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The function f is defined for all nonzero x by the equation f(x) = x 
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Updated on: 21 Nov 2016, 21:12
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The function f is defined for all nonzero x by the equation \(f(x) = x  \frac{1}{x}\). If \(x\neq{0}\), which of the following equals \(f(\frac{1}{x})\)? A. \(f(x)\) B. \(f(x)\) C. \(f(\frac{1}{x})\) D. \(\frac{1}{f(x)}\) E. \(\frac{1}{f(x)}\)
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Originally posted by felippemed on 21 Nov 2016, 14:28.
Last edited by Bunuel on 21 Nov 2016, 21:12, edited 1 time in total.
Edited the question.




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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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21 Nov 2016, 16:54
felippemed wrote: the function \(f\) is defined for all nonzero x by the equation \(f(x) = x  \frac{1}{x}\). If \(x\neq{0}\), which of the following equals \(f(\frac{1}{x})\)?
A.\(f(x)\) B. \(f(x)\) C. \(f(\frac{1}{x})\) D. \(\frac{1}{f(x)}\) E. \(\frac{1}{f(x)}\) Dear felippemed, This is a great question. I'm happy to respond. \(f(x) = x  \frac{1}{x}\) When we plug 1/x into the argument of the function, the first time, x, becomes simply 1/x. The second term, 1/x, becomes 1/(1/x) = x. Thus, \(f(\frac{1}{x}) = \frac{1}{x}  x = (x  \frac{1}{x}) = f(x)\) Notice that f(x) would be a valid choice, but that's not an answer. As it happens, when we simply put a negative sign, x, into the argument, each term becomes negative, so in this instance, f(x) = f(x). (In Precalculus, this would tell us that the function's graph has odd symmetry, but this is beyond what you need to know for the GMAT). Answer = (B) Does all this make sense? Mike
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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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21 Nov 2016, 17:17
Quote: Does all this make sense? Mike No hahaha If there is an example plugging number, it could clarify a little more. Thanks anyway



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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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17 Apr 2017, 19:55
mikemcgarry wrote: felippemed wrote: the function \(f\) is defined for all nonzero x by the equation \(f(x) = x  \frac{1}{x}\). If \(x\neq{0}\), which of the following equals \(f(\frac{1}{x})\)?
A.\(f(x)\) B. \(f(x)\) C. \(f(\frac{1}{x})\) D. \(\frac{1}{f(x)}\) E. \(\frac{1}{f(x)}\) Dear felippemed, This is a great question. I'm happy to respond. \(f(x) = x  \frac{1}{x}\) When we plug 1/x into the argument of the function, the first time, x, becomes simply 1/x. The second term, 1/x, becomes 1/(1/x) = x. Thus, \(f(\frac{1}{x}) = \frac{1}{x}  x = (x  \frac{1}{x}) = f(x)\) Notice that f(x) would be a valid choice, but that's not an answer. As it happens, when we simply put a negative sign, x, into the argument, each term becomes negative, so in this instance, f(x) = f(x). (In Precalculus, this would tell us that the function's graph has odd symmetry, but this is beyond what you need to know for the GMAT). Answer = (B) Does all this make sense? Mike Hi Mike, I determined f(1/x) as (1/x)x and then worked through the functions in answer choices, landing at B. Do you see any pitfalls to this approach?



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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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23 May 2017, 16:43
Cez005 wrote: Hi Mike, I determined f(1/x) as (1/x)x and then worked through the functions in answer choices, landing at B. Do you see any pitfalls to this approach? Dear Cez005, I'm happy to respond. You ask if there were any pitfalls. No and yes. I am sure your algebra was superb. The drawback is on a larger scale. To discuss this, I will introduce the following distinction. Left brain thinking = rulebased; good at stepbystep recipes and procedures; operates with logic and analysis; proceeds stepbystep Right brain thinking = patternbased; good at seeing complex connections and larger patterns; operates with analogy and association; proceeds by nonlinear leaps You can read more on this blog: How to do GMAT Math FasterYou see, leftbrain thinkers love to algebra, and they look for any opportunity to use algebra in a stepbystep solution. Even if all the algebra is flawless, the problem with this approach is that it often takes too long. In fact, the GMAT Quant, on higher level questions, loves to create question that are complete traps for someone who opts for the straightforward algebraic solution. This GMAT Prep problem is along these lines. When I looked at the problem, I solved it in my head in under 10 seconds by observing the patterns. I tried to make this clear in my explanation. I don't know you, so I don't know whether you are predominately a leftbrain thinker. I will say the solution method you described was an extreme leftbrain lengthy solution for a problem that can be completed quite quickly with a rightbrain approach. This is the paradox of growth. You are not really preparing for the GMAT if you keep doing the things that already come naturally to you. You improve by stretching yourself to get at least a little better in the areas that are completely unfamiliar. Does all this make sense? Mike
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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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03 Jun 2017, 14:50
Bump
Reading the method of plugging in makes a lot of sense and I understand that. I am having an issue going the algebraic route.
Can anyone explain why D is incorrect algebraically? (I sense this is a brain fart on my end but I can't figure it out)



