45 Graphing Extrema
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View 45 Graphing Extrema as PDF for free.
More details
- Words: 497
- Pages: 8
How Not to Graph f (x) = sin 63x: sin 63x
sin 63x
Applications of the Mean Value Theorem:
If f 0(x) = 0 for all x in an interval, then f is constant there.
If f 0(x) > 0 for all x in an interval, then f is increasing there.
If f 0(x) < 0 for all x in an interval, then f is decreasing there.
More Applications of the Mean Value Theorem: If f 00(x) > 0 for all x in an interval, then f 0 increases there.
The curve “cups up”, or is “concave upward”.
If f 00(x) < 0 for all x in an interval, then f 0 decreases there.
The curve “cups down”, or is “concave downward”.
Wherever f 00(x) changes sign, we have a Point of Inflection.
f 00(x) < 0
f 00(x) < 0 f 00(x) > 0
f 00(x) > 0
At an inflection point, f 00(x) usually equals 0 or doesn’t exist.
f <0
f (x) = x3 − x4
= x (1 − x) Roots at 0 and 1 2
f (x) = 3x − 4x
3
f0 > 0
1
f0 > 0 0
2
f (x) = 6x − 12x
2
f 00 < 0 f 00 > 0 ∩ ∪ 0
= 6x(1 − 2x) 1 ) Inflection points at (0, 0) and at ( 12 , 16
Important x’s: 0, 12 , 34 , 1.
f0 < 0 3 4
= x (3 − 4x) 27 ) Leveling at (0, 0), Maximum at ( 34 , 256 00
f <0
0
3
0
f >0
1 2
f 00 < 0 ∩
1 27 Important y’s: 0, 16 , 256 .
Initial Steps:
27 ( 34 , 256 )
1 ) ( 12 , 16
0 Completed Graph:
1 27 ( 34 , 256 )
1 ) ( 12 , 16
x
0
1
x
What might happen with a misdrawn inflection point:
1 ) ( 12 , 16
27 ( 34 , 256 )
0
1
x
1
√ f <0 f (x) = x − x 0 4 Equals 0 only at 0 0 0 1 1 f < 0 f >0 0 f (x) = − √ 0 4 4 2 x √ x−2 = √ 4 x Endpoint Maximum and Absolute Maximum at (0, 0)
9
9
Absolute Minimum at (4, −1), Endpoint Maximum at (9, − 34 ) 00
1
− 32
f (x) = − x > 0, 4 Important x’s: 0, 4, 9.
so graph is always concave upward. Important y’s: 0, − 34 , −1.
0 1 2 3 4 5 6 7 8 9 Initial Steps:
(4, −1)
(9, − 34 )
0 1 2 3 4 5 6 7 8 9
Completed Graph:
(4, −1)
(9, − 34 )
sin 63x
Applications of the Mean Value Theorem:
If f 0(x) = 0 for all x in an interval, then f is constant there.
If f 0(x) > 0 for all x in an interval, then f is increasing there.
If f 0(x) < 0 for all x in an interval, then f is decreasing there.
More Applications of the Mean Value Theorem: If f 00(x) > 0 for all x in an interval, then f 0 increases there.
The curve “cups up”, or is “concave upward”.
If f 00(x) < 0 for all x in an interval, then f 0 decreases there.
The curve “cups down”, or is “concave downward”.
Wherever f 00(x) changes sign, we have a Point of Inflection.
f 00(x) < 0
f 00(x) < 0 f 00(x) > 0
f 00(x) > 0
At an inflection point, f 00(x) usually equals 0 or doesn’t exist.
f <0
f (x) = x3 − x4
= x (1 − x) Roots at 0 and 1 2
f (x) = 3x − 4x
3
f0 > 0
1
f0 > 0 0
2
f (x) = 6x − 12x
2
f 00 < 0 f 00 > 0 ∩ ∪ 0
= 6x(1 − 2x) 1 ) Inflection points at (0, 0) and at ( 12 , 16
Important x’s: 0, 12 , 34 , 1.
f0 < 0 3 4
= x (3 − 4x) 27 ) Leveling at (0, 0), Maximum at ( 34 , 256 00
f <0
0
3
0
f >0
1 2
f 00 < 0 ∩
1 27 Important y’s: 0, 16 , 256 .
Initial Steps:
27 ( 34 , 256 )
1 ) ( 12 , 16
0 Completed Graph:
1 27 ( 34 , 256 )
1 ) ( 12 , 16
x
0
1
x
What might happen with a misdrawn inflection point:
1 ) ( 12 , 16
27 ( 34 , 256 )
0
1
x
1
√ f <0 f (x) = x − x 0 4 Equals 0 only at 0 0 0 1 1 f < 0 f >0 0 f (x) = − √ 0 4 4 2 x √ x−2 = √ 4 x Endpoint Maximum and Absolute Maximum at (0, 0)
9
9
Absolute Minimum at (4, −1), Endpoint Maximum at (9, − 34 ) 00
1
− 32
f (x) = − x > 0, 4 Important x’s: 0, 4, 9.
so graph is always concave upward. Important y’s: 0, − 34 , −1.
0 1 2 3 4 5 6 7 8 9 Initial Steps:
(4, −1)
(9, − 34 )
0 1 2 3 4 5 6 7 8 9
Completed Graph:
(4, −1)
(9, − 34 )
Related Documents
45 Graphing Extrema
November 2019 12
Graphing
October 2019 33
Graphing
October 2019 29
Graphing
May 2020 23
Pobreza Extrema
May 2020 23