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Re: Question of the Day  II
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05 Jun 2012, 06:22
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 doable. Most people will probably not get any mods question with more than 2 terms.
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Re: Question of the Day  II
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06 Dec 2012, 05:06
I love this question. One must understand that f(x) = 4x1 + x3 + 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 4x1 = 0 x+1 = 1 1/4 x3 = 2 3/4 Answer: 4
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16 Dec 2012, 21:57
Karishma, So from your method, I infer That minimum of a function will always be at either its critical points or zero. Await your valued response.
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17 Dec 2012, 04:30
eaakbari wrote: Karishma,
So from your method, I infer
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 + 2x3 + x  7 For what value(s) of x will f(x) have the minimum value?
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Re: Question of the Day  II
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12 May 2013, 12:09
What if the question asks for max value of f(x) ??? disregarding the answer choices ??



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13 May 2013, 09:52
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.
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Re: Question of the Day  II
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19 May 2013, 17:08
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
Thanks, Rohit



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20 May 2013, 07:07
eaakbari wrote: Karishma,
What about a function with evenly spaced numbers?
for instance f(x) =  4x + 1 +  2x + 1 +  4x + 3 +  x 
What will be the min. value of x for this? The function will take a minimum value for a range of values of x (between the second and the third values). Think 'Why?'
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Re: Question of the Day  II
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20 May 2013, 07:14
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.
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Re: Question of the Day  II
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28 May 2013, 14:04
Hello all.
I am wondering why we cannot take the positive and negative cases of f(x) = 4x  1 + x3 + x + 1 and solve for x that way?
In other words, f(x) = 4x  1 + x3 + x + 1
I. f(x) = (4x1) + (x3) + (x+1)
II. f(x) = (4x1) + (x3) + (x1)
Thanks!



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28 May 2013, 14:56
WholeLottaLove wrote: Hello all.
I am wondering why we cannot take the positive and negative cases of f(x) = 4x  1 + x3 + x + 1 and solve for x that way?
In other words, f(x) = 4x  1 + x3 + x + 1
I. f(x) = (4x1) + (x3) + (x+1)
II. f(x) = (4x1) + (x3) + (x1)
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 \(4x1>0\) and \(x3>0\) and \(x+1>0\) \(x>\frac{1}{4}\) and \(x>3\) and \(x>1\) so 4x1 is positive for x>1/4, x3 is +ve for x>3 and x1 is +ve for x>1 Now you have to split the original function into the areas defined above: \(x<1\) all functions are negative \(f(x) = (4x1) + (x3) + (x1)\) if \(1<x<\frac{1}{4}\) the third term is positive,the others negative \(f(x) = (4x1) + (x3) +(x1)\) 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
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28 May 2013, 15:06
Ok, I get that for f(x) = (4x1) + (x3) + (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!



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28 May 2013, 15:20
WholeLottaLove wrote: Ok, I get that for f(x) = (4x1) + (x3) + (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) = (4x1) + (x3) + (x+1) if 1/4<x<3 for example you see that 4x1 is positive, x+1 is positive BUT x3 is negative => f(x) = (4x1) + ()(x3) + (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 intervalWhat I mean is that f(x) = (4x1) + ()(x3) + (x+1) is valid only in the 1/4<x<3 area. Each area has its own function
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28 May 2013, 15:31
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) (4x1) and (x+1) would be positive but (x3) 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.



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28 May 2013, 15:45
WholeLottaLove wrote: 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) (4x1) and (x+1) would be positive but (x3) 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) = (4x1) + (x3) + (x+1) II. f(x) = (4x1) + (x3) + (x1) 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.
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Re: Question of the Day  II
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03 Jul 2013, 07:31
VeritasPrepKarishma wrote: Q. If f(x) = 4x  1 + x3 + 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.) keep x = 0 and eliminate d, e. now put x = 1, 1/2 we end up with some dis satisfing valye now keep x =1/4 u get ur answer what i did was, i took even 1/8 but pattern reverses and u get to know 1/4 is correct and confirmed logic and basic = magic at gmat



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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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09 Jul 2013, 17:39
If f(x) = 4x  1 + x3 + x + 1, what is the minimum value of f(x)?
x=1/4, x=3, x=1
x<1, 1<x<1/4, 1/4<x<3, x>3
x<1 4x  1 + x3 + x + 1 (4x1) + (x3) + (x+1) 4x+1 + x+3 + x1 6x+3 If x<1 then minimally, 6x+3 would be greater than 9
1<x<1/4 4x  1 + x3 + x + 1 (4x1) + (x3) + (x+1) 4x+1 + x+3 + x + 1 4x+5 If x is between 1 and 1/4...try three values of 1, 0, 1/4 a)4x+5 4(1)+5 = 9 4(0)+5 = 5 4(1/4)+5 = 4
1/4<x<3 4x  1 + x3 + x + 1 (4x1) + (x3) + (x+1) 4x1 + x+3 + x+1 4x+3
For 4x+3, any value 1/4<x<3 will produce a value greater than 4
x>3 4x  1 + x3 + x + 1 (4x1) + (x3) + (x1) 6x5
For 6x5, any value x>3 will produce a value greater than 13
The smallest number possible for this particular function is f(x) = 4
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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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25 Jul 2013, 12:36
hI Gurpreet,
Why did you change the value of x3 to 3x,
For x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x1 + 3x + x+1 = 4x+3 > 3
Kriti



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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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25 Jul 2013, 21:42
Kriti2013 wrote: hI Gurpreet,
Why did you change the value of x3 to 3x,
For x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x1 + 3x + x+1 = 4x+3 > 3
Kriti Because when x is less than 3, (x  3) is negative. Therefore, x3 =  (x  3) = 3  x
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Re: Question of the Day  II
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06 Aug 2013, 21:46
VeritasPrepKarishma wrote: eaakbari wrote: Karishma,
So from your method, I infer
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 + 2x3 + 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!
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