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The function f is defined for all nonzero x by the equation f(x) = x 
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05 Jun 2017, 11:38
gmathopeful19 wrote: Bump
Reading the method of plugging in makes a lot of sense and I understand that. I am having an issue going the algebraic route.
Can anyone explain why D is incorrect algebraically? (I sense this is a brain fart on my end but I can't figure it out) Dear gmathopeful19, I'm happy to respond. My friend, I believe you have fallen into what is known as a robust mistake. In pedagogical research, a robust mistake is one to which students return even after they have had a moment of fully understanding why it is wrong. Here is this particular mistake pattern: \(\dfrac{1}{A + B} \neq \dfrac{1}{A} + \dfrac{1}{B}\)Others include \((A \pm B)^2 \neq A^2 \pm B^2\)\(\sqrt{A \pm B} \neq \sqrt{A} \pm \sqrt{B}\)All of these are incorrect overgeneralizations of the Distributive Law. The Distributive Law says that multiplication & division distribute over addition & subtraction. \(C \times (A \pm B) = C \times A \pm C \times B\)
\(\dfrac{A \pm B}{C} = \dfrac{A}{C} \pm \dfrac{A}{C}\)Multiplication & division distribute, but basically, nothing else distributes. All of these mistake are incorrect extensions of the pattern of the Distributive Law to things that don't distribute over addition & subtraction. Be very careful: even when you fully understand why the above mistakes are in fact mistakes, when you are tired or under stress, you mind almost on automatic pilot will think the mistake pattern is correct. It takes quite a bit of effort to root out a robust mistake. Does all this make sense? Mike
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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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05 Jun 2017, 12:16
mikemcgarry wrote: gmathopeful19 wrote: Bump
Reading the method of plugging in makes a lot of sense and I understand that. I am having an issue going the algebraic route.
Can anyone explain why D is incorrect algebraically? (I sense this is a brain fart on my end but I can't figure it out) Dear gmathopeful19, I'm happy to respond. My friend, I believe you have fallen into what is known as a robust mistake. In pedagogical research, a robust mistake is one to which students return even after they have had a moment of fully understanding why it is wrong. Here is this particular mistake pattern: \(\dfrac{1}{A + B} \neq \dfrac{1}{A} + \dfrac{1}{B}\)Others include \((A \pm B)^2 \neq A^2 \pm B^2\)\(\sqrt{A \pm B} \neq \sqrt{A} \pm \sqrt{B}\)All of these are incorrect overgeneralizations of the Distributive Law. The Distributive Law says that multiplication & division distribute over addition & subtraction. \(C \times (A \pm B) = C \times A \pm C \times B\)
\(\dfrac{A \pm B}{C} = \dfrac{A}{C} \pm \dfrac{A}{C}\)Multiplication & division distribute, but basically, nothing else distributes. All of these mistake are incorrect extensions of the pattern of the Distributive Law to things that don't distribute over addition & subtraction. Be very careful: even when you fully understand why the above mistakes are in fact mistakes, when you are tired or under stress, you mind almost on automatic pilot will think the mistake pattern is correct. It takes quite a bit of effort to root out a robust mistake. Does all this make sense? Mike Yes I definitely know those rules ha. Long day I guess. Appreciate the help Posted from my mobile device



