Respostas

2013-05-05T14:53:32-03:00

y = (2x + 1)^3

y' = 3(2x + 1)^2 . (2x - 1)'

y' = 3(2x + 1)^2 . 2

y' = 6(2x + 1)^2

y' = 6(4x^2 +2x + 2x + 1)

y' = 6(4x^2 + 4X + 1)      => y' = 0

 

4x^2 + 4x + 1 = 0

 

A = b^2 - 4ac

A = (4)^2 - 4(4)(1)

A = 16 - 16

A = 0

 

x1 = (- 4 + 0) / 8

x1 = -1/2

 

y" = 6(8x + 4)    =y" = 0

y"(x) = 8x + 4 = 0

y"(-1/2) = 8(-1/2) + 4 = 0

-8/2 + 4 = 0

-4+4 = 0

 

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