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

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
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f(x) = f(x - 1)

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

|x + 3| = |x + 1|

The above equation satisfies only if x = -2

Ans B

Posted from my mobile device

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.
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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

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.

Is there a quicker way than testing each of the values?

|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

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)

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

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

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

chetan2u Bunuel

Please help solving this graphically?

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.

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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.

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


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