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Re: Question of the Day  II
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06 Aug 2013, 22:52
vaishnogmat wrote: 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! f(x) = 3x + 1 + 2x3 + x  7 f(x) = 3x + 1/3 + 2x3/2 + x  7 1/3  3/27 (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.
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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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10 Sep 2013, 21:12
In the interval (1,1/4) f(x)=(4x1)(x3)+(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?



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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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10 Sep 2013, 21:45
alphabeta1234 wrote: In the interval (1,1/4) f(x)=(4x1)(x3)+(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? f(x) = (4x1)(x3)+(x+1)=4x+ 5 (calculation mistake above) f(0) = 5
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Re: Question of the Day  II
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24 Nov 2013, 07:16
VeritasPrepKarishma wrote: ficklehead wrote: I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : x+6x1 ? 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.



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Re: Question of the Day  II
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24 Nov 2013, 20:20
arunspanda wrote: VeritasPrepKarishma wrote: ficklehead wrote: I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : x+6x1 ? 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
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Re: Question of the Day  II
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01 Dec 2013, 23:49
VeritasPrepKarishma wrote: ficklehead wrote: In example : x  1 + x3 + 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 + x3 + x + 1 + x + 6 = 11 put x = 1, you get x  1 + x3 + x + 1 + x + 6 = 11 put x = 0, you get x  1 + x3 + 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 + x3 + 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.
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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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25 Mar 2014, 14:06
VeritasPrepKarishma wrote: 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.) Here is the easiest solution
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If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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14 Feb 2016, 06:41
Is it safe to say that for this type of questions one just needs to test checkpoints (1; 1/4; 3) to find out the min value? Just because by utilizing one of those points we will get rid of one "trip" and hence bascially the total length of trips would be shorter (when one person stays at home as Karishma exemplified).
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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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16 Feb 2016, 23:20
shasadou wrote: Is it safe to say that for this type of questions one just needs to test checkpoints (1; 1/4; 3) to find out the min value? Just because by utilizing one of those points we will get rid of one "trip" and hence bascially the total length of trips would be shorter (when one person stays at home as Karishma exemplified). Yes, the game changer will be at the transition point.If one or more people stay at home, the length of the trip shortens. These two posts will help you solidify the concept: http://www.veritasprep.com/blog/2011/01 ... edoredid/http://www.veritasprep.com/blog/2011/01 ... spartii/
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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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12 Oct 2016, 01:20
Here's my way of looking at it, let me know if you see loop holes 
Since we need to minimize the value of f(x), we need to minimize the factor that contributes most to the value of f(x). In this case, it would be 4x. Therefore, making this term 0 should give us the smallest value of f(x). Substituting 1/4 for x gives f(x) to be equal to 4. Option B.



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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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04 Dec 2016, 19:05
Don't know if this method is already discussed in the forum in the above posts:
F(x) has three different expressions inside mods. None of them can be negative if the mods are removed. Hence, the three mods can take the value of 'zero' to be closest towards the minimum value. This gives us three different values of x = 1/4,3,1. Putting, three values in the expression, once at a time, yields the minimum value of f(x) when x = 1/4. Hence, F(x) = 4



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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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04 Dec 2016, 22:52
VeritasPrepKarishma wrote: 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
Apply two inequilities below (1) For all \(x,y \in R\) we have \(x+y \geq x+y\). Sign "=" occurs \(\iff xy \geq 0\) (2) For all \(x \in R\) we have \(x \geq 0\) Now, let's apply these inequilities into \(f(x)\) \(f(x)=4x  1 + x3 + x + 1=4x1+(3x+x+1) \geq 4x1+3x+x+1=4x1+4 \geq 0+4=4\). \(min f(x)=4 \iff \Bigg\{\begin{split} 4x1=0 \\ (x+1)(3x) \geq 0 \end{split} \iff \Bigg\{\begin{split} x=\frac{1}{4} \\ 1 \leq x \leq 3 \end{split} \implies x=\frac{1}{4}\)
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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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Re: If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII)
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