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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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(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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That's right! I can count on you guys to always give the correct answer and some innovative ways of solving...
Let me give you the method I would use to solve it too...
Where mods are concerned, the fact that mod signifies the distance from 0 on the number line is my best friend...
So |x-3| is the distance from 3, |x + 1| is the distance from -1 and |4x - 1| is 4 times the distance from 1/4
So |4x - 1| + |x-3| + |x + 1| is the sum of distances from 3, -1 and four times the distance from 1/4 e.g. if x takes the value at point A, f(x) will be sum of length of red line, green line and blue line. the question here is, what is the minimum such sum possible?
Ques.jpg [ 4.56 KiB | Viewed 34474 times ]
Can I say that minimum total distance will be covered from point 1/4?
Ques1.jpg [ 3.43 KiB | Viewed 34370 times ]
The logic being that the distance from 3 to -1, which is 4 units has to be covered. Why to cover any distance from 1/4 at all?
Check this video discussing basics of how to use absolute values as distances on the number line: https://youtu.be/oqVfKQBcnrs
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Let me give you the method I would use to solve it too...
Where mods are concerned, the fact that mod signifies the distance from 0 on the number line is my best friend...
So |x-3| is the distance from 3, |x + 1| is the distance from -1 and |4x - 1| is 4 times the distance from 1/4
So |4x - 1| + |x-3| + |x + 1| is the sum of distances from 3, -1 and four times the distance from 1/4 e.g. if x takes the value at point A, f(x) will be sum of length of red line, green line and blue line. the question here is, what is the minimum such sum possible?
Attachment:
Ques.jpg [ 4.56 KiB | Viewed 34474 times ]
Can I say that minimum total distance will be covered from point 1/4?
Attachment:
Ques1.jpg [ 3.43 KiB | Viewed 34370 times ]
The logic being that the distance from 3 to -1, which is 4 units has to be covered. Why to cover any distance from 1/4 at all?
Check this video discussing basics of how to use absolute values as distances on the number line: https://youtu.be/oqVfKQBcnrs
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Let g(x) = |4x - 1| & h(x)=|x-3| + |x + 1| ... So f(x)=g(x)+h(x)
Now h(x) is equal to 4 between -1 and 3 and is higher everywhere else, so it minimizes between -1 & 3
g(x) is minimum at x=(1/4) where it is equal to 0.
Since -1<(1/4)<3 & f(x)=g(x)+h(x) ... It is straight forward to imply that f(x) will be minimum at 1/4, since both g(x) & h(x) are minimum at that point
f(1/4)=g(1/4)+h(1/4)=0+4
Hence answer is 4
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Now h(x) is equal to 4 between -1 and 3 and is higher everywhere else, so it minimizes between -1 & 3
g(x) is minimum at x=(1/4) where it is equal to 0.
Since -1<(1/4)<3 & f(x)=g(x)+h(x) ... It is straight forward to imply that f(x) will be minimum at 1/4, since both g(x) & h(x) are minimum at that point
f(1/4)=g(1/4)+h(1/4)=0+4
Hence answer is 4
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General Discussion
gurpreetsingh
Joined: 12 Oct 2009
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VeritasPrepKarishma
put x=0 we get f(x) = 5..rule out D and E
put x=1/4 we get f(x) = 0 + 11/4 + 5/4 = 16/4 = 4.. rule out C D and E
Now the answer is either 3 or 4.
Reason for above checking of values:
for every value of x> 3 the f(x) is quite big because of 4x-1
for every value of x < -1 the f(x) if bigger than 3 and 4.
for x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x-1 + 3-x + x+1 = 4x+3 > 3 => only 4 is the probable answer.
Thus we only need to check x>=-1 and x<= 1/4
f(x) in this domain is = 1-4x + 3-x + x+1 = 5-4x => for f(x) to be minimum the x should be +ve
=> for x = 1/4 , f(x) = 4.
This can be solved using graph as well by plotting the f(x) in different domains.
In other words, as my mentor says, assume there is one guy on point 3, one on point -1 and 4 guys on point 1/4. If they have to meet up, but cover minimum distance, they should meet at point 1/4. What happens when there are 4 such points? Lets add another term (2x - 3) to f(x)...
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would the answer still remain 1/4? based on your 'distance' method?
The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance
The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance
gmat1011
Yes, that is right! The answer still remains 1/4.
Since |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4.
Attachment:
Ques.jpg [ 5.2 KiB | Viewed 33426 times ]
Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6|
f(x) = |2x - 3| + |4x + 7| etc
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Posts: 9
VeritasPrepKarishma
Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?
In example : |x - 1| + |x-3| + |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.
ficklehead
Forget these numbers. Think logically.
My house is 10 miles away from your house. If we have to meet up, how much distance do we need to cover together? In any case, we need to cover 10 miles together at least, right? Either you come down to my place (you cover 10 miles) or I come down to yours (I cover 10 miles) or we meet mid way (10 miles covered together) or we meet up at a nice coffee place 2 miles further down from my house in the opposite direction in which case we will need to cover more than 10 miles (i.e. we cover 2 + 12 = 14 miles)
Now say, another friend is at my place. In which case will people cover minimum distance together? If two of us come down to your place, we cover 10+10 = 20 miles together but if you come down to our place, you cover only 10 miles. If instead, we meet midway, we cover 5+5 and you cover 5 miles so in all 15 miles. So less number of people should travel the entire distance.
If there are 4 people at point A and 2 at point B, minimum distance will be covered if people at point B travel to point A. So people at 3/2 should come down to 1/4.
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ficklehead
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| + |x-3| + |x + 1| + |x + 6| = 11
put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
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Posts: 9
VeritasPrepKarishma
Thanks Karishma for this detailed explanation.
I got it now.
I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.
To seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?
ficklehead
Look at the diagram below. Make a number line in such questions.
Attachment:
Ques3.jpg [ 5.1 KiB | Viewed 31938 times ]
For x and y to meet, they have to cover a distance of 9 together. For p and q to meet, they have to cover a distance of 2 together. They can meet anywhere between -1 and 1 and they will cover a total distance of 11 only. So x can take any value -1 < x < 1 and the value of the expression will be 11.
|2x-3|+|4x+7| = 2|x-3/2| + 4|x+7/4|
Attachment:
Ques4.jpg [ 5.35 KiB | Viewed 31933 times ]
There are 4 people at -7/4 and 2 people at 3/2. Distance between the two points is 7/4 + 3/2 = 13/4
For these people to meet covering the minimum distance, the 2 people X and Y should travel to point -7/4. (Make minimum people travel). So minimum distance that needs to be covered = 2*13/4 = 13/2 (because 2 people travel 13/4 each) which is the minimum value of the expression. The value of x when the expression takes minimum value is -7/4.
Check by putting x = -7/4. You get |2x-3|+|4x+7| = 13/2
Also see that when you put x = 0 or 3/2 etc, the value of the expression is higher.
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Posts: 9
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| ?
Ex: minimum value of : |x+6|-|x-1| ?
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?
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?
kuttingchai
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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ficklehead
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.
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devinawilliam83
Joined: 25 Aug 2011
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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
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
devinawilliam83
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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koro12
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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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?
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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