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Re: Question of the Day  II [#permalink]
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20 Mar 2012, 22:20
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 + 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.)



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26 May 2012, 12:00
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.) for any value of X , term 4x1 must give maximum value, so anything that gives lowest of 4x1 will give lowest for f(x), so x=1/4 and F(x) is 4, B



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04 Jun 2012, 19:48
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 + 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.) 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 + x3 + 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.
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04 Jun 2012, 20:06
koro12 wrote: 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.) for any value of X , term 4x1 must give maximum value, so anything that gives lowest of 4x1 will give lowest for f(x), so x=1/4 and F(x) is 4, B Beware of using this logic in other similar questions e.g. f(x) = 2x  1 + x3 + x  1 + x  5 or f(x) = 3x + 1 + 2x3 + 2x  7
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Re: Question of the Day  II [#permalink]
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04 Jun 2012, 23:09
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?



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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 [#permalink]
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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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16 Dec 2012, 22:06
Karishma, For my better understanding of the subject, lets find the minimum of the function below. f(x) = x  1  x  5 x = 0 :: f(x) = 4 x = 1 :: f(x) = 4 x = 5 :: f(x) = 4 Is this correct?
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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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17 Dec 2012, 08:25
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?  1/3  0  3/2 7  Method that I prev used to solve. x = 1/3 => f(x) = 11 x = 3/2 => f(x) = 0 x = 7 => f(x) = 33 So x = 3/2 is point of where f(x) = 0 Am I right. Your method which seems ingenious Since 3/2 is the middle value in between 1/3 & 7. The distance will be the least at 3/2... Have I inferred correctly?
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17 Dec 2012, 08:29
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?
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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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19 May 2013, 17:08
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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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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 [#permalink]
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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
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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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