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The function f is defined for all nonzero x by the equation f(x) = x 
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Updated on: 13 Feb 2018, 18:28
Hi All, This question can be solved by TESTing VALUES. We're told that f(X) = X  (1/X) and that X does not equal 0. We're asked for the value of f(1/X). IF.... X = 2 f(1/2) = 1/2  (1/(.5) = 1/2  2 = 3/2 So we're looking for an answer that equals 3/2 when we plug X=2 into the answers... Answer A: f(2) = 2  1/2 = 3/2 NOT a match Answer B: f(2) = 2  (1/2) = 3/2 This IS a match Answer C: f(1/2) = .5  (1/.5) = +3/2 NOT a match We can use the calculation from Answer A to deal with Answers D and E.... Answer D: 1/f(2) = 1/(3/2) = 2/3 NOT a match Answer E: 1/f(2) = 1/(3/2) = 2/3 NOT a match Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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05 Feb 2018, 11:24
EMPOWERgmatRichC wrote: Hi All, This question can be solved by TESTing VALUES. We're told that f(X) = X  (1/X) and that X does not equal 0. We're asked for the value of f(1/X). IF.... X = 2 f(1/2) = 1/2  (1/(.5) = 1/2  2 = 3/2 So we're looking for an answer that equals 3/2 when we plug X=2 into the answers... Answer A: f(2) = 2  1/2 = 3/2 NOT a match Answer B: f(2) = 2  (1/2) = 3/2 This IS a match Answer C: f(1/2) = .5  (1/.5) = 5/2 NOT a match We can use the calculation from Answer A to deal with Answers D and E.... Answer D: 1/f(2) = 1/(3/2) = 2/3 NOT a match Answer E: 1/f(2) = 1/(3/2) = 2/3 NOT a match Final Answer: GMAT assassins aren't born, they're made, Rich Why did you put (.5) for x: (1/x) ?



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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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05 Feb 2018, 11:52
Hi cman2010, This question asks us for the value of f(1/X). If we TEST X=2, then we have to plug 2 wherever an "X" appears. To start, that's in the value of 1/X (which would make that value 1/2), which we then plug into the given function. GMAT assassins aren't born, they're made, Rich
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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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12 Feb 2018, 21:21
EMPOWERgmatRichC wrote: Hi All, This question can be solved by TESTing VALUES. We're told that f(X) = X  (1/X) and that X does not equal 0. We're asked for the value of f(1/X). IF.... X = 2 f(1/2) = 1/2  (1/(.5) = 1/2  2 = 3/2 So we're looking for an answer that equals 3/2 when we plug X=2 into the answers... Answer A: f(2) = 2  1/2 = 3/2 NOT a match Answer B: f(2) = 2  (1/2) = 3/2 This IS a match Answer C: f(1/2) = .5  (1/.5) = 5/2 NOT a match We can use the calculation from Answer A to deal with Answers D and E.... Answer D: 1/f(2) = 1/(3/2) = 2/3 NOT a match Answer E: 1/f(2) = 1/(3/2) = 2/3 NOT a match Final Answer: GMAT assassins aren't born, they're made, Rich I got 3/2 for both B and C... Answer C: f(1/2) = 1/2  (1/1/2) = 3/2 What did I do wrong?



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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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13 Feb 2018, 18:28
Hi OCDianaOC, It looks like we both made errors in our calculations (I've edited my explanation accordingly). With Answer C, we're dealing with.... 0.5  (1/0.5) = 1/2  (2) = 1/2 + 2 = 1/2 + 4/2 = +3/2 That's NOT a match for what we're looking for (we're looking for an answer that equals 3/2). GMAT assassins aren't born, they're made, Rich
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Re: The function f is defined for all nonzero x by the equation f(x) = x 
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13 Feb 2018, 19:50
EMPOWERgmatRichC wrote: Hi OCDianaOC,
It looks like we both made errors in our calculations (I've edited my explanation accordingly).
With Answer C, we're dealing with.... 0.5  (1/0.5) = 1/2  (2) = 1/2 + 2 = 1/2 + 4/2 = +3/2
That's NOT a match for what we're looking for (we're looking for an answer that equals 3/2).
GMAT assassins aren't born, they're made, Rich Oh, you're right! Okay, I get it now Thanks for showing this method!




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