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.

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

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