What is the minimum value of |x - 4| + |x + 3| + |x - 5|?

Question

A. -3
B. 3
C. 5
D. 7
E. 8

Self Made

chetan2u · Accepted Answer

Good Explanation  Engr2012 .

The Q tests us on the understanding of Property of a Modulus..
 CONCEPT: -
when we have two modulus, the value will be the minimum between the two Critical Points and will increase on either side of CP.

the two extremeties are -3 and 5, so |x+3| + |x-5| will remain constant, 3+5=8, within the range from -3 to 5.. 
so we have to keep x-4 as minimum to have the minimum value for |x-4| + |x+3| + |x-5| ..
x-4 will be 0 at x=4..
so our min value will OCCUR at x=4 and will be 8
E

Kurtosis · Answer

A cannot be the answer as all the three terms are in modulus and hence the answer will be non negative.

|x-4| >= 0 --> Minimum occurs at x = 4
|x+3| >= 0 --> Minimum occurs at x = -3
|x-5| >= 0 --> Minimum occurs at x = 5

x = -3 --> Result = 7 + 0 + 8 = 15. Also any negative value will push the combined value of |x-4| + |x-5| to a value > 9.
x = 4 --> Result = 0 + 7 + 1 = 8
x = 5 --> Result = 1 + 8 + 0 = 9
x = 3 --> Result = 1 + 6 + 2 = 9

So minimum value of the expression occurs at x = 4 and the resultant value = 8

Answer: E

ENGRTOMBA2018 · Answer

The ranges that this question needs to be looked at are: 1. x<-3 2. -3 \leq x < 4 3. 4 \leq x < 5 4. x \geq 5 Taking each one of them one by one: 1. x<-3 , the given expression will become, f(x) = |x-4| + |x+3| + |x-5| = -(x-4)-(x+3)-(x-5) = -3x+6 . Slope of the line f(x) = -3x+6 is <0 and thus this line will not have a defined minimum (= -inf !) 2. In the range, -3 \leq x < 4, the equation takes the form, f(x) = 12-x, again negative slope, potential of a minimum value in the given range. The minimum value will be at x=-3 ---> 12+3=15. Not in the options, move on. 3. 4 \leq x < 5 , equation becomes f(x) = x+4, positive slope ---> minimum value will be when x=4 ---> 4+4 =8 . 4. x \geq 5, equation becomes, f(x)= 3x-6, positive slope and as such the minimum will be at x=5 , giving you f(x)=9 as the minimum value. Hence E, 8 is the correct answer. Method 2: |x-a| is the distance of x from a. Thus f(x) = |x-4| + |x+3| + |x-5| = distance of x from 4 + distance of x from -3 + distance of x from 5. The ranges remain the same as the ones mentioned above, but now let us assume some values in the given ranges, (ignore the first one as it should be pretty obvious that this range will not give a minimum value!) Ranges # 2 and 3, -3 \leq x < 4, take x=-2, you get f(x) =13, move to x=0, you get f(x) = 12, the values are decreasing, good. Now take x= 3, you get f(x) = 9, for x=4, you get f(x)=8, when you take x=5, f(x) = 9, starts to increase again. Thus 8 , E is the correct answer. This method though depends on the fact that the options given are integers and the nature of the values (decreasing to increasing) does not vary inconsistently. If the options would have been decimals or the nature of the values would have been varying, then this method will not be a recommended course of action. Hope this helps. FYI, the graph of the given functions look like this: 2016-03-12_23-27-34.jpg

Youraisemeup · Answer

Hello,

We can start here by defining the Critical point limits for potential sign changes at may occur.
1. x<-3:

x-4=-ve, x+3=-ve and x-5=-ve==>AFTER OPENING MODs==> -x+4-x-3-x+5==>-3x+6==>infintely positive.

2. -3=<x<0

x-4=-ve, x+3=+ ve and x-5=-ve==> After mod. are opened==> -x+4+x+3-x+5==>12-x==> min. poss value would be if x=-3===> 15.

