Solucin

() =3

3

2

2 1 () = 2

Haciendo () = 0 2 = 0 ( 1) = 0 = 0 = 1

X=0 x=1

< 0; = 1 (1) = 2

0 < < 1; = 0,5 (0,5) = 0,25

> 1; = 2 (2) = 2

Evaluando en

() =4

12

3

6

(0) =04

12

03

6 0 = 0

() =14

12

13

6 1 =

13

12

(0,0) (1, 13

12)

) () = 2 3 () = 0 =3

2

(3

2) = (

3

2)

2

3 (3

2) 10 =

49

4

Hay un mnimo en el punto (3

2,

49

4)

b)

x=3/2

3

2; = 2 (2) = 2(2) 3 = 4 (

3

2, )

c)

() = 2 () > 0 () (, )

d) La funcin no posee puntos de inflexin

e)

x -1 0 1 2

y -6 -10 -12 -12

a) lim2

2+53

24=

0

0 lim

2

2+5

2= lim

2

1

22+5=

1

6

b) lim1

31

21=

0

0 lim

1

32

2=

3

2

c) lim0

1

2=

0

0 lim

0

2=

1

2

d) lim

=

lim

1/

1= 0

) () = 32 + 12 9 () = 0 32 + 12 9 = 0 = 1 = 3

() = 6 + 12 ; (1) = 6(1) + 12 = 6 = 1; (3) = 6(3) + 12 = 6

= 3 (1) = (1)3 + 6(1)2 9(1) + 8 = 4 (1,4) (3) = (3)3 + 6(3)2 9(3) + 8 = 8 (3,8)

)

X=1 x=3

< 1; = 0 (0) = 9 () (, 1)

1 < < 3; = 2 (2) = 3 () (1,3)

> 3; = 4 (4) = 9 () (3, )

c) () = 6 + 12 () = 0 = 2

x=2

< 2; = 0 (0) = 12 (, 2)

> 2; = 4 (4) = 12 (2, )

e) Hay un punto de inflexin en x=2

(2) = (2)3 + 6(2)2 9(2) + 8 = 6 (2,6)

f) Grafica

() = 300(1 2) () = 0 =1

2

0 < 0 0 < 0

1

2< < 1