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 500,000 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:

Ok, I get that for f(x) = (4x-1) + (x-3) + (x+1), x must be greater than 1/4, 3 and -1 respectively. But that's where I get lost.

I'm sorry for being so dense on this topic!

Study of the absolute value:

1)take each term into the "| |" and define where it's positive 2)draw a number line with each "edge value" (where each term changes sign) 3)Split the function according to those intervals

Refer to the image

So if x>3 all terms are positive =>f(x) = (4x-1) + (x-3) + (x+1) if 1/4<x<3 for example you see that 4x-1 is positive, x+1 is positive BUT x-3 is negative => f(x) = (4x-1) + (-)(x-3) + (x+1)

Repeat this operation for each interval and you'll have all possible combinations

Remeber that the each function is valid only in that interval

What I mean is that f(x) = (4x-1) + (-)(x-3) + (x+1) is valid only in the 1/4<x<3 area. Each area has its own function

Attachments

Untitled.png [ 1.55 KiB | Viewed 1425 times ]

_________________

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

Haha - I am lost (not your fault...I am just very slow with mathematical concepts, unfortunately)

I understand that for certain values of x,(say if x =2) (4x-1) and (x+1) would be positive but (x-3) would be negative. But why would we bother finding what is positive and what is negative? I almost feel as if testing a series of integers and fractions, both positive and negative, would be a quicker way to figure out the right answer. Still, I am trying to get the concepts buttoned down.

Haha - I am lost (not your fault...I am just very slow with mathematical concepts, unfortunately)

I understand that for certain values of x,(say if x =2) (4x-1) and (x+1) would be positive but (x-3) would be negative. But why would we bother finding what is positive and what is negative? I almost feel as if testing a series of integers and fractions, both positive and negative, would be a quicker way to figure out the right answer. Still, I am trying to get the concepts buttoned down.

What I explained above is how to study an abs value from a theoretical point of view, because you original methos is wrong.

Of course this is not required to answer the question, you can try with real number and see what you find out. But am I trying to explain how an abs function works, for instance your original method

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

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

does not work to find the answer. The function cannot be reduced to that form!

"But why would we bother finding what is positive and what is negative? " This is required to study an abs value. In my original post you see the graph of F(X), and you notice that is defined into sections. Each section is one of the intervals above, and in each one of those the fucntion has a different equation.

The concept that I apply here is the same as the one that you would apply to solve \(y=|x|\) How would you study this? for x>0 => y=x for x<0 => y=-x

Define where the funct is positive, treat each part as a separate equation. The concept in the question is the same, only involves more intervals. _________________

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

Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink]

Show Tags

09 Jul 2013, 19:58

Here is my idea: f(x) = f1(x) + f2(x) + f3(x) with all the functions are first-order ones. Suppose that: f1(x1)=0 --> x1=1/4 f2(x2)=0 --> x2=3 f3(x3)=0 --> x3=-1 Always does f(x) reach its minimum value with x(1) or (x2) or (x3) in this problem, having absolute value | | Therefore, we can calculate the value of f(1/4), f(3), f(-1) f(1/4)=4 f(3) and f(-1) are both bigger than 4 surely without further calculation

That minimum of a function will always be at either its critical points or zero.

Await your valued response.

The minimum could also be in an entire range. Take this question for example.

f(x) = |3x + 1| + |2x-3| + |x - 7|

For what value(s) of x will f(x) have the minimum value?

Thank you for your intuitive explanation and clearing the concept for us. So using your analogy of people at every point, shouldn't -1/3 have the minimum value for x (or distance so to speak)?

To be precise, you said 1/4 point has the minimum distance for x or gives the minimum value because of the denominator 4 (my assumption - as you did not state it specifically). This is why we chose 1/4 over 3/2! _________________

Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us.

Your playing small does not serve the world. There's nothing enlightened about shrinking so that other people won't feel insecure around you.

It's not just in some of us; it's in everyone. And as we let our own light shine, we unconsciously give other people permission to do the same.

As we are liberated from our own fear, our presence automatically liberates others.

That minimum of a function will always be at either its critical points or zero.

Await your valued response.

The minimum could also be in an entire range. Take this question for example.

f(x) = |3x + 1| + |2x-3| + |x - 7|

For what value(s) of x will f(x) have the minimum value?

Thank you for your intuitive explanation and clearing the concept for us. So using your analogy of people at every point, shouldn't -1/3 have the minimum value for x (or distance so to speak)?

