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

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.

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?

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.

Missed the last step. Got it thanks

Posted from my mobile device

See the attachment -> D

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<x<=b
F(x)=-x+a+0+x-c
F(x)=a-c
D:)

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

Hi,

Can anyone explain why |b-a| becomes -b+a?

I also ended up with 2b -a - c

Thanks a lot

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 ?

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

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

GMATinsight would appreciate your contribution as well

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(+)

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

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

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

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