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\(f(x)=|x+4|+|x+1|+|x−2|\)

­In this type of question, we generally look for critical values.
These critical values are where the sign of equation inside absolute values will change.
These values will be \(x = -4\), \(x = -1\), and \(x = 2\).

We need to analyze interval to determine where we can find minimum values.
There are 4 cases: \((-∞, -4), (-4, -1), (-1, 2) and (2, ∞).\)

For \(-∞ < x < -4\):
\(| x + 4 | = -(x + 4) = - x - 4\)
\(| x + 1 | = -(x + 1) = - x - 1\)
\(| x − 2 | = -(x - 2) = - x + 2\)

\(f(x) = - x - 4 - x - 1 - x + 2 = - 3x - 3\)

\(- 3x - 3\) will decrease as \(x\) increases, so on \(f(-4) = 9\) would be minimum value for this interval. 

For \(-4 < x < -1\):
\(| x + 4 | = (x + 4) = x + 4\)
\(| x + 1 | = -(x + 1) = - x - 1\)
\(| x − 2 | = -(x - 2) = - x + 2\)

\(f(x) = x + 4 - x - 1 - x + 2 = - x + 5\)

\(-x + 5\) will decreases as \(x\) increases, so on \(f(-1) = 6\) would be minimum value for this interval. 

For \(-1 < x < 2\):
\(| x + 4 | = (x + 4) = x + 4\)
\(| x + 1 | = (x + 1) = x + 1\)
\(| x − 2 | = -(x - 2) = - x + 2\)

\(f(x) = x + 4 + x + 1 - x + 2 = x + 7\)

\(x + 7\) will increases as \(x\) increases, so on \(f(-1) = 6\) would be minimum value for this interval. 

For \(2 < x < ∞\):
\(| x + 4 | = (x + 4) = x + 4\)
\(| x + 1 | = (x + 1) = x + 1\)
\(| x − 2 | = (x - 2) = x - 2\)

\(f(x) = x + 4 + x + 1 + x - 2 = 3x + 3\)

\(3x + 3\) will increases as \(x\) increases, so on \(f(2) = 9\) would be minimum value for this interval.

Comparing all 4, we can say that minimum value would occur when x = -1.
\(f(-1) =  |-1+4|+|-1+1|+|-1−2| = 3 + 3 = 6\)

Answer = D. 6





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Minimum value of f(x) at x=-1 results in f(x)=6.

Also f(x)=7 at x=0
f(x)=8 at x=1
f(x)=7 at x=-2
f(x)=8 at x=-3
Thus it is verified that at x=-1, f(x) is minimum
Thus option D is our answer
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Since value is in mod we need to consider closest value to 4,1, -2

This is to avoid adding up anything more to existing values.  Also, large negative will increase |-x-2| value


Considering -1, we get 3+0+3=6

Consider 0, 4+1+2=7

Ans.D
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­Given,

f(x) = |x+4| + |x+1| + |x−2|;

Assume x= -4;

f(x) = 0 + 3 + 6 = 9; ---(1)

Assume x= -1;

f(x) = 3 + 0 + 3 = 6; ---(2)

Assume x= 2;

f(x) = 6 + 3 + 0 = 9; ---(3)

Therefore, by comparing (1), (2), and (3), f(x) has the minimum value of 6.

Answer - D
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hi , i could not understand why at some cases eg it is less than or grater than(eg. -4<x<-1) V/S less than equal to and greater than equal to (eg. x<=-4)
Bunuel

­
­

GMAT Club Official Explanation:



APPROACH 1:

To find the minimum value of f(x) = |x + 4| + |x + 1| + |x - 2|, we need to analyze the critical points where the absolute values change, which are at x = -4, x = -1, and x = 2. We will evaluate the function around these points.

1. For x ≤ -4:

f(x) = -(x + 4) - (x + 1) - (x - 2) =
= -3x - 3
In this range, the minimum value of f(x) occurs when x = -4: f(-4) = -3(-4) - 3 = 12 - 3 = 9

2. For -4 < x < -1:

f(x) = (x + 4) - (x + 1) - (x - 2) =
= -x + 5
In this range, the minimum value of f(x) occurs as x approaches -1 from the left: f(x) approaches -(-1) + 5 = 1 + 5 = 6

3. For -1 ≤ x < 2:

f(x) = (x + 4) + (x + 1) - (x - 2) =
= x + 7
In this range, the minimum value of f(x) occurs when x = -1: f(-1) = -1 + 7 = 6

4. For x ≥ 2:

f(x) = (x + 4) + (x + 1) + (x - 2) =
= 3x + 3
In this range, the minimum value of f(x) occurs when x = 2: f(2) = 3(2) + 3 = 6 + 3 = 9

Comparing these values, the minimum value of f(x) is 6, which occurs at x = -1.

Answer: D.­

APPROACH 2:

|x + 4| + |x + 1| + |x - 2| represents the sum of the distances of x from -4, -1, and 2.

--------(-4)------------(-1)------------(2)--------
We can notice that the further we go from -4 to the left, or from 2 to the right, the total distance increases. Hence, the minimum distance should occur when x is somewhere between -4 and 2. In that range, when we move from -1 in either direction, the distance from two of the points increases while the distance from the third one decreases, resulting in a net increase in the total distance. For example, at x = 0, the distances are 4, 1, and 2, totaling 7 units. At x = 1, the distances are 5, 2, and 1, totaling 8 units. Thus, exactly at x = -1, we'd get the smallest sum of the distances: 3, 0, and 3, which total 6 units.

--------(-4)------------(-1)------------(2)--------


Therefore, the minimum value of the function f(x) occurs at x = -1, and the minimum value is 6.

Answer: D.­
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ektalakra
hi , i could not understand why at some cases eg it is less than or grater than(eg. -4<x<-1) V/S less than equal to and greater than equal to (eg. x<=-4)

When determining the intervals around the critical points, it's important to include each critical point in one of the intervals. It doesn't matter which interval includes the equal sign for a critical point, as long as all critical points are accounted for. The key is to ensure every critical point is included without any gaps or overlaps.

Thus, as long as you include the equal sign at each critical point, it doesn't matter which interval it’s assigned to
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