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Re: Find the maximum value of f(x) = 18-|3+x|, x belongs to R [#permalink]

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14 May 2013, 20:47

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This post was BOOKMARKED

Vamshiiitk wrote:

Find the maximum value of f(x) = 18-|3+x|, x belongs to R a) 12 b)18 c)20 d)15

Answer: b) 18

Reason: We have to find the max. value of f(x). Max value of 18-|3+x| will be 18 because the modulus will make |3+x|>=0. Lower the value of |3+x|, higher the value of 18-|3+x|=f(x). The lowest value of a modulus expression is 0, which implies f(x)=18-0=18.

Re: Find the maximum value of f(x) = 18-|3+x|, x belongs to R [#permalink]

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30 May 2013, 14:21

With numerous problems involving absolute value, we flip the signs inside the absolute value function if we know it to be a negative #. For example, |3-x|...if we know that X>3 then we: -(3-x) = -3+x. I'm guessing we don't do that here because in f(x) x cannot be less than 0?

psychout wrote:

Vamshiiitk wrote:

Find the maximum value of f(x) = 18-|3+x|, x belongs to R a) 12 b)18 c)20 d)15

Answer: b) 18

Reason: We have to find the max. value of f(x). Max value of 18-|3+x| will be 18 because the modulus will make |3+x|>=0. Lower the value of |3+x|, higher the value of 18-|3+x|=f(x). The lowest value of a modulus expression is 0, which implies f(x)=18-0=18.

Re: Find the maximum value of f(x) = 18-|3+x|, x belongs to R [#permalink]

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30 May 2013, 15:17

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This post received KUDOS

WholeLottaLove wrote:

With numerous problems involving absolute value, we flip the signs inside the absolute value function if we know it to be a negative #. For example, |3-x|...if we know that X>3 then we: -(3-x) = -3+x. I'm guessing we don't do that here because in f(x) x cannot be less than 0?

x can be negative, negative numbers belong to R.

We can read the question as:

Find the maximum value of \(f(x) = 18-|3+x|\), x belongs to R Find the maximum value of \(f(x) = 18-(num\geq{0})\)

Remeber that an abs value is a number positive or equal to zero (never negative).

So what in which case the operation \(18-(num\geq{0})\) has the max-value? when the \(num\geq{0}\) is 0. Clearly \(18-0>18-1\) for example, so the max value is \(18\).

For the sake of clarity: This case corespond to x=-3, |3-3|=0. For any other value of x the quantity |3-x| will result in a positive value that you will subtract to 18, obtaining a lesser value (of course).
_________________

It is beyond a doubt that all our knowledge that begins with experience.

With numerous problems involving absolute value, we flip the signs inside the absolute value function if we know it to be a negative #. For example, |3-x|...if we know that X>3 then we: -(3-x) = -3+x. I'm guessing we don't do that here because in f(x) x cannot be less than 0?

psychout wrote:

Vamshiiitk wrote:

Find the maximum value of f(x) = 18-|3+x|, x belongs to R a) 12 b)18 c)20 d)15

Answer: b) 18

Reason: We have to find the max. value of f(x). Max value of 18-|3+x| will be 18 because the modulus will make |3+x|>=0. Lower the value of |3+x|, higher the value of 18-|3+x|=f(x). The lowest value of a modulus expression is 0, which implies f(x)=18-0=18.

Actually f(x) can be less than 0. For example, if x=20, then f(20)=18-|3+20|=-5 or if x=-25, then f(-25)=18-|3-25|=-4.

Now, the question asks about the maximum value of f(x)=18-|3+x| (f(x) is equal to 18 minus some non-negative value). To maximize f(x) we need to minimize |3+x|. The minimum value of |3+x| is 0, thus the maximum value of f(x)=18-|3+x|=18-0=0.

Hope it helps.

P.S. Notice that f(x) reaches its minimum for x=-3 --> f(-3)=18-|3-3|=18-0=0.
_________________

Re: Find the maximum value of f(x) = 18-|3+x|, x belongs to R [#permalink]

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25 Mar 2015, 00:51

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Find the maximum value of f(x) = 18-|3+x|, x belongs to R [#permalink]

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23 Feb 2017, 13:49

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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