Post on 31-Dec-2015
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f ’(x)
-10 16/3 6 8 10
+ 0 - 0 +
f (x)
-10
16/3
20/3
8
10
(-10)^3 – 20(-10)^2 +128(-10) – 280 = -560
(16/3)^3 – 20(16/3)^2 +128(16/3) – 280 = -14.518
(10)^3 – 20(10)^2 +128(10) – 280 = 0
(8)^3 – 20(8)^2 +128(8) – 280 = -24
(20/3)^3 – 20(20/3)^2 +128(20/3) – 280 = -19.258 See Slide 9 for further details
Endpoints [-10, 10]
f ’(x)
-10 16/3 6 8 10
+ 0 - 0 +
There is a max at x= 16/3 because f ‘ (x) changes from positive 0 negative
There is a min at x= 8 because f ‘ (x) changes from negative 0 positive
Endpoints [-10, 10]
f (x)
-10
16/3
20/3
8
10
(-10)^3 – 20(-10)^2 +128(-10) – 280 = -560 GLOBAL MIN
(10)^3 – 20(10)^2 +128(10) – 280 = 0 GLOBAL MAX
(8)^3 – 20(8)^2 +128(8) – 280 = -24 MIN
(16/3)^3 – 20(16/3)^2 +128(16/3) – 280 = -14.518 MAX
(20/3)^3 – 20(20/3)^2 +128(20/3) – 280 = -19.258 POINT OF INFLECTION
x= -10 is an endpoint on the left and f ‘(x) to its right is positive :. x= -10 is a min
x= 10 is an endpoint on the right and f ‘(x) to its left is positive :. x= 10 is a max
Endpoints [-10, 10]
f “(x) = 6x – 40
f “(x)
16/3
8
6(16/3) – 40 = -8 < 0
6(8) – 40 = 8 > 0
f “(x)
-10 20/3 10
- 0 +
To Find the Point(s) of Inflection6x – 40 = 0x= 20/3
Endpoints [-10, 10]