The function f is defined for all numbers x by f(x) = |x + 3| + |x + 2

Question

The function f is defined for all numbers x by f(x) = |x + 3| + |x + 2|. For which value of x, does f(x) = f(x - 1) ?

A. -3
B. -2
C. -1
D. 1
E. 2

M37-69

Bunuel · Accepted Answer

Official Solution:

The function  f  is defined for all numbers  x  by  f(x) = |x + 3| + |x + 2| . For which value of  x , does  f(x) = f(x - 1)  ?

A.  -3 
B.  -2 
C.  -1 
D.  1 
E.  2

f(x) = |x + 3| + |x + 2| . 
 
 f(x-1) = |(x-1) + 3| + |(x-1) + 2|= |x + 2| + |x + 1| . 
 
We should find such  x  that  |x + 3| + |x + 2|=|x + 2| + |x + 1| :
 
Cancel  |x + 2|  to get  |x + 3| = |x + 1| .
 
The above mean that, on the number line,  x  is equidistant from -3 and -1, so  x=\frac{-3 + (-1)}{2}=-2 .

Answer: B

AyushCrackVerbal · Answer

Given

The function f(x) = |x + 3| + |x + 2| ......  Equation A

To Find

For which value of x, does f(x) = f(x - 1) ?

Let us replace "x" by "x-1" in the function f(x) to get the function f(x-1):

f(x-1) = | (x-1) + 3 | + | (x-1) + 2 |

f(x-1) = |x+2| + |x+1| ......  Equation B

To find the value of x:

Equation A = Equation B

|x + 3| + |x + 2| = |x+2| + |x+1|

|x+3| = |x+1| ......  Equation C

Now, let us start substituting the values from the options to check if it fits the Equation c

1) When x = -3

|-3+3| = |-3+1|

|0| = |-2|

0 = 2 (Absurd)

So x cannot be -3

2) When x = -2

|-2+3| = |-2+1|

|1| = |-1|

1 = 1 ( Makes sense )

Since LHS = RHS , x=-2 is a sloution of the  Equation C .

We do not need to check for the rest of the options as we have already got our answer, x=-2

So the correct answer is  B .

Archit3110 · Answer

equate for values f(x) = |x + 3| + |x + 2|

f(x) = f(x - 1)

test with answer values we see at -2 we get f(-2)=f(-3)
IMO B

LeoN88 · Answer

I started with 2 in a jiffy, putting values I realized that if we put x -ve then we can cancel out terms to get f(x) = f(x - 1) Substituting x= -2 we get it. It all boils down to how fat we can get to the answer. I'm sharing a snapshot of my notepad. IMG_20190522_100059.jpg

chandan1988 · Answer

|something|=|something|
√(Something)^2= √(something)^2

After putting the values 
|X+3|=|X+1| 
(X+3)^2=(X+1)^2

So , X =-2

Posted from my mobile device

exc4libur · Answer

f(x)=f(x-1):
f(x)=|x+3|+|x+2|
f(x-1)=|x-1+3|+|x-1+2|=|x+2|+|x+1|
|x+3|+|x+2|=|x+2|+|x+1|…|x+3|-|x+1|=|x+2|-|x+2|…|x+3|-|x+1|=|x+2|(1-1)…|x+3|=|x+1|
|x+3|≥0 (positive) x≥-3 and |x+3|<0 (negative) x<-3
|x+1|≥0 (positive) x≥-1 and |x+1|<0 (negative) x<-1
range x≥-1: |x+3|=|x+1|…x-x=1-3…0=-2 invalid;
range -3≤x<-1: |x+3|=|x+1|…x+3=-x-1…2x=-1-3…x=-2 valid (inside range);
range x<-3: |x+3|=|x+1|…-x-3=-x-1…0=2 invalid;

Ans (B)

Asariol · Answer

Can you do this quicker with any other method other than plugging in?

