If a b c 0, then what is the value of the minimum value of the

Question

If a > b > c > 0, then what is the value of the minimum value of the function f(x) = |x - a| + |x - b| + |x - c| ?

A. 2a - b - c
B. a + b - 2c
C. a + b + c
D. a - c
E. a + c

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ScottTargetTestPrep · Accepted Answer

Solution:

If x = a, then f(a) = 0 + (a - b) + (a - c) = (a - b) + (a - c).

If x = b, then f(b) = -(b - a) + 0 + (b - c) = a - c.

If x = c, then f(c) = -(c - a) +   + 0 = (a - c) + (b - c).

We see that choice A is f(a), choice B is f(c) and choice D is f(b). Furthermore, we see that both f(a) and f(c) are greater than f(b) because they both have a positive term besides a - c. Since both choices C and E are also greater than a - c, we see that f(b) = a - c must be the minimum value of f(x).

Answer: D

chetan2u · Answer

Always remember that when you have x between the two extreme points, the value of the modulus will be the difference between these two extreme points. So |x-a|+|x-c| will have minimum value as a-c whenever cb>c>0 . Now minimum value of |x - a| + |x - b| + |x - c| will be a-c+|x-b|, (as |x-a|+|x-c| will have minimum value as a-c whenever c

NitishJain · Answer

Given: f(x) = |x - a| + |x - b| + |x - c| & a > b > c > 0

f(x) will be minimum when x=b

so f(x)min = |b-a| + |b-b| + |b-c| = 2b-a-c but that is not a option.

I also tried assuming a=3, b=2, c= 1 and gets min value when x= b or x = 2.

Can somebody guide.

PrankurD · Answer

ankurjuit 
if you observe here """ so f(x)min = |b-a| + |b-b| + |b-c| = 2b-a-c but that is not a option. """"
|b-a| = a-b as a>b
so your expression will be a-b+b-c which is a-c option d.
Does it clear your doubt?

GMATWhizTeam · Answer

Solution 
 • a > b > c > 0
 o It means a, b, and c are distinct positive numbers. 
• Now, f(x) = |x – a| + |x – b| + |x – c|
 o Since, three absolute values are added, so f(x) ≥ 0
  f(x) = 0 ⟹ |x- a| = |x – b| = |x – c|= 0 ⟹ x = a = b= c, which is not possible.
 So f(x) cannot be 0, but we will try to make at least one term 0.   
• Since, b will come in between a and c on the number line. (a – b) or (b-c ) < (a – c)
 o Therefore, f(x) will be minimum at x = b
o f(x) = | b-a| + |b – b| + |b-c| = - (b- a) + 0 + b – c = a- c  
Thus, the correct answer is  Option D.

sambitspm · Answer

See the attachment -> D

satya2029 · Answer

On a number line from left to right critical points are c....b.....a ..(-).......c.........(+)........ b.........(-)...........a.....(+)........ Since F(x) is positive, for values of F(x) we will look into (+) interval Greater than "a" or Between c and b For X>=a F(X)=0+x-b+x-c =2x-b-c=2a-b-c=(a-b)+(a-c)=(a-c)+ positive value For c

prathyushaR · Answer

a > b > c > 0, minimum value of f(x) = |x - a| + |x - b| + |x - c| ?
Visualize this on the number line
|x - a| + |x - b| + |x - c| this is sum of distances of a number x from a , b and c, if x is greater than a or less than c the sum of distances will keep increasing.
and sum of distances from b will be the least, as it lies between the other two numbers,

we can check by substitution:
distance from a would be : 0+ |a-b| + | a-c| = a-b+a-c=2a-b-c
distance from b : |b-a|+0+|b-c| = a-b+b-c = a-c
distnace from c: |c-a|+|c-b|+0 = a-c+b-c = a+b - 2c
and from these we know a-b >0 (since a>b) so 2a-b-c = a-c+ (a-b) > a-c
similarly with b-c>0 (since b>c) so a+b-2c = a-c + (b-c) > a-c
thus we see the minimum value is  a-c  (when x=b )
D

zoechrys · Answer

Hi, Can anyone explain why |b-a| becomes -b+a? I also ended up with 2b -a - c Thanks a lot

dave13 · Answer

hey ScottTargetTestPrep

Would you please take your time to RE-explain your solution

i don`t get why you put minus in from of parentheses .... so here is my question: why ?

