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# Question of the Day - II

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Re: Question of the Day - II [#permalink]  20 Mar 2012, 21:07
kuttingchai wrote:
I got C = 5

f(x) = |4x-1| + |x-3| + |x+1|

=(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1)
so, x= 1/2

then i used the value of x=1/2

f(x) = |4x-1| + |x-3| + |x+1|

f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1|
= |1| + |-5/2| + |3/2|
= 1 + 5/2 + 3/2
= 5

Is that correct?

Put x = 1/4 and the minimum value you will get is 4.
How did you get x = 1/2?
I would suggest you to check out one of the approaches mentioned above.
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Re: Question of the Day - II [#permalink]  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| + |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.)
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Re: Question of the Day - II [#permalink]  26 May 2012, 12:00
VeritasPrepKarishma wrote:
Q. 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.)

for any value of X , term |4x-1| must give maximum value, so anything that gives lowest of |4x-1| will give lowest for f(x), so x=1/4 and F(x) is 4, B
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Re: Question of the Day - II [#permalink]  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| + |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.)

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| + |x-3| + |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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Re: Question of the Day - II [#permalink]  04 Jun 2012, 20:06
koro12 wrote:
VeritasPrepKarishma wrote:
Q. 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.)

for any value of X , term |4x-1| must give maximum value, so anything that gives lowest of |4x-1| 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| + |x-3| + |x - 1| + |x - 5|
or
f(x) = |3x + 1| + |2x-3| + |2x - 7|
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Re: Question of the Day - II [#permalink]  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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Re: Question of the Day - II [#permalink]  06 Dec 2012, 05:06
I love this question. One must understand that f(x) = |4x-1| + |x-3| + |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
|4x-1| = 0
|x+1| = 1 1/4
|x-3| = 2 3/4

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Re: Question of the Day - II [#permalink]  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.

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Re: Question of the Day - II [#permalink]  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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Re: Question of the Day - II [#permalink]  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.

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?
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Re: Question of the Day - II [#permalink]  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.

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?

----------- -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.

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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Re: Question of the Day - II [#permalink]  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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Re: Question of the Day - II [#permalink]  12 May 2013, 12:09
What if the question asks for max value of f(x) ??? disregarding the answer choices ??
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Re: Question of the Day - II [#permalink]  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 [#permalink]  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!

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
Re: Question of the Day - II   [#permalink] 19 May 2013, 17:08
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