3. 0=<x<4:

x-4 -ve, x+3 +ve, and x-5 -ve==>-x+12==>min poss. value would be when x=0==>min. value of function in this limit=12.

4. 4=<x<5:
x-4 +ve, x+3 +ve, x-5 -ve.===>x+4 (+ve) and minimum poss. value as per this limit=8.

5. x>=5:
x-4 +ve, x+3 +ve and x+5 +ve.===>3x+6==> infinitely positive.

Hence minimum poss. value of function overall=8=Option E.

I understand i shouldn't have put 0 in between for extra limit. but personally i'm kind of scared of zero. it sometimes leads to disasters when ignored.

Also, do let me know Engr2012 and chetan2u how much time a person should take solving this. i took 4:56 minutes.

ENGRTOMBA2018 · Answer

3 mods questions are rare in GMAT (didnt see any across my 3 attempts) and as such if they do come in the GMAT, try to remember what  chetan2u  has mentioned above that minimum value will be between the 2 extreme critical points. But your detailed method is correct and should have taken you 2-3 minutes.

In this case, x=0 should not be taken into account as it is NOT a critical point. If you had a mod |x| in the given expression, then yes, x=0 would have become a critical point.

Additionally, for absolute value questions, try to plot a rough graph of what it should look like and it will help you in honing onto the range that you need to worry about. As shown in the graph in my post, the only value that you were supposed to worry about would have been in the range  4 \leq x < 5 . This can reduce the time taken to solve this question to less than 2 minutes.

Hope this helps.

Youraisemeup · Answer

Thanks! will def try graphs

sjavvadi · Answer

I tried using the extreme points and a couple of integers around them to see the trend.

The three points I used are x = -3, 4 and 5

With x = -3 the value is 15
With x = 4 the value is 8
With x = 5 the value is 9

because of the mod (x+3) term any value greater than 5 will lead to a sum greater than 8.

I think the minimum value is 8.

umabharatigudipalli · Answer

I tried the same way!

One question, from where did you get x=3?? (x=-3 or 4 or 5 sounds good though!) Why did you try with x=3? Thanks in advance!

akshanka1 · Answer

The question can be solved by plotting the origin of the three modulus sign on the number line.
Please note |x-4| actually means distance from the point 4. So, 4 is origin.
Similarly, |x+3| denotes distance from -3.

So, in the question we need to find a point whose sum of distances from 4 , -3 and 5 is minimum.

Point -3,4 and 5 on the number line and try to figure out a point whose sum is minimum.

Please note it will lie always in the middle point when odd number of origins are there.
So, the minimum will lie at x=4.

Put x=4 in the equation, you will 8 as the answer

CounterSniper · Answer

value of a mod is min when it is equal to 0

when x = 4 
0 + 7 + 1 = 8

when x= -3
7 + 0 + 8 = 15

when x = 5
1 + 8 + 0 =9

so minimum = 8

brains · Answer

Well i was just wondering whether in only this case , mod of (x+3) and mod of (x-5) will remain constant for any value of x(of course i tried only integer value) , or is it general that between two extremities the sum of mod will be always constant for any value of x in between. Well , ofcourse you said that the value will be minimum between the C.P

chetan2u · Answer

Yes it is for all values, even fraction..
Reason :- you are adding in one and equivalent is getting subtracted from other..

|x+3|+|x-5|....
x as 0...3+5=8
x as 1 |4|+|-4|=8
x as 1/2...|3+1/2|+|1/2-5|=|7/2+9/2=16/2=8

isaperelli · Answer

Hi! 
Could you please explain why you are also considering x=3? 
Thank you!

bumpbot · Answer

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What is the minimum value of |x - 4| + |x + 3| + |x - 5|?

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What is the minimum value of |x - 4| + |x + 3| + |x - 5|?

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