dave13 · Answer

hey satya2029 can you please explain what method are you using ?

by ..(-).......c.........(+)........ b.........(-)...........a.....(+)........ is it wavy curve ? or what i am kinda confused thanks

dave13 · Answer

hey sambitspm man, looking at your fantastic painting how did you arrive at the answer ?

is wayvy curve or pyramid method ...

i kinda cant wrap my mind around your solution, would appreciate your answer

p.s. Just imagine in 100 years this painting will cost 5 million EUR

satya2029 · Answer

Sure.
Consider each modulus element as a separate function
Since at X=a , lX-al=0, a is a critical points, so are b and c
plot them on a number line.
From extreme right here "a" when x>=a, assign a positive sign and then change the sign alternatively
as here a>b>c
......(-)........c.....(+).......b.......(-)...........a(+)

GMATinsight · Answer

dave13

We know that f(x) = |x - a| + |x - b| + |x - c|

i.e .f(x) = Distance of x from a + Distance of x from b + Distance of x from c

for minimizing this expression, x must be equal to b because b is between a and c and x will be closest to both a, b and c

For x = b

f(x) = |x - a| + |x - b| + |x - c| = -x+a + 0 + x-c = a-c

Answer: Option D

Kinshook · Answer

Asked: If a > b > c > 0, then what is the value of the minimum value of the function f(x) = |x - a| + |x - b| + |x - c| ?

f(x) = |x - a| + |x - b| + |x - c|

Case 1: a>b>c>={x,0}
f(x) = |x - a| + |x - b| + |x - c| = (a-x) + (b-x) + (c-x) = (a+b+c-3x)
f(0) = a + b + c
f(c) = a + b - 2c

Case 2: a> b>= x > c > 0
f(x) = |x - a| + |x - b| + |x - c| = (a-x) + (b-x) + (x-c) = (a+b-x-c)
f(b) = a - c

Case 3: a>=x>b>c>0
f(x) = |x - a| + |x - b| + |x - c| = (a-x) + (x-b) + (x-c) = (a-b+x-c)
f(a) = 2a - b - c

a + b - 2c > a - c ; Since a + b - 2c = a + (b-c) - c > a -c ; Since b-c>0
2a - b - c = a + (a-b) - c > a - c ; Since (a-b)>0

a-c is the minimum value of the function f(x) = |x - a| + |x - b| + |x - c|

IMO D

dave13 · Answer

many thanks GMATinsight

appreciate your taking time to answer my question

just one more question why did you put minus sign in front of x ?

(see highlighted part) are checking one case only ? i mean to break down by cases openening modulus and etc

GMATinsight · Answer

dave13 Because a > b > c and x = b i.e. a > x then lx-al = a-x e.g. @a=5, x=2 lx-al = l2-5l = l-3l = 3 which is same as a-x or written as -x+a

ScottTargetTestPrep · Answer

Response: If x = b, then f(b) = |b - a| + |b - b| + |b - c|. Obviously, |b - b| = 0. For |b - a|, we are told that a > b > c > 0, thus b - a < 0. For an expression less than zero, absolute value is found by multiplying the expression by -1, hence |b - a| = -(b - a). That’s why I put minus in front of the parentheses for the first highlighted expression. Also, b - c > 0 and thus, |b - c| = b - c. That’s why I did not put any minus signs in front of the parentheses in front of the corresponding expression. For the second and third highlights, using the same arguments, c - a < 0 and thus, |c - a| = -(c - a). Similarly, c - b < 0 and thus, |c - b| = -(c - b).

Jr2000wharton · Answer

f(x) will be minimum when x=b is the perfect intuition to have. When you get here, it becomes easier in my opinion!

If x=b, then the first term, which is the distance btw a and x, becomes: a-b
If x=b, then the second term, which is the distance btw b and x, becomes: 0
If x=b, then the third term, which is the distance btw c and x, becomes: b-c

(a-b)+(b-c) = a-c

If a > b > c > 0, then what is the value of the minimum value of the

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