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)y=f(x) be defined on the domain DD.

x=cx=c is an ABSOLUTE MAXIMUM if →

f(x)f(c)f(x) \leq f(c) for all xx in DD

x=cx=c is an ABSOLUTE MINIMUM if

f(c)f(x)f(c) \leq f(x) for all xx in DD

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

EXAMPLE

Find the global min of y=x2y=x^{2} on D=(,)D =(-\infty,\infty)

We know x20x^{2} \geq 0 thus x=0x=0 is the minimum

Example

If possible find the global min/max of y=x2y=x^{2} on the domain D=[1,2]D=[1,2]

We have x=1x=1  is a minimum.However, x=2x=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)y=sin(x)

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

f(c) ≤ f(x) for all x near c

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

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

OR

xc<tf(x)f(c)\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][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=cx=c is a local min or max and f(c)f'(c) is defined then f(c)=0f'(c)=0

Example - Find the local min/max of f(x)=x33xf(x)=x^{3}-3x

# Find the derivative of f(x)f(x)

f(x)=3x23f'(x)=3x^{2}-3

# Find the roots of the derivative

f(x)=3x23=0x21=0±1f'(x)=3x^{2}-3=0 \Rightarrow x^{2}-1=0 \Rightarrow \pm 1

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

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

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

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

Example

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

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

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

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

3. Thus

x=0x=0 is the global min

x=5x=5 is the global max.

Example

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

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

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

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

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

2. f(1)=1e1=ef(-1)=1e^{1}=-e

f(1)=1e1=1ef(1)=1\cdot e^{-1}=\frac{1}{e}

3. Thus,

x=1x=-1 is the min

x=1x=1 is the max

Example

Find all the critical points of f(x)=6x2x3f(x)=6x^{2}-x^{3}

f(x)=12x3x2=0f'(x)=12x-3x^{2}=0

=3x(4x)=3x(4-x)

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

Example

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

f(x)=1x+11x2+1f'(x)=\frac{1}{x+1} -\frac{1}{x^{2}+1}

=(x2+1)(x+1)(x2+1)=\frac{(x^{2}+1)-(x+1)}{(x^{2}+1)}

=x2x(x+1)(x2+1)=\frac{x^{2}-x}{(x+1)(x^{2}+1)}

=x(x1)(x+1)(x2+1)=\frac{x(x-1)}{(x+1)(x^{2}+1)}

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

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

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

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


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