To be precise, you said 1/4 point has the minimum distance for x or gives the minimum value because of the denominator 4 (my assumption - as you did not state it specifically). This is why we chose 1/4 over 3/2!

f(x) = |3x + 1| + |2x-3| + |x - 7|

f(x) = 3|x + 1/3| + 2|x-3/2| + |x - 7|

-1/3 ------------------ 3/2-----------------------------------7 (3) ...........................(2)...........................................(1) Say, there are 3 people at -1/3, 2 people at 3/2 and 1 person at 7. They need to meet while covering the least distance. Where should they meet?

Obviously, the person at 7 should travel to 3/2. The distance covered will be 7 - 3/2 = 11/2 Now there are 3 people at -1/3 and 3 people at 3/2. They can meet anywhere between -1/3 and 3/2. The distance covered will be the same in each case.

The point is not whether it is 1/4, the point is the constant outside i.e. how many people need to travel from that point. _________________

For what value(s) of x will f(x) have the minimum value?

Thank you for your intuitive explanation and clearing the concept for us. So using your analogy of people at every point, shouldn't -1/3 have the minimum value for x (or distance so to speak)?

To be precise, you said 1/4 point has the minimum distance for x or gives the minimum value because of the denominator 4 (my assumption - as you did not state it specifically). This is why we chose 1/4 over 3/2!

f(x) = |3x + 1| + |2x-3| + |x - 7|

f(x) = 3|x + 1/3| + 2|x-3/2| + |x - 7|

-1/3 ------------------ 3/2-----------------------------------7 (3) ...........................(2)...........................................(1) Say, there are 3 people at -1/3, 2 people at 3/2 and 1 person at 7. They need to meet while covering the least distance. Where should they meet?

Obviously, the person at 7 should travel to 3/2. The distance covered will be 7 - 3/2 = 11/2 Now there are 3 people at -1/3 and 3 people at 3/2. They can meet anywhere between -1/3 and 3/2. The distance covered will be the same in each case.

The point is not whether it is 1/4, the point is the constant outside i.e. how many people need to travel from that point.

Thank you for you prompt reply. As we use the people analogy, you mentioned that 2 people are at point 3/2 and 3 are at -1/3. Wouldn't the distance covered be more if 3 people had to travel instead of 2: because the constant outside the mod is 3, true?

Alternatively, if I use the traditional approach i.e. calculate the distance among these three points, I get to the conclusion that point 3/2 is where they cover the least distance.

So what am I missing here? Please assist me in getting the concept right using your holistic approach to Mods. Thanks. _________________

Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us.

Your playing small does not serve the world. There's nothing enlightened about shrinking so that other people won't feel insecure around you.

It's not just in some of us; it's in everyone. And as we let our own light shine, we unconsciously give other people permission to do the same.

As we are liberated from our own fear, our presence automatically liberates others.

Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink]

Show Tags

10 Sep 2013, 21:12

In the interval (-1,1/4) f(x)=-(4x-1)-(x-3)+(x+1)=-4x+1 f(0)=1 However, when I plug in 0 in the original f(x), I get it it to equal 5. f(0)=5. What am I doing wrong?

In the interval (-1,1/4) f(x)=-(4x-1)-(x-3)+(x+1)=-4x+1 f(0)=1 However, when I plug in 0 in the original f(x), I get it it to equal 5. f(0)=5. What am I doing wrong?

Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink]

Show Tags

14 Oct 2013, 05:18

May I ask a couple of silly questions : why do we find critical values when solving for such questions (with multiple mod expressions, either equalities or inequalities)?

And why do we find critical values equating the expressions inside the mod to zero? Is this because we know that the expression must be equal to or greater than zero?

Can't quite explain it to myself. Please help me out to straighten these out. _________________

There are times when I do not mind kudos...I do enjoy giving some for help

I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : |x+6|-|x-1| ?

You can do it with a negative sign too. You want to find the minimum value of (distance from -6) - (distance from 1)

Make a number line with -6 and 1 on it. (-6)..........................(1)

Think of a point in the center of -6 and 1. Its distance from -6 is equal to distance from 1 and hence (distance from -6) - (distance from 1) = 0 .

What if instead, the point x is at -6? Distance from -6 is 0 and distance from 1 is 7 so (distance from -6) - (distance from 1) = 0 - 7 = -7

If you keep moving to the left, (distance from -6) - (distance from 1) will remain -7 so the minimum value is -7.

In the above example, the absolute value function f(x), which is sum of absolute functions, can not be negative for any value of x. Kindly clarify whether the minimum value of f(x) is 7 or -7.

If f(x) = | 1 - x | + | x - 1 |, then minimum value of f(x) is 0 for x = 1. Kindly comment.

I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : |x+6|-|x-1| ?

