyle {\frac {dy}{dx}}=0}$  to find local extrema.
${\displaystyle 6x+4=0\ }$
${\displaystyle 6x=-4\ }$
${\displaystyle x=-{\frac {2}{3}}}$
• Determine whether there is a local minimum or maximum at ${\displaystyle x=-{\frac {2}{3}}}$ .
Choose an x value smaller than ${\displaystyle -{\frac {2}{3}}}$ :
${\displaystyle f'(-1)=6(-1)+4=-2<0\ }$
Choose an x value larger than ${\displaystyle -{\frac {2}{3}}}$ :
${\displaystyle f'(1)=6(1)+4=10>0\ }$

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

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


## Second Derivative Test

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

• If ${\displaystyle {\frac {d^{2}y}{dx^{2}}}|_{x=c}=f''(c)=0}$  and ${\displaystyle f''(c)\ }$  changes signs at ${\displaystyle x=c\ }$ , then there is a point of inflection (change in concavity) at ${\displaystyle x=c\ }$ .
• If ${\displaystyle f''(x)<0\ }$ , then the graph of ${\displaystyle f(x)\ }$  is concave down.
• If ${\displaystyle f''(x)>0\ }$ , then the graph of ${\displaystyle f(x)\ }$  is concave up.

Example 2:

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

• Find ${\displaystyle {\frac {d^{2}y}{dx^{2}}}}$ .
${\displaystyle f(x)=x^{3}+2x+7\ }$
${\displaystyle f'(x)=3x^{2}+2\ }$
${\displaystyle f''(x)=6x\ }$
• Set ${\displaystyle {\frac {d^{2}y}{dx^{2}}}=0}$ .
${\displaystyle 6x=0\ }$
${\displaystyle x=0\ }$
• Determine whether ${\displaystyle f''(x)\ }$  changes signs at ${\displaystyle x=0\ }$ .
Choose an x value that is smaller than 0:
${\displaystyle f''(-1)=6(-1)=-6<0\ }$
Choose an x value that is larger than 0:
${\displaystyle f''(1)=6(1)=6>0\ }$

Therefore, there exists a point of inflection at ${\displaystyle x=0\ }$  because ${\displaystyle f''(0)=0\ }$  and ${\displaystyle f''(x)\ }$  changes signs at ${\displaystyle x=0\ }$ .

Answer: point of inflection: ${\displaystyle x=0\ }$ .