Sabby · Answer

Asariol

Hi,

Yes you can do it faster, in few seconds, and without plugging in numbers :
Ix+3I=Ix+1I

now drop the brackets, you can see that it can't be x+3=x+1, therefore, one side of the equation has to be negative

-x-3=x+1
-2x=4
x=-2

Answer B

Danraf · Answer

MUCH quicker way:

|x+3| + |x+2| find the zeros of each module: x= -3 and -2

substitute x-1 into x:

|x-1+3| + |x-1+2| = |x+2| + |x+1|

|x+2| + |x+1| find the zeros of each module: x = -2 or 1

x = -2 is common, plug it in and it works

wishmasterdj · Answer

chetan2u   Bunuel

Please help solving this graphically?

chetan2u · Answer

If  f(x) = |x + 3| + |x + 2| , then  f(x-1)= |x-1 + 3| + |x-1 + 2|=|x+2|+|x+1|

Given that   f(x)=f(x-1)   =>  |x + 3| + |x + 2|=|x + 1| + |x + 2| 
 |x + 3|=|x + 1|

We can square both sides as both sides are non-negative
 |x + 3|^2=|x + 1|^2 
 x^2+6x+9=x^2+2x+1 
 4x=-8 
 x=-2

B

wishmasterdj 
It would be better to follow approach as above rather than graphical.

If you wish to do it graphically, plot the graphs as following
1) Take y=|x+3|+|x+2|....brown line
2) Take y=|x-1+3|+|x-1+2|=|x+2|+|x+1|....blue line
They intersect at x=-2, where y=1 in each case.

JITESHSINGH · Answer

we can always solve such question using conventional methods but by looking at the question and going through the options i could easily figure out that we can use option to arrive at the solution in less time.
To begin with x must be a negative number so option D and E are ruled out now trying with option A,B & C i easily found that -2 is the correct answer.

Hope this helps for those looking for tricks to solve questions in less time.

BrushMyQuant · Answer

Given that f(x) = |x + 3| + |x + 2| and we need to find the value of x for which f(x) = f(x - 1)

To find f(x - 1) we need to compare what is inside the bracket in f(x - 1) and f(x)

=> We need to substitute x with x-1 in f(x) = |x + 3| + |x + 2| to get the value of f(x - 1)

=> f(x-1) = |x-1 + 3| + |x-1 + 2| = |x + 2| + |x + 1| = f(x) (given)

=> |x + 2| + |x + 1| = |x + 3| + |x + 2|

Cancelling |x +2| from both the sides we get

|x +1| = |x + 3|

Squaring both the sides we get ( Watch this video  to learn about the  Basics of Absolute Values )

(x +1)^2  =  (x +3)^2  
=>  x^2 + 2*x*1 + 1^2  =  x^2 + 2*x*3 + 3^2 
=> 2x + 1 = 6x + 9
=> 6x - 2x = 1-9 = -8
=> 4x = -8
=> x =  \frac{-8}{4}  = -2

So,  Answer will be B 
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

kaveree · Answer

hey everything is understanble, but why did we square both sides?

Bunuel · Answer

Squaring is a common technique to eliminate absolute value signs because (|x|)^2 = x^2. For example, if we have |x| = |y|, squaring both sides gives x^2 = y^2, removing the modulus signs to simplify.

10. Absolute Value 
 
    Theory    
 
 Math: Absolute value (Modulus) 
 Absolute Value: Tips and hints 
 Properties of Absolute Values on the GMAT 
 Absolute modulus : A better understanding 
 GMAT Question Patterns: Abolute Value 
 The Reason Behind Absolute Value Questions on the GMAT 
 
    Questions    
 
 Inequality and absolute value questions from my collection 
 The E-GMAT Question Series on ABSOLUTE VALUE 
 DS Questions 
 PS Questions

For more check  Ultimate GMAT Quantitative Megathread

Hope it helps.

ankushsambare · Answer

Manual Notice:
Nice question for concept application, and give kudos to this thread, it shows my efforts are useful. 
I’ve bumped it to the top so more people can benefit. 
Great solution by an expert in the comments.

The function f is defined for all numbers x by f(x) = |x + 3| + |x + 2

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