You can do it with a negative sign too. You want to find the minimum value of (distance from -6) - (distance from 1)

Make a number line with -6 and 1 on it. (-6)..........................(1)

Think of a point in the center of -6 and 1. Its distance from -6 is equal to distance from 1 and hence (distance from -6) - (distance from 1) = 0 .

What if instead, the point x is at -6? Distance from -6 is 0 and distance from 1 is 7 so (distance from -6) - (distance from 1) = 0 - 7 = -7

If you keep moving to the left, (distance from -6) - (distance from 1) will remain -7 so the minimum value is -7.

In the above example, the absolute value function f(x), which is sum of absolute functions, can not be negative for any value of x. Kindly clarify whether the minimum value of f(x) is 7 or -7.

If f(x) = | 1 - x | + | x - 1 |, then minimum value of f(x) is 0 for x = 1. Kindly comment.

Thanks.

Arun.

Sum of two absolute functions cannot be negative but difference can be. The original post discusses the sum of absolute functions. ficklehead asked about f(x) which is difference between two absolute functions. The '-7' is the minimum value of f(x) in case of difference.

f(x) = |a| - |b| can easily be negative e.g. if a = 2 and b = 5 _________________

If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink]

Show Tags

25 Nov 2013, 03:01

VeritasPrepKarishma wrote:

arunspanda wrote:

VeritasPrepKarishma wrote:

You can do it with a negative sign too. You want to find the minimum value of (distance from -6) - (distance from 1)

Make a number line with -6 and 1 on it. (-6)..........................(1)

Think of a point in the center of -6 and 1. Its distance from -6 is equal to distance from 1 and hence (distance from -6) - (distance from 1) = 0 .

What if instead, the point x is at -6? Distance from -6 is 0 and distance from 1 is 7 so (distance from -6) - (distance from 1) = 0 - 7 = -7

If you keep moving to the left, (distance from -6) - (distance from 1) will remain -7 so the minimum value is -7.

In the above example, the absolute value function f(x), which is sum of absolute functions, can not be negative for any value of x. Kindly clarify whether the minimum value of f(x) is 7 or -7.

If f(x) = | 1 - x | + | x - 1 |, then minimum value of f(x) is 0 for x = 1. Kindly comment.

Thanks.

Arun.

Sum of two absolute functions cannot be negative but difference can be. The original post discusses the sum of absolute functions. ficklehead asked about f(x) which is difference between two absolute functions. The '-7' is the minimum value of f(x) in case of difference.

f(x) = |a| - |b| can easily be negative e.g. if a = 2 and b = 5

The linear sum of absolute value function f(x) = |3x + 1| + |2x-3| + |x - 7| is minimum either at x = -1/3 or x = 3/2.

Rewriting the function f(x) as 3 | x + 1/3 | + 2 | x - 3/2 | + | x - 7 |, we have values of x as -1/3, 3/2 and 7 where constituent absolute functions |3x +1 |, | 2x - 3| and |x-7| are minimum respectively.

Let us write these values of x in ascending order: (the values are repeated based on individual weights or coefficients of absolute value functions)

-1/3, -1/3, -1/3, 3/2, 3/2 and 7

The median value of the above list, -1/3 or 3/2 evaluates f(x) as minimum.

Extending this example further, f'(x) = 4 | x + 1/3 | + 2 | x - 3/2 | + | x - 7 | is minimum for x = -1/3. [ -1/3, -1/3, -1/3, -1/3, 3/2, 3/2 and 7; the median is -1/3] Similarly, f"(x) = 3 | x + 1/3 | + 3 | x - 3/2 | + | x - 7 | is minimum for x = 3/2. [ -1/3, -1/3, -1/3, 3/2, 3/2, 3/2 and 7; the median is 3/2]

Thus we may conclude the median value of x evaluates f(x) as minimum.

Kindly comment.

Arun.

Last edited by arunspanda on 08 Jul 2014, 06:18, edited 1 time in total.

In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?

Please correct me if I am wrong.

x is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance. Distance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together. If -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet. If they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11.

To check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11

Responding to a pm:

Quote:

Just so am clear the minimum value of f(x) for the below Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6| f(x) = |2x - 3| + |4x + 7| etc would be 11 and 13/2 respectively? Am kinda confused over which value x will take.Will it be the total shortest distance covered or the point to which they meet

The 11 and 13/2 that you obtained are the minimum values of the respective functions f(x). This is the total minimum distance covered. The point at which they meet is the value of x i.e. For first question, whenever x is in this range: -1 <= x <= 1, f(x) will take the value 11. For second question, when x = -4/7, f(x) will take the minimum value 13/2. _________________

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...