The function f is defined for all nonzero x by the equation f(x) = 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 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

No hahaha

If there is an example plugging number, it could clarify a little more.

Thanks anyway

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 = rule-based; good at step-by-step recipes and procedures; operates with logic and analysis; proceeds step-by-step

Right brain thinking = pattern-based; good at seeing complex connections and larger patterns; operates with analogy and association; proceeds by non-linear leaps

You can read more on this blog:
How to do GMAT Math Faster

You see, left-brain thinkers love to algebra, and they look for any opportunity to use algebra in a step-by-step 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 left-brain thinker. I will say the solution method you described was an extreme left-brain lengthy solution for a problem that can be completed quite quickly with a right-brain 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
_________________

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

Why did you put (.5) for x: (1/x) ?

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

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?

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!

THIS IS NOT HARD. Here's how you solve this any similar questions

We are given the function \(f(x) = x - \frac{1}{x}\)
We are then asked what does the function spit out when you input a certain value - in this case \(\frac{1}{x}\)

We can do just that
\(f(\frac{1}{x}) = \frac{1}{(1/x)}\)

Simplifies to \(\frac{1}{x}- x\)
then to \(\frac{1}{x} -\frac{x^2}{x}\)
then finally to \(\frac{(1-x^2)}{x}\)

Now, what happens if you simplify the original function f(x)?
\(f(x) = x - \frac{1}{x}\)
\(= \frac{(x^2 -1)}{x}\)

Don't you see that if we substitute \(-1\) out of \(f(1/x)\) i.e. \(\frac{(1-x^2)}{x}\) we get f(x) or \(= \frac{(x^2 -1)}{x}\)?

The trick with these questions is to see that the answer is some sort of different input that would produce a similar result

Look for the answer choice that would be the most likely contender. This is of course f(-x)

\(f(-x) = -x \frac{-1}{-x}\)
\(f(-x) = \frac{-x^2}{x} + \frac{1}{x}\)
\(f(-x) = \frac{(1-x^2)}{x}\)

Therefore\(f(-x) = f(1/x)\)

Thanks for the splendid explanation, Rich!

Just wanna make sure - we can use this technique because it's stated in the question that the function f is defined for all nonzero x, right? What if later we stumble onto a similar question, but the function f is not stated to be defined for all nonzero x? Thanks!

Hi tagheueraquaracer,

TESTing VALUES is a remarkably useful Tactic on the GMAT - and you can use it in a variety of ways in the Quant section (and sometimes in IR and even in some rare CR prompts). When a Quant question includes Number Property 'vocabulary' (such as odd, even, prime, positive, negative, etc.), then that's typically a clue that you can TEST VALUES.

GMAT assassins aren't born, they're made,
Rich

Dear Bunuel ,

Please tag this question --- EP6


Best !!
_________________

_________________
Done. Thank you.
_________________

Given: 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 \(f(\frac{1}{x})\)?

\(f(\frac{1}{x})= 1/x - x = -(x-1/x) = f(-x)\)

IMO B

Posted from my mobile device