## Calcolare il dominio,studio del segno,intersezioni, asintoti orizzontali,verticali e obliqui, e punti di scontinuità (1 specie,2 specie, 3 s

Question

Calcolare il dominio,studio del segno,intersezioni, asintoti orizzontali,verticali e obliqui, e punti di scontinuità (1 specie,2 specie, 3 specie).
lim(x^4-x^2-9)
x=>-∞

in progress 0
12 hours 2021-09-12T17:38:52+00:00 1 Answer 0

Step-by-step explanation:

f( x ) = ln( | 2 x + 1 | – 2 )x – 4:

a) determine the domain;

b) study the behavior at the ends of the domain;

c) shows that in the intervals and[

3

[ – 3 ; – 74]is canceled at least once;[ 34; 3 ]

d) compute the solutions of ;f( x ) = 0

e) draw the possible graph.

2) After finding for which values ​​of the real parameter the functionk

f( x ) = { sink xxx < 0x2- k + 1x – 2x ≥ 0

presents a discontinuity of the first kind with a jump of in x=0 ,l = 1x = 0

a) determine the domain;

b) classifies any other points of discontinuity;

c) search for asymptotes.

s.

in the first case, the function exists for those values ​​of such that |2x+1|>2 and x≠4 , that is, for x∈]-∞;-

3

x| 2 x + 1 | > 2x ≠ 4x ∈ ] – ∞ ; – 32[ ∪ ] 12; + ∞ [ – { 4 }

f( x ) = ⎧⎩⎨ln( 2 x – 1 )x – 4x > 12,x ≠ 4ln( – 2 x – 3 )x – 4x < – 32.

xx → + ∞

limx → ± ∞f( x ) = 0limx → – 32-f( x ) = – ∞- 11 / 2= + ∞

limx → 12+f( x ) = – ∞- 7 / 2= + ∞limx → 4∓f( x ) = ln70∓= ∓ ∞.

f( – 3 ) = – ln37< 0f( – 74) =4 ln223> 0

f( 34) =4 ln313> 0f( 3 ) = – ln5 < 0

ln( 2 x – 1 ) = 0 → 2 x – 1 = 1 → x = 1ln( – 2 x – 3 ) = 0 → – 2 x – 3 = 1 → x = – 2.

l = 1x = 0

| limx → 0-f( x ) – limx → 0+f( x ) | =1

|||k – k – 12|||= 1 → k = 1 ∨ k = – 3.

D = R – { 2 }k

k = – 1x = 0x = 2

k = 5x = 0x = 2

k ≠ – 1 ∨ k ≠ 5x = 0x = 2

ky= 0y= x – 2limx → + ∞f( x )x= 1limx → + ∞( f( x ) – x ) =2x = 2k ≠ 5