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Bosquejo de Graficas 2

Module by: DIEGO DIAZ LEON. E-mail the author

Summary: Bosquejo de Graficas

GRAFICA DE FUNCIONES2

1. encuentre los cortes, puntos criticos, intervalos, puntos de inflexion, concavidad.

y = x 4 - 2 x 2 Cortes con ejes : Con x . cuando y = 0 0 = x 4 - 2 x 2 0 = x 2 ( x 2 - 2 ) x 1 = 0 x 2 - 2 = 0 x = 2 x = ± 1 . 4 x 2 = 1 . 4 & x 3 = - 1 . 4 Con y . Cuando x = o y = 0 Puntos Criticos : y ' = 4 x 3 - 4 x = 0 x 3 - x = 0 x ( x 2 - 1 ) x 1 = 0 ( x + 1 ) ( x - 1 ) x 2 = - 1 & x 3 = 1 Para hallar los y se reemplazan en la original . y 1 = 0 & y 2 = - 1 & y 3 = - 1 Puntos . ( 0 , 0 ) ( - 1 , - 1 ) ( 1 , - 1 ) Intervalos : y ' = 4 x 3 - 4 x y ' ( - 2 ) = 4 ( - 2 ) 3 - 4 ( - 2 ) = - ' ' negativo ' ' y ' ( - 0 . 1 ) = 4 ( - 0 . 1 ) 3 - 4 ( - 0 . 1 ) = + ' ' positivo ' ' y ' ( 0 . 1 ) = 4 ( 0 . 1 ) 3 - 4 ( 0 . 1 ) = - ' ' negativo ' ' y ' ( 2 ) = 4 ( 2 ) 3 - 4 ( 2 ) = + ' ' positivo ' ' Puntos de Inflexion y ' ' = 12 x 2 - 4 = 0 12 x 2 = 4 x = 1 3 x 1 = 0 . 5 & x 2 = - 0 . 5 Reemplaza en la ecuacion original para hallar los y . y = 4 x 3 - 2 x 2 y 1 = ( 0 . 5 ) 4 - 2 ( 0 . 5 ) 2 = 0 . 062 - 0 . 5 y 2 = ( - 0 . 5 ) 4 - 2 ( - 0 . 5 ) 2 = 0 . 062 - 0 . 5 y 1 = - 0 . 43 y 2 = - 0 . 43 Puntos . ( 0 . 5 , - 0 . 43 ) ( - 0 . 5 , - 0 . 43 ) Concavidad : y ' ' = 12 x 2 - 4 y ' ' ( - 1 ) = 12 - 4 = 8 > 0 y ' ' ( 0 ) = - 4 < 0 y ' ' ( - 1 ) = 8 > 0 y = x 4 - 2 x 2 Cortes con ejes : Con x . cuando y = 0 0 = x 4 - 2 x 2 0 = x 2 ( x 2 - 2 ) x 1 = 0 x 2 - 2 = 0 x = 2 x = ± 1 . 4 x 2 = 1 . 4 & x 3 = - 1 . 4 Con y . Cuando x = o y = 0 Puntos Criticos : y ' = 4 x 3 - 4 x = 0 x 3 - x = 0 x ( x 2 - 1 ) x 1 = 0 ( x + 1 ) ( x - 1 ) x 2 = - 1 & x 3 = 1 Para hallar los y se reemplazan en la original . y 1 = 0 & y 2 = - 1 & y 3 = - 1 Puntos . ( 0 , 0 ) ( - 1 , - 1 ) ( 1 , - 1 ) Intervalos : y ' = 4 x 3 - 4 x y ' ( - 2 ) = 4 ( - 2 ) 3 - 4 ( - 2 ) = - ' ' negativo ' ' y ' ( - 0 . 1 ) = 4 ( - 0 . 1 ) 3 - 4 ( - 0 . 1 ) = + ' ' positivo ' ' y ' ( 0 . 1 ) = 4 ( 0 . 1 ) 3 - 4 ( 0 . 1 ) = - ' ' negativo ' ' y ' ( 2 ) = 4 ( 2 ) 3 - 4 ( 2 ) = + ' ' positivo ' ' Puntos de Inflexion y ' ' = 12 x 2 - 4 = 0 12 x 2 = 4 x = 1 3 x 1 = 0 . 5 & x 2 = - 0 . 5 Reemplaza en la ecuacion original para hallar los y . y = 4 x 3 - 2 x 2 y 1 = ( 0 . 5 ) 4 - 2 ( 0 . 5 ) 2 = 0 . 062 - 0 . 5 y 2 = ( - 0 . 5 ) 4 - 2 ( - 0 . 5 ) 2 = 0 . 062 - 0 . 5 y 1 = - 0 . 43 y 2 = - 0 . 43 Puntos . ( 0 . 5 , - 0 . 43 ) ( - 0 . 5 , - 0 . 43 ) Concavidad : y ' ' = 12 x 2 - 4 y ' ' ( - 1 ) = 12 - 4 = 8 > 0 y ' ' ( 0 ) = - 4 < 0 y ' ' ( - 1 ) = 8 > 0
(1)

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