Answer:
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