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

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KarishmaB
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Re: Question of the Day - II [#permalink] New post   24 Nov 2013, 20:20
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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+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
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Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink] New post   25 Mar 2014, 14:06
VeritasPrepKarishma wrote:
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.)



Here is the easiest solution
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If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink] New post   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| + |x-3| + |x + 1| (Question of the Day-II) [#permalink] New post   23 Jul 2019, 19:01
The min value of the mod is 0

make individual mods equal to zero.

We can get
x=1/4
x=3
x=-1

Then, substitute these values one by one to see which value of x gives you the least
You will get it for x=1/4
answer:4

VeritasKarishma Any wrong here?
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Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink] New post   23 Jul 2019, 23:25
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ShivamoggaGaganhs wrote:
The min value of the mod is 0

make individual mods equal to zero.

We can get
x=1/4
x=3
x=-1

Then, substitute these values one by one to see which value of x gives you the least
You will get it for x=1/4
answer:4

VeritasKarishma Any wrong here?


With 3 points, and co-efficient of x as 1 for each term, the minimum will be at the middle point on the number line.
With 4 points, the minimum would be in a range.
It's best to understand the process and why it is followed to make adjustments as needed.
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Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink] New post   23 Apr 2020, 05:26
VeritasKarishma wrote:
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?'


Hi VeritasKarishma
Regarding the question given above, using your approach, shouldn't the minimum value be at x = -1/2 ?
Taking critical points as -1/4, -1/2, -3/4 and 0. So, there are four people each at -1/4 and -3/4, and 2 people at -1/2 and one person at 0.
Representing on number line as follows.

(4)........(2)...........(4)........(1)
-3/4......-1/2........-1/4..........0

So, 4 guys at -3/4 and 4 guys at -1/4 come to -1/2. Now there are 10 people at -1/2 and 1 person at 0. So minimum f(x) should be at x = -1/2.
Had they been not equidistant then the value of x could have been in a range for f(x) to be minimum?
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Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink] New post   23 Apr 2020, 22:48
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ultimateashish wrote:
VeritasKarishma wrote:
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?'


Hi VeritasKarishma
Regarding the question given above, using your approach, shouldn't the minimum value be at x = -1/2 ?
Taking critical points as -1/4, -1/2, -3/4 and 0. So, there are four people each at -1/4 and -3/4, and 2 people at -1/2 and one person at 0.
Representing on number line as follows.

(4)........(2)...........(4)........(1)
-3/4......-1/2........-1/4..........0

So, 4 guys at -3/4 and 4 guys at -1/4 come to -1/2. Now there are 10 people at -1/2 and 1 person at 0. So minimum f(x) should be at x = -1/2.
Had they been not equidistant then the value of x could have been in a range for f(x) to be minimum?



Yes, the function will be minimum at x = -1/2 here. The co-efficients are not the same. With a 2 at extreme right, it will become a range again.
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Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink] New post   12 Jul 2023, 11:01
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