If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be

Question

If  y = \frac{|3x - 5|}{-x^2 - 3}  for what value of x will the value of y be greatest?

A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3

PS29580.02

BrentGMATPrepNow · Accepted Answer

Key idea:  x^2  is always greater than or equal to 0
This means  -x^2  is always less than or equal to 0
And this means  -x^2 - 3  is always NEGATIVE  
In other words, the denominator of  y  is most definitely NEGATIVE

From here, there are two possible cases: 
 case i : If the numerator of  y  is NEGATIVE, then y is POSITIVE
 case ii : If the numerator of  y  is POSITIVE, then y is NEGATIVE

Since we're trying to MAXIMIZE the value of y, we need the denominator to be negative (which would make y POSITIVE)
However, the numerator, |3x - 5|, will always be greater than or equal to 0. 
Since we can't make the numerator NEGATIVE, the only way to maximize the value of y is make the numerator ZERO 
In other words:  |3x - 5|=0 
Which means  3x -5=0  
So  3x =5 
Ergo  x = \frac{5}{3}

Answer: E

Cheers, 
Brent

nick1816 · Answer

|3x-5| is always non-negative and -(x^2+3) is alway negative

\frac{|3x - 5|}{(-x^2 - 3)} ≤ 0

Hence maximum value of  \frac{|3x - 5|}{-x^2 - 3}  is 0.

|3x-5| = 0

x= \frac{5}{3}

GMATinsight · Answer

We have  y = \frac{|3x - 5|}{-x^2 - 3}

The denominator,  -x^2 - 3  MUST be Negative

The Numerator,  |3x - 5|  can  NOT  be Negative

i.e.  y = \frac{|3x - 5|}{-x^2 - 3}  MUST be Non-Positive

i.e. Maximum Value of a non-Positive function can be ZERO when Numerator is Zero

i.e. 3x-5 = 0
i.e. y is maximum at  x = 5/3  and  y_{max}=0

Answer: Option E

itspCv2 · Answer

Confused of signs ? just substitute X= options

arrive at Solution which gives greatest takes 1 to 2 mins. But helps if u are confused

E.5/3 gives greatest value when substituted

PhantomAY · Answer

Segment the numerator and denominator to look for extreme combination: |3x-5|>=0 and -x^2-3<=-3 , and hence y<=0. Thus the greatest value of y is 0 --> 3x-5 = 0 --> x = 5/3

unraveled · Answer

y = \frac{+ numerator }{ - denominator} 
Hence y will always be negative except when the numerator is  zero which is the largest value y can take .

\implies  |3x-5| = 0
 \implies x = \frac{5}{3}

Answer E.

dave13 · Answer

Hi BrentGMATPrepNow Brent, the explanation is nice

i just have a few questions: When i saw

|3x - 5| i thought i should consider two case 3x-5 or -3x+5 so I kinda confused... 3x=5 or -3x=-5 x = 5/3 or x = 5/3 (negative/negative =positive) both options give the same results

thats strange to me Why ?

General question: if the absolute value by definition is distance from 0 and the distance is always positive or zero. why do we consider two cases positive and negative

p.s. i think you meant -x^2 - 3 in the highlighted part

BrentGMATPrepNow · Answer

You have taken the equation |3x - 5| = 0 and concluded that EITHER 3x - 5 = 0 OR -(3x - 5) = 0
Noticed that both of these equations are the same. 
If we take -(3x - 5) = 0 and multiply both sides by -1, we got the equivalent equation 3x - 5 = -0, which is identical to the equation 3x - 5 = 0

Thanks for the heads-up about  -x^2 - 3

dave13 · Answer

BrentGMATPrepNow this time i picked A

because i thought since Absolute value is always positive i can pick -5 so and in denominator since it it is -- -x^2 after plugging in -5, the sign will change into positive - *-5^2 = 25 So y = \frac{|3 (-5) - 5|}{- -5^2 - 3} = 10/22 greater than 0 why its wrong with it?

BrentGMATPrepNow · Answer

Think of -x² as -(x²)
So, -x² - 3 = -(x²) - 3
So, when we plug in x = -5, we get: -(x²) - 3 = -((-5)²) - 3 = -(25) - 3 = -28

Does that help?

avigutman · Answer

Video solution from Quant Reasoning starts at 22:23 Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=P7HIS8v1k4g

100mitra · Answer

IMO - E
I did it by subsituting the given value of X in option and found the largest value for Y is obtained by option E (5/3)
also realized, is denominator is negative and numerator is positive, 
to obain highest value Y when the numerator is zero

rvgmat12 · Answer

Denominator is always negative -x^2 - 3

Max value can be 0

x = 5/3

E

Sud2021 · Answer

Hi Brent,

Thank you for the great explanation. I understood the concept based on your walk-through.

Although, I did catch one issue with your one statement and please correct me if I am wrong.

Shouldn't the statement above read "Numerator" instead of "Denominator"?

Regards,
Sud

GMATking94 · Answer

Hi Brent,

I have a question regarding the modulus of the numerator.....since this is an absolute value function, the sign of the numerator will depend on the range of X. In this case, if X<5/3 the function should be written as -(3X-5), which makes the numerator negative for all range of values of X<5/3. So there can be values greater than 0 now.

BrushMyQuant · Answer

y = \frac{|3x - 5|}{-x^2 - 3}

=  \frac{|3x - 5|}{ -(x^2 + 3)} 
= -  \frac{|3x - 5|}{x^2 + 3}

Now, both |3x - 5| and  x^2 + 3  are non-negative values and we have a negative sign outside
=> Value can be maximum only when the numerator is either zero or is equal to denominator so that we get a small negative value overall

|3x - 5|=0
=> x =  \frac{5}{3}

So, y = 0 when  \frac{5}{3}

So,  Answer will be E 
Hope it helps!

Watch the following video to learn the Basics of Absolute Values

https://www.youtube.com/watch?v=huFiBk8PaGY

onlymalapink · Answer

why is denomunatir always negative

Bunuel · Answer

x^2 is the square of a number, hence it cannot be negative, thus -x^2 - 2 = -(positive or 0) - positive = (negative or 0) - positive = negative.

shubhim20 · Answer

@why can't 3/5 be right?

If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be

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If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be

Prep Toolkit

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