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Páginas: 5 (1032 palabras) Publicado: 22 de noviembre de 2010
PUNTOS DE INFLEXION

F X= 9-3X^2 => F(0)= 1 + 9(0) – (0)^3
F X= -6X = 1
-6X= 0 ( 0, 1)
X= 0
PUNTOS CRITICOS ( maximos y minimos)
Para
X=3=1.73 => F(3)= 149(3)- (3)^3
=11.39 (1.73,11.35)
Para x = -3 = -1.73
* F (-3)=-1.73
* F (3)= 1 + 9 (-3)- (3)=-9.39 ( -1.73 , -9.39)
*

F(X)=3X+5X
F(X)= 15X^4+15^2
15X^4+15X^2=0
15X^2(X^2+1)=0
15X^2=0 X^2+1 = 0
X^2=0/15 X^2= -1
X=0 X^2=±-1
X=-1 O X=+1
NO EXISTEN VALORES CRITICOS(-) ------------------------------ > (+)
<--------------------------------
------------------l--------------------l---------------------------l---------------------
0INTERVALO -∞<X<O O<X<∞
NUMERO DE PRUEBA -1 1
SIGNO DE F(-1)015(-1)^4 + 15(-1)^2 F(1) =15(1)^4+15(1)^2
F(X) =15+15 =15+15=30 =30
PENDIENTES M(+) M(+)
F CRECIENTE SI SI
F DECRESIENTE NO NO

PUNTO DE INFLEXION
F’’(X) = 60 X^3+30X 60X^3X= 0 30X(-2X^2 +1)=0
3X=0 2X^2 + 1 =0 X=±-1/2
X= 0/3 X^2=-1/2 ENTONSES X= 0

Y= X^2 SEN^3 (3X34a)

Y’=X^2 ddx( sen3X34a)^3 + sen^3(3X34a) ddx(x^2)

Y’=x^2[3(sen3X34a)^2ddx(sen3X34a)] + sen^3(3X34a)[2x]

Y’=x^2[3(sen3X34a)^2[cos 3X34addx(3X34a)]+ 2x sen^3(3X34a)

Y’= x^2[3(sen3X34a)^2[cos 3X34a[9x^24a]+2x sen3(3X34a)]

Y’= 27x^44a(sen3X34a) cos3X34a + 2x sen^3 3X34a

Y’= x sen^2(3X34a)[ 27x^44a cos 3X34a+ 2sen (3X34a)]

Y=In [33x(3x+8)^3(3x+8)^2]= In [3x(3x+8)^3]

y=In3x + In (3x + 8 )

y= 1/3In x + In (3x+8)

y’= 1/3 [ddx(x)/x]+[ddx(3x+d)/3x+8]

y’=1/3[1x] + 3(3)3x+8=13x + 33x+8

Y’=3x+8+27x3x(3x+8) = 30x+83x(3x+8)

F(X) =1+9X-X^3
F’= 9-3X^2
9-3X^2=0
-3(-3 + X^2)=0
-3=0 -3+X^2
X^2=3
X= ±3
X=3 0’ X =-3------------------l--------------------l---------------------------l---------------------
-3 0 3

INTERVALOS -∞<X<-3 -3<X<3 3<X<∞
NUMEROS DE PRUEVA -2 0 2

SIGNO DE =9-12 =9 =9-12
F’(X)=-3 =-3
-3<0 9>0 -3<0
PENDIENTES M(-) M(+) M(-)

F CRECIENTE NO SI NO
F DECRECIENTE SI NO SI