5.4 Absolute Maxima and Minima - Warren Hills …...5.4 Absolute Maxima and Minima.pptx Author Ilona DiCosmo Created Date 1/20/2014 6:30:43 PM ...

Absolute Maxima and Minima

Absolute Extrema on Close Intervals [ a, b ]

Find the absolute extrema of f (x) = 6x43 −3x

13

on the interval [ -1, 1 ], and determine where these values occur.

f '(x) = 8x13 − x

−23 = x

−23 8x −1( )

f '(x) =8x −1( )

x23

Points to consider:

critical points

endpoints of the interval

1/8 0

-1 1

f (−1) = 9

f (0) = 0

f 18!

"#$

%&= −

98

f (1) = 3

abs. max

abs. min

Absolute Extrema on Infinite Intervals LIMITS

CONCLUSION abs min no max

abs max no min

no max no min

no max no min

GRAPH

limx→−∞

f (x) = +∞

limx→+∞

f (x) = +∞

limx→−∞

f (x) = −∞

limx→+∞

f (x) = −∞

limx→−∞

f (x) = −∞

limx→+∞

f (x) = +∞limx→−∞

f (x) = +∞

limx→+∞

f (x) = −∞

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-0.8

0.8

1.6

2.4

3.2

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-2.4

-1.6

-0.8

0.8

1.6

-0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2

-2.4

-1.6

-0.8

0.8

1.6

-2.8 -2.4 -2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2

-3.2

-2.4

-1.6

-0.8

0.8

1.6

2.4

Determine by inspection whether p(x) = 3x4 + 4x3 has any absolute extrema. If so, find them and state where they occur.

limx→−∞

3x4 + 4x3 = +∞

limx→+∞

3x4 + 4x3 = +∞

min exists

p '(x) =12x3 +12x2 = 0

12x3 +12x2 = 0

12x2 x +1( ) = 0

cp = −1,0

p(−1) = −1

p(0) = 0abs min

-2.8 -2.4 -2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2

-1.6

-0.8

0.8

1.6

2.4

3.2

•

Absolute Extrema on Open Intervals LIMITS

CONCLUSION abs min no max

abs max no min

no max no min

no max no min

GRAPH

limx→a+

f (x) = +∞

limx→b−

f (x) = +∞

limx→a+

f (x) = −∞

limx→b−

f (x) = −∞

limx→a+

f (x) = −∞

limx→b−

f (x) = +∞

limx→a+

f (x) = +∞

limx→b−

f (x) = −∞

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-0.8

0.8

1.6

2.4

3.2

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-2.4

-1.6

-0.8

0.8

1.6

-0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2

-2.4

-1.6

-0.8

0.8

1.6

-2.8 -2.4 -2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2

-3.2

-2.4

-1.6

-0.8

0.8

1.6

2.4

Determine by inspection whether p(x) = 1

x2 − xhas any absolute extrema

On the interval ( 0, 1 ). If so, find them and state where they occur.

limx→0+

1x2 − x

= limx→0+

1x x −1( )

= −∞max exists

f '(x) = − 2x −1x2 − x( )

2 = 0

cp = 0, 12,1

limx→1−

1x2 − x

= limx→1−

1x x −1( )

= −∞

f (0) = DNE

f 12!

"#$

%&= −4

f (1) = DNE

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

-7.5

-5

-2.5

2.5

•

How about some practice?

1. f (x) =1+ 1x

Find the absolute extrema for the following:

0,+∞( )

2. f (x) = x3e−2 x 1, 4[ ]

3. f (x) = sin x − cos x 0,π( ]