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)


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


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


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


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)


# 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


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.


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)



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


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



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


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}




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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