Domain of definitions

[Aladdin]

1 - \(\displaystyle Log|x+1|\)

| x + 1|>0

For :

x+1>0

x > -1

2 - \(\displaystyle Log\frac{|x+1|}{|x-1|}\)

D: \(\displaystyle \frac{|x+1|}{|x-1|}>0\)

|x+1|>0

x > -1

3 - \(\displaystyle Log|x^2 -5x +4 |\)

|x^2 -5x +4 |>0 --- a+b+c=0 --- x'=1 , x''=4

Now would I look for the values where the values of x are positive (in the table of signs).

Is my work correct. Thanks for all the help .

 

[fasteddie65]

1 - \(\displaystyle Log|x+1|\)

| x + 1|>0

For :

x+1>0

x > -1

2 - \(\displaystyle Log\frac{|x+1|}{|x-1|}\)

D: \(\displaystyle \frac{|x+1|}{|x-1|}>0\)

|x+1|>0

x > -1

3 - \(\displaystyle Log|x^2 -5x +4 |\)

|x^2 -5x +4 |>0 --- a+b+c=0 --- x'=1 , x''=4

Now would I look for the values where the values of x are positive (in the table of signs).

Is my work correct. Thanks for all the help .

2. [(x + 1)/(x - 1)] > 0 if x < -1 or x > 1.

3. |x^2 - 5x + 4| > 0 for all x, except x = 1 and x = 4, since it has absolute values around it.
Without the absolute value signs, x^2 - 5x + 4 > 0 if x < 1 or x > 4.

 

[Aladdin]

2. [(x + 1)/(x - 1)] > 0 if x < -1 or x > 1.--> Eddie, can you please be more specific .Thank you

3. |x^2 - 5x + 4| > 0 for all x, except x = 1 and x = 4, since it has absolute values around it.
Without the absolute value signs, x^2 - 5x + 4 > 0 if x < 1 or x > 4.

 

[stapel]

2. [(x + 1)/(x - 1)] > 0 if x < -1 or x > 1.--> Eddie, can you please be more specific .Thank you

To learn how to solve rational inequalities, try here. :wink:

 

[Aladdin]

Re:

2. [(x + 1)/(x - 1)] > 0 if x < -1 or x > 1.--> Eddie, can you please be more specific .Thank you

To learn how to solve rational inequalities, try here. :wink:

Thanks Stapel_Eliz , What makes me preserved is the absolute value ? -- What happened to the absolute.

 

[stapel]

I would guess that the helper assumed that the absolute-value bars were being used in place of, but with the same meaning as, regular grouping symbols. If the original expression has absolute value specified, then the domain is all x-values which do not cause equality with zero or division by zero.

 

[Aladdin]

Re:

I would guess that the helper assumed that the absolute-value bars were being used in place of, but with the same meaning as, regular grouping symbols. If the original expression has absolute value specified, then the domain is all x-values which do not cause equality with zero or division by zero.

Then the domain is for all x except 1 and -1

 

[Deleted member 4993]

Re: Re:

Then the domain is for all x except 1 and -1

I am not sure what you exactly mean by that statement.

For example, for #3 (Log|x[sup:32480i66]2[/sup:32480i66]-5x+4|) all the values of x is acceptable (-? < x < ?) except at x = 1 and x=4 correction

 

[Aladdin]

Re: Re:

Then the domain is for all x except 1 and -1

I am not sure what you exactly mean by that statement.

For example, for #3 (Log|x[sup:1v1oo5sc]2[/sup:1v1oo5sc]-5x+4|) all the values of x is acceptable (-? < x < ?)

So Mr khan, #3 the domain is all values of x are acceptable --So Eddie's work is wrong ?.

What I meant is \(\displaystyle \ ] -\infty ,-1-1,11,+\infty[\\)

 

[pka]

To clear things up.
#1 All \(\displaystyle x\not=-1\).

#2 All \(\displaystyle x\not= \pm1\)

#3 ALL \(\displaystyle x\not=1\text{ and }x\not=4\) This a correction.

 

[Aladdin]

To clear things up.
#1 All \(\displaystyle x\not=-1\).

#2 All \(\displaystyle x\not= \pm1\)

#3 \(\displaystyle (-\infty,1) \cup (4, \infty)\)

Thanks pka, for your Neat summary. :wink: --

 

[pka]

Please see my correction above.
You reply was too quick for me.