Absolute value Problems

Hermitage

New member
Joined
Jul 30, 2014
Messages
14
Hello,
Im completely new here and i am very sorry if similar topic exist, though search did not bring up any reasonable answers.

I seem to be unable to solve an inequation involving absolute value. That is to say that i get different answer from that provided in the book. I uploaded a picture with my problem + solution. I hope it makes sense there. Im sorry if it is messy, im just getting used to the graphic pad. The problem is, that the book says answer should be from -infinity to -1, but i get from -infinity to -1/4. Many thanks in advance :)
Ül 69.jpg
 

Attachments

  • Ül 69.jpg
    Ül 69.jpg
    7.4 KB · Views: 7
I'm sorry, but I can't read much of the text in the image. Kindly please provide something larger, or else simply type out the text from the image.

When you reply, kindly please clarify how this is a "calculus" exercise. Thank you! ;)
 
I think the problem is to determine where \(\displaystyle \left|\frac{2x- 1}{x+ 1}\right|\ge 2\).

If x= -1/2 then 2x- 1= -1- 1= -2 and x+ 1= 1/2 so that \(\displaystyle \frac{2x- 1}{x+ 1}= \frac{-2}{1/2}= -4\). It's absolute value is greater than 2 so x= -1/2 certainly does satisfy the inequality. Either you have copied the problem incorrectly or the answer in the book is wrong.
 
Last edited by a moderator:
I think the problem is to determine where \(\displaystyle \left|\frac{2x- 1}{x+ 1}\right|\ge 2\).

If x= -1/2 then 2x- 1= -1- 1= -2 and x+ 1= 1/2 so that \(\displaystyle \frac{2x- 1}{x+ 1}= \frac{-2}{1/2}= -4\). It's absolute value is greater than 2 so x= -1/2 certainly does satisfy the inequality. Either you have copied the problem incorrectly or the answer in the book is wrong.


Thank you for your time and answer:
I will try to clarify:

"In problems 66-70 find the solution set for inequalities"

The inequality:
MSP6721d75d7840agb02ga00003g3a3d32a8441433
So subsequently i find where left hand side is eqal to 0. x1=1/2 and x2=-1 (but x cant be 0, but thats beside the point)

Next thing i do, is separate the x-axis in the regions of interest> 1st: x<-1 ; 2nd -1<x<1/2 ; 3rd x>1/2

Next: rewrite equation and solve it with respect to region of intetrest and find overall solution:
I - both absolute values are negative if i plug in number lesser than -1.
Eq>
MSP112720927827e0ha846900001fca4b8ichiaf19b

Solution i get: 3>=0 (idk what im supposed to here, probably dismiss it)
II - if i plug in 0 1st absolute value is still negative second is positive
eq:
MSP16121hf61dafb28cih600000580d23e990e26h05

Solution i get: x<=- 1/4 eg: X=]-infinity; -1/4]

III- all absolute values are positive if i plug in anything greater than 1/2
Eq:


MSP4249202e4c20d6gc08c700003hb13hc8gf366578

solution i get: -3>=0 (i dismiss it, because it is impossible)

Therefore overall solution is X=]-infinity; -1/4] but the book says it should be X=]-infinity; -1]

Why is that, and how should i be able to get that book answer?
 
I'm sorry, but I can't read much of the text in the image. Kindly please provide something larger, or else simply type out the text from the image.

When you reply, kindly please clarify how this is a "calculus" exercise. Thank you! ;)

I just wrote a long post (above), as soon as it gets approved you can view the problem and my thoughts on it.

In regard to calculus question, it wery well may be that i just misunderstand the calculus word itself. English is not my 1st language. Thing is i have courses mathemaical analysis 1 and 2 and currilulum in this analysis and calculus correlated when i was on the web and srearched for something so i assumed its calculus. If its not, im sure the topic could be moved to appropriate place and you could provide the right name for this.
 
Last edited by a moderator:
Assume the problem is as stated by HallsofIvy:
Determine where [FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Main](2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1) / ([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1)[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Main]≥[/FONT][FONT=MathJax_Main]2[/FONT].

For problems like this you have to partition the answers in order to get rid of the absolute value sign. For notation sake let
p = (2x-1)/(x+1)
We also have
p' = 3 / (x+1)^2

The first partition is when p is positive, i.e. abs(p) = p. This occurs when both the numerator & denominator are positive, that is both 2x-1 is positive and x+1 is positive or when x > 1/2 or when both the numerator & denominator are negative, i.e. x < -1. Since abs(p) = p in these regions and p' is always positive (p is always increasing) the value of p is greater than the value at -inf in the interval (-inf, -1) and less than the value at +inf in the interval (1/2, +inf). The limiting value of p in both cases is 2.

