Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Question of the Day - II [#permalink]
04 Jun 2012, 18:48

Expert's post

devinawilliam83 wrote:

Karishma, Is there a way to solve this by setting boiundary conditions. ie. x<-1,-1<x<1/4,1/4<x<3,x>3?? I tried but wasnt able to make sense

VeritasPrepKarishma wrote:

Q. If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)?

(A) 3 (B) 4 (C) 5 (D) 21/4 (E) 7

(Still high on mods! Next week, will make questions on some other topic.)

If you mean whether you can make equations using positive and negative values, you cant do that with minimum/maximum questions. You don't really have a value to equate them to. e.g. |4x - 1| + |x-3| + |x + 1| = 10 is workable but minimum value of f(x) isn't. You will need to find the value at the critical points and then figure how f(x) changes or just use the number line. _________________

Re: Question of the Day - II [#permalink]
04 Jun 2012, 22:09

Hi Karishma...

Can you please explain the example which has negative between the modulus (could not understand fully the explaination given by you earlier). Also if in a question we have a combination of + and - then how to go about it?

Re: Question of the Day - II [#permalink]
05 Jun 2012, 05:22

1

This post received KUDOS

Expert's post

ankitbansal85 wrote:

Hi Karishma...

Can you please explain the example which has negative between the modulus (could not understand fully the explaination given by you earlier). Also if in a question we have a combination of + and - then how to go about it?

When you add two mods, you try to add up the distances e.g. |x - 1| + |x - 5| = 10 you try to find the point where distance from 1 and distance from 5 adds up to give you 10.

When you subtract two mods, you subtract out the distances e.g. |x - 1| - |x - 5| = 3 you try to find the point where distance from 1 and distance from 5 have a difference of 3. You know that at x = 3, distances from 1 and from 5 are equal (distance of 1 from 3 is 2 and distance of 5 from 3 is also 2). At x = 4, the difference between the distances will be 2. At x = 4.5, the difference between the distances will be 3.

The subtraction is a little less intuitive and will take more practice. Questions with both + and - would be too complicated though do-able. Most people will probably not get any mods question with more than 2 terms. _________________

Re: Question of the Day - II [#permalink]
06 Dec 2012, 04:06

I love this question. One must understand that f(x) = |4x-1| + |x-3| + |x+1| means the sum of the distances of x to 1/4, 3 and -1. The best way to minimize is to zero out the distance in the middle.

==========(-1)==========(0)======(1/4)=========(3)======= So if x = 1/4 |4x-1| = 0 |x+1| = 1 1/4 |x-3| = 2 3/4

Re: Question of the Day - II [#permalink]
13 May 2013, 08:52

Expert's post

yezz wrote:

What if the question asks for max value of f(x) ??? disregarding the answer choices ??

Not every function will have a minimum and a maximum value. The greater the value of x, the greater the function will become. It is an infinitely increasing function. _________________

Re: Question of the Day - II [#permalink]
19 May 2013, 16:08

Many Thanks to all of you for sharing such amazing techniques. I was overwhelmed with mod questions when I started, but your explanations and techniques have helped me build confidence. Bunuel, Karishma, Gurpreet, Shrouded1....Awesome!

How about this approach:

F(x) will be minimum when each individual term in the function has the lowest possible value. So, I get x = 3, -1 and 1/4. Now, substituting each value of x in F(x), I can easily see that x=1/4 gives me the smallest possible value for F(x) = 4

Re: Question of the Day - II [#permalink]
20 May 2013, 06:14

1

This post received KUDOS

Expert's post

rohitd80 wrote:

Many Thanks to all of you for sharing such amazing techniques. I was overwhelmed with mod questions when I started, but your explanations and techniques have helped me build confidence. Bunuel, Karishma, Gurpreet, Shrouded1....Awesome!

How about this approach:

F(x) will be minimum when each individual term in the function has the lowest possible value. So, I get x = 3, -1 and 1/4. Now, substituting each value of x in F(x), I can easily see that x=1/4 gives me the smallest possible value for F(x) = 4

Thanks, Rohit

The technique is fine but the logic is not sound. Why should we say that the function will take minimum value only when it takes one of these three values? For one of these values, sure one mod will be 0 but the other two could be much greater. The reason why this works is because the minimum value will be at one of the transition points - the middle point (logic explained in the post on previous page) in case there are odd number of terms OR at two points (and for every value in between) in case there are even number of terms. _________________

Re: Question of the Day - II [#permalink]
28 May 2013, 13:56

1

This post received KUDOS

WholeLottaLove wrote:

Hello all.

I am wondering why we cannot take the positive and negative cases of f(x) = |4x - 1| + |x-3| + |x + 1| and solve for x that way?

In other words, f(x) = |4x - 1| + |x-3| + |x + 1|

I. f(x) = (4x-1) + (x-3) + (x+1)

II. f(x) = -(4x-1) + -(x-3) + -(x-1)

Thanks!

It's not that easy!

if you wanna study the absolute value, more math is required. You have to study each abs>0 so 4x-1>0 and x-3>0 and x+1>0 x>\frac{1}{4} and x>3 and x>-1 so 4x-1 is positive for x>1/4, x-3 is +ve for x>3 and x-1 is +ve for x>-1

Now you have to split the original function into the areas defined above: x<-1 all functions are negative f(x) = -(4x-1) + -(x-3) + -(x-1) if -1<x<\frac{1}{4} the third term is positive,the others negative f(x) = -(4x-1) + -(x-3) +(x-1) and so on...

You cannot take all positive or all negative, you have to study each function in all possible intervals Below there is the graph of F(x) that I hope will make thing clear. As you see there are 4 functions, each one defined in the intervals above, so your way of studying the abs value (reducing all to 2 functions) is incomplete

Hope it's clear

Attachments

Untitled.png [ 5.15 KiB | Viewed 959 times ]

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...

Almost half of MBA is finally coming to an end. I still have the intensive Capstone remaining which started this week, but things have been ok so far...