4. Find the absolute extrema of f (x) = 6x
4
3
-3x
1
3
on the interval [ -1, 1 ], and determine where
these values occur.
f '(x) = 8x
1
3
- x
-2
3
= x
-2
3
8x -1
( )
f '(x) =
8x -1
( )
x
2
3
Points to consider:
critical points
endpoints of the interval
1/8
0
-1
1
f (-1) = 9
f (0) = 0
f
1
8
æ
è
ç
ö
ø
÷ = -
9
8
f (1) = 3
abs. max
abs. min
5. Absolute Extrema on Infinite Intervals
LIMITS
CONCLUSION abs min
no max
abs max
no min
no max
no min
no max
no min
GRAPH
lim
x®-¥
f (x) = +¥
lim
x®+¥
f (x) = +¥
lim
x®-¥
f (x) = -¥
lim
x®+¥
f (x) = -¥
lim
x®-¥
f (x) = -¥
lim
x®+¥
f (x) = +¥
lim
x®-¥
f (x) = +¥
lim
x®+¥
f (x) = -¥
6. Determine by inspection whether
p(x) = 3x4
+4x3
has any absolute extrema.
If so, find them and state where they occur.
lim
x®-¥
3x4
+4x3
= +¥
lim
x®+¥
3x4
+4x3
= +¥
min
exists
p'(x)=12x3
+12x2
= 0
12x3
+12x2
= 0
12x2
x+1
( )= 0
cp = -1,0
p(-1) = -1
p(0) = 0
abs min
·
7. Absolute Extrema on Open Intervals
LIMITS
CONCLUSION abs min
no max
abs max
no min
no max
no min
no max
no min
GRAPH
lim
x®a+
f (x) = +¥
lim
x®b-
f (x) = +¥
lim
x®a+
f (x) = -¥
lim
x®b-
f (x) = -¥
lim
x®a+
f (x) = -¥
lim
x®b-
f (x) = +¥
lim
x®a+
f (x) = +¥
lim
x®b-
f (x) = -¥
8. Determine by inspection whether
p(x) =
1
x2
- x
has any absolute extrema
On the interval ( 0, 1 ). If so, find them and
state where they occur.
lim
x®0+
1
x2
- x
= lim
x®0+
1
x x -1
( )
= -¥
max
exists
f '(x) = -
2x -1
x2
- x
( )
2
= 0
cp = 0,
1
2
,1
lim
x®1-
1
x2
- x
= lim
x®1-
1
x x -1
( )
= -¥
f (0)= DNE
f
1
2
æ
è
ç
ö
ø
÷ = -4
f (1) = DNE
·
9. How about some
practice?
1. f (x) =1+
1
x
Find the absolute extrema for the following:
0,+¥
( )
2. f (x) = x3
e-2x 1,4
[ ]
3. f (x)=sinx-cosx 0,p
( ]