The other partition is when p is negative, i.e. abs(p) = -p. The derivative is -p'. Now -p' is always negative (p always decreasing) which implies abs(p) is greater than 2 to the left of where -p = 2 or in the interval (-1, a) where p(a) = -2.
 
Assume the problem is as stated by HallsofIvy:
Determine where [FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Main](2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1) / ([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1)[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Main]≥[/FONT][FONT=MathJax_Main]2[/FONT].

For problems like this you have to partition the answers in order to get rid of the absolute value sign. For notation sake let
p = (2x-1)/(x+1)
We also have
p' = 3 / (x+1)^2

The first partition is when p is positive, i.e. abs(p) = p. This occurs when both the numerator & denominator are positive, that is both 2x-1 is positive and x+1 is positive or when x > 1/2 or when both the numerator & denominator are negative, i.e. x < -1. Since abs(p) = p in these regions and p' is always positive (p is always increasing) the value of p is greater than the value at -inf in the interval (-inf, -1) and less than the value at +inf in the interval (1/2, +inf). The limiting value of p in both cases is 2.

The other partition is when p is negative, i.e. abs(p) = -p. The derivative is -p'. Now -p' is always negative (p always decreasing) which implies abs(p) is greater than 2 to the left of where -p = 2 or in the interval (-1, a) where p(a) = -2.


Im sorry, im not going to read and get into it at this moment, as i have been up and at work 16 hours already.

I do have 1 question thoug: on my second post, are all the pictures gone? As you said "assuming the problem is..."
 
...I do have 1 question thoug: on my second post, are all the pictures gone? As you said "assuming the problem is..."

Your images are still there but I have a hard time reading them.

Before continuing, one thing about notation for intervals: I was taught (a long time ago) that a square bracket meant you should include the point [a closed end] and a round bracket (parenthesis) meant you should exclude the point [an open end]. Thus (u, v] would be the interval from u to v excluding u but including v. I was also taught that infinity was never included in an interval so an interval including -infinity was always open on the left and an interval including infinity was always open on the right.

As a point of interest re your statement "Solution i get: 3>=0 (idk what im supposed to here, probably dismiss it)", in cases like this you would conclude something like, since 3 is always greater than or equal to zero, the absolute value is always greater than 2 in the region (-inf, -1). Similarly the other situation leads to something like since -3 is never greater than zero, the absolute value is never greater than 2. As you indicated, the a in my equations is -1/4 so we only differ in our answers in that you include the point -1 and I exclude the point -1. That is your interval is (-inf, -1/4] and my intervals are (-inf, -1) and (-1, -1/4]. The reason I exclude the point -1 is that it would lead to division by zero and that is not generally defined as a number. Thus the result has no comparison to 2, i.e. the result is neither greater than nor equal to nor less than 2.
 
Your images are still there but I have a hard time reading them.
on my second post i see only broken images icons for some reason. Nvm that, just will make them differently next time.
Before continuing, one thing about notation for intervals: I was taught (a long time ago) that a square bracket meant you should include the point [a closed end] and a round bracket (parenthesis) meant you should exclude the point [an open end]. Thus (u, v] would be the interval from u to v excluding u but including v. I was also taught that infinity was never included in an interval so an interval including -infinity was always open on the left and an interval including infinity was always open on the right.
Correct
As a point of interest re your statement "Solution i get: 3>=0 (idk what im supposed to here, probably dismiss it)", in cases like this you would conclude something like, since 3 is always greater than or equal to zero, the absolute value is always greater than 2 in the region (-inf, -1). Similarly the other situation leads to something like since -3 is never greater than zero, the absolute value is never greater than 2. As you indicated, the a in my equations is -1/4 so we only differ in our answers in that you include the point -1 and I exclude the point -1. That is your interval is (-inf, -1/4] and my intervals are (-inf, -1) and (-1, -1/4]. The reason I exclude the point -1 is that it would lead to division by zero and that is not generally defined as a number. Thus the result has no comparison to 2, i.e. the result is neither greater than nor equal to nor less than 2.
[/QUOTE]

yes youre right about the brackets, that was an error on my part, i typed them here wrong. I was more concerned about the problem and missed that.

And thankyou for the explanation of the 3>=0 part. That makes sense, and that probably is true because i can vaguelly recall something similar.

Thankyou again!
 
on my second post i see only broken images icons for some reason.

Those image links are broken because their URLs point to pages at WolframAlpha that do not exist.

You may upload saved-images directly to the freemathhelp server, as post attachments. See the FAQ. :cool:
 
Top