Chapter 22: Extreme Values

Extreme Values

We have the detailed study of the structure of graphs of functions. Flow are they shaped? What are their key properties?

Definition → Let a function $y=f(x)$ be defined on the domain $D$.

$x=c$ is an ABSOLUTE MAXIMUM if →

$f(x) \leq f(c)$ for all $x$ in $D$

$x=c$ is an ABSOLUTE MINIMUM if →

$f(c) \leq f(x)$ for all $x$ in $D$

We say there are EXTREME VALUES of $f(x)$

EXAMPLE

Find the global min of $y=x^{2}$ on $D =(-\infty,\infty)$

We know $x^{2} \geq 0$ thus $x=0$ is the minimum

Example

If possible find the global min/max of $y=x^{2}$ on the domain $D=[1,2]$

We have $x=1$ is a minimum.However, $x=2$ is not in the domain and thus, there is no max.

Discuss → Fill in the table with domains for the function

$y=sin(x)$

Definition → A function $y=f(x)$ has a LOCAL MAX $f(x)\leq f(c)$ for all $x$ near $c$. $f(x)$ has a LOCAL MIN at $x=c$

f(c) ≤ f(x) for all x near c

"Near" - Means there is a small number t so that -

$\mid x-c\mid < t \Rightarrow f(c)\leq f(x)$ for minima

OR

$\mid x-c\mid < t \Rightarrow f(x)\leq f(c)$ for maxima

Discuss → Label the local/global min/max of -

Fact → (The Extreme Value Theorem)

Any continuous function defined on a closed interval $[a,b]$ has a global maximum and a global minimum.

This fact like the IVT, is not easy to prove. It is tricky and topological.

Fact → If $x=c$ is a local min or max and $f'(c)$ is defined then $f'(c)=0$

Example - Find the local min/max of $f(x)=x^{3}-3x$

# Find the derivative of $f(x)$

$f'(x)=3x^{2}-3$

# Find the roots of the derivative

$f'(x)=3x^{2}-3=0 \Rightarrow x^{2}-1=0 \Rightarrow \pm 1$

Thus, $x=-1$ and $x=1$ may be local max/min

Definition → A point where $f'(x) =0$ or is undefined is a CRITICAL POINT of $y=f(x)$

Algorithm → Find the global min/max of $f(x)$ on a finite closed intervals

1. Find the critical points of $f(x)$
2. Evaluate $f(x)$ at all endpoints and critical points
3. Take the largest and smallest values

Example

Find the global min/max of $y=\mid x\mid$ on $[-2,5]$

1. $f'(x)=0$ or undef $\Rightarrow x=0$
2. $f(-2)=\mid -2\mid =2$

$f(0)=\mid 0\mid =0$

$f(5)=\mid 5\mid =5$

3. Thus

$x=0$ is the global min

$x=5$ is the global max.

Example

Find the global min/max of $f(x)=xe^{-x}$ on $[-1,1]$

1. $f'(x)=e^{-x}+xe^{-x}(-1)$

$=e^{-x}-xe^{-x}$

$=(1-x)e^{-x}$

$f'(x)=0 \Rightarrow x=1$

2. $f(-1)=1e^{1}=-e$

$f(1)=1\cdot e^{-1}=\frac{1}{e}$

3. Thus,

$x=-1$ is the min

$x=1$ is the max

Example

Find all the critical points of $f(x)=6x^{2}-x^{3}$

$f'(x)=12x-3x^{2}=0$

$=3x(4-x)$

Thus, $x=0$ OR $x=4$

Example

Find all critical points of $f(x)=in(x+1)-arctan(x)$

$f'(x)=\frac{1}{x+1} -\frac{1}{x^{2}+1}$

$=\frac{(x^{2}+1)-(x+1)}{(x^{2}+1)}$

$=\frac{x^{2}-x}{(x+1)(x^{2}+1)}$

$=\frac{x(x-1)}{(x+1)(x^{2}+1)}$

Thus, $f(x)$ has critical points at -

$x=1 \Rightarrow f'(x)$ undef

$x=0 \Rightarrow f'(x)=0$

$x=1 \Rightarrow f'(x)=0$