How To Solve Them The Right Way

Endpoints [-10, 10]

Take the 1st derivative

Endpoints [-10, 10]

Endpoints [-10, 10]

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)

f(x)

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]

Endpoints [-10, 10]