Download 2.7 NotesThe Extreme Value Theorem (EVT)

2.7 The Extreme Value Theorem (EVT)
The Extreme Value Theorem (EVT) states:
If f is continuous on [a,b] then f has both a max and a min value on [a,b].
fmax
fmax
*Not continuous from [A,B]; No min value.
fmin
A
c
B
A
c
B
*Continuous from [A,B], and has both a max and a min value on [A,B].
____________________________________________________________________________________________________________
Fmax
fmax
Continuous with a max and a min.
fmin
A
B
A
B
Not continuous and no min.
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fmax
Not continuous with both a max and a min.
fmin
fmin
A
c
B
A
c
B
____________________________________________________________________________________________________________
fmax
f(a)=f(b)
“Cusp”
fmin
f(a)=f(b)
A
c
B
A
c
B
Where can maxes and mins occur???
Either at “a” or “b” (one of the end points), or at a “c” value where f’(c)=0 or f’(c) is undefined.
How to find Extrema on a given interval:
1) Find f’(x)
2) Find all “critical numbers”- Where f’(c)=0 or f’(c) is undefined. c 1, c2, c3 …
3) Evaluate f(x) at each critical number and at a and b.
f(a)=
f(c1)=
f(c2)=
f(c3)=
f(b)=
4) Whichever “Y” is biggest in this list will be the max of f on [a,b] and whichever “Y” is smallest in this list will be the min of
f on [a,b].
Example:
Find the extrema of
Step 1:
Step2:
3x 4 - 4x 3 on [-1,2]
f '(x) = 12x 3 - 12x 2
0 = 12x 3 - 12x 2
0 = 12x 2 ( x - 1)
x0
x  1 (Both of the critical numbers are on the interval [-1,2]
f  a   f  1  7
f  c1   f  0   0
Step 3:
f  c2   f 1  1
min
f  b   f  2   16
max
This shows us that at f(1) we have a min and at f(2) we have a max.
Example:
Find the extrema of
f  x   2sin x  cos 2 x on 0, 2 
f '  x   2cos x  2sin 2 x
0  2cos x  2sin 2x
2 0  cos x  sin 2 x
0  cos x  2sin x cos x
0  cos x 1  2sin x 
cos x  0
1  2sin x  0
 3
1
x ,
sin x  
2 2
2
7 11
x
,
2 6
f  0  1
 
f    3 max
2
3
 7 
f
min
 
2
 6 
 3 
f
  1
 2 
3
 11 
f
 
2
 6 
f  2   1
min