[email protected] 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

Ans-B

|a-b| = the distance between a and b.
|a+b| = |a-(-b)| = the distance between a and -b.

f(x) = |4x - 1| + |x-3| + |x + 1| = the SUM of 3 distances on the number line.
To MINIMIZE the value of f(x), we must MINIMIZE the 3 distances.

|x+1| + |x-3|:
|x-3| = the distance between x and 3:
|x+1| = the distance between x and -1.
To minimize the sum of these 2 distances, x must be AS CLOSE AS POSSIBLE to -1 and 3.
In other words, x must be BETWEEN -1 and 3, inclusive.
-1<--|x+1|-->x<--|x-3|-->3
Here, since the total distance between the two endpoints is 4, the sum of the 2 distances in red is 4.
To illustrate:
If x=-1, then |x+1| + |x-3| = 4.
If x=0, then |x+1| + |x-3| = 4.
If x=3, then |x+1| + |x-3| = 4.
In each case, since x is between -1 and 3, inclusive, the sum of the two distances is 4:
|x+1| + |x-3| = 4.

If x is positioned OUTSIDE -1 and 3, the sum of the two distances will INCREASE.
If x=-2, then |x+1| + |x-3| = 6.
If x=4, then |x+1| + |x-3| = 6.
Thus:
The LEAST possible value of |x+1| + |x-3| is yielded when x is between -1 and 3, inclusive:
|x+1| + |x-3| = 4.

|4x-1|:
Since |4x-1| cannot be negative, the least possible value of this term is 0.
4x-1 = 0
x=1/4.
Since x=1/4 is between -1 and 3, inclusive, it will also minimize the value of |x+1| + |x-3|.

Thus, when x=1/4, the least possible value for f(x) is yielded:
f(x) = |4x-1| + |x-3| + |x+1| = 0 + 11/4 + 5/4 = 4.

The correct answer is B.

[email protected] 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

Ans-B

|4x-1| = 4x-1 for x>=1/4; 1-4x for x<1/4
|x-3| = x-3 for x>=3; 3-x for x<3
|x+1| = x+1 for x>=-1; -x-1 for x<-1

if we arrange in a num line, the above values will appear as ....-1....1/4....3...

for x<-1 : f(x) = 3-5x --- for this the minimum value will be greater than 8.
for -1<=x<1/4 : f(x) = 5-4x --- for this the minimum value will be 9.
for 1/4<=x< 3 : f(x) = 4x+3 --- for this the minimum value will be 4.
for x>=3 : f(x) = 6x-3 --- for this the minimum value will be 15.

Therefore, across the ranges we have, the minimum value of f(x) = 4

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

Let's try to make each term to 0 and compare value

4x - 1 = 0
x = 1/4
Equation = 0 + 2.75 + 1.75 = 4

x - 3 = 0
x = 3
Equation = 11 + 0 + 4 = 15

x + 1 = 0
x = -1
Equation = 5 + 4 + 0 = 9

Answer [spoiler]{B}[/spoiler]

theCodeToGMAT wrote:f(x) = |4x - 1| + |x-3| + |x + 1|

We see that in the first term we have 4x and we know that all the terms will be greater than or equal to 0.

So,
Let x = 0 ==> 1 + 3 + 1 = 4
Let x = 1 ==> 3 + 2 + 2 = 7
If we multiply further then 4x will result in larger value

So, [spoiler]{B}[/spoiler]

I'm not quite sure of your reasoning here.
The portion in red yields a sum of 5, which is not the least possible value of f(x).

GMATGuruNY wrote:

theCodeToGMAT wrote:f(x) = |4x - 1| + |x-3| + |x + 1|

We see that in the first term we have 4x and we know that all the terms will be greater than or equal to 0.

So,
Let x = 0 ==> 1 + 3 + 1 = 4
Let x = 1 ==> 3 + 2 + 2 = 7
If we multiply further then 4x will result in larger value

So, [spoiler]{B}[/spoiler]

I'm not quite sure of your reasoning here.
The portion in red yields a sum of 5, which is not the least possible value of f(x).

Silliest Mistake

potentially there are 3 lowest values of x:
Either:
4x-1 = 0 -> x = 1/4 -> f = 4
x-3 = 0 -> x = 3 -> f = 12 + ... > 4
x+1 = 0 -> x = -1 -> f = 5 + ... > 4

Therefore f minimum = 4

Just to elaborate a bit on the shorter, non-Mitch answers provided here ...

When you're minimizing |4x - 1| + |x - 3| + |x + 1|, the first thing to realize is that each of the terms is going to be either 0 or positive. If they're all positive, that sum is likely going to be greater than if one of them is zero and the other two are positive, so you "set" each summand equal to 0 and see what you find (e.g. plug in x = 1/4, x = 3, and x = -1, respectively).

Once you find that |4x - 1| + |x - 3| + |x + 1| = 4 when x = 1/4, so you can eliminate most of the answers: the only question left is whether 3 is a valid answer too. Here you can work with |x - 1| + |x + 3|, and notice, as Mitch details, that the minimum value of this sum is 4. Since that's the case, 3 is an impossible answer, as the sum of two of our terms is already greater than 3, and the third term we have to add to them cannot be negative.