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If y = |3x - 5|/(-x^2 - 3) for what value of x will the valu
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23 Jul 2020, 07:29
23 Jul 2020, 07:29
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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
A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3
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Given Kudos: 136
Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the valu
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23 Jul 2020, 07:55
23 Jul 2020, 07:55
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Carcass wrote:
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
A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3
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 denominator, |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
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the valu
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09 Dec 2023, 02:58
09 Dec 2023, 02:58
\(y = \frac{|3x - 5|}{-x^2 - 3}\)
= \(\frac{|3x - 5|}{ -(x^2 + 3)}\)
= - \(\frac{|3x - 5|}{x^2 + 3}\)
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
= \(\frac{|3x - 5|}{ -(x^2 + 3)}\)
= - \(\frac{|3x - 5|}{x^2 + 3}\)
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
