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Re: M19-14 [#permalink]
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Tanyanagra wrote:
I have reached the substitution but unable to get the process after that...


Hi Tanyanagra,

For me using substitution is better than trying to manipulate the variables, especially under pressure....here's how I solve it !

Substitute x=2 into this 1st equation: f(x) = x divided by x+1 f(2) = 2 divided by 2+1 = 2/3

Then, the question asks what is f(1/x) in terms of f(x)?

You'll need to convert the f(x) to f(1/x) that will be 1/x divided by 1/x+1

Substitute x=2 into this 2nd equation f(1/x): 1/2 divided by 1/2+1 = 1/2 divided by 3/2 = 1/2 * 2/3 = 1/3

Now, you have to find 1/3in the answer choices:

A) f(x) = 2/3 not a match

B) - f(x) = - 2/3 not a match

C) 1/f(x) = 1 divided by 2/3 = 3/2 not a match

D) 1 - f(x) = 1 - 2/3 = 1/3 BINGO!! That's a match

E) none of the above = wrong

I hope it helps!

Thanks Ale! :-)
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Re M19-14 [#permalink]
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Official Solution:

If \(f(x) = \frac{x}{x + 1}\), what is \(f(\frac{1}{x})\) in terms of \(f(x)\)?

A. \(f(x)\)
B. \(-f(x)\)
C. \(\frac{1}{f(x)}\)
D. \(1 - f(x)\)
E. none of the above


\(f(\frac{1}{x})= \frac{\frac{1}{x} }{\frac{1}{x} + 1} =\)

\(= \frac{\frac{1}{x} }{(\frac{1+x}{x})} =\)

\(=\frac{1}{x} *\frac{x}{x+1} =\)

\(=\frac{1}{1 + x} =\)

\(=\frac{(1 + x) - x}{1 + x} =\)

\(=\frac{1 + x}{1 + x}-\frac{x}{1+x}=\)

\(=1 - \frac{x}{1 + x} =\)

\(=1 - f(x)\)


Answer: D
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Re: M19-14 [#permalink]
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Hi,

How did you get from 1/(1+x) to (1+x-x)/(1+x), and then from this to the final step?

Thanks
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Re: M19-14 [#permalink]
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I have reached the substitution but unable to get the process after that...
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Re: M19-14 [#permalink]
Expert Reply
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M19-14 [#permalink]
Bunuel wrote:
Official Solution:

If \(f(x) = \frac{x}{x + 1}\), what is \(f(\frac{1}{x})\) in terms of \(f(x)\)?

A. \(f(x)\)
B. \(-f(x)\)
C. \(\frac{1}{f(x)}\)
D. \(1 - f(x)\)
E. none of the above


\(f(\frac{1}{x})= \frac{\frac{1}{x} }{\frac{1}{x} + 1} =\)

\(= \frac{\frac{1}{x} }{(\frac{1+x}{x})} =\)

\(=\frac{1}{x} *\frac{x}{x+1} =\)

\(=\frac{1}{1 + x} =\)

\(=\frac{(1 + x) - x}{1 + x} =\)

\(=\frac{1 + x}{1 + x}-\frac{x}{1+x}=\)

\(=1 - \frac{x}{1 + x} =\)

\(=1 - f(x)\)


Answer: D


------------------

Hi Bunuel,
It's not given to us that 'x' is not equal to 0 so how can we cancel 'x' when solving for f(1/x)??
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Re: M19-14 [#permalink]
Expert Reply
Rishikavats wrote:
Bunuel wrote:
Official Solution:

If \(f(x) = \frac{x}{x + 1}\), what is \(f(\frac{1}{x})\) in terms of \(f(x)\)?

A. \(f(x)\)
B. \(-f(x)\)
C. \(\frac{1}{f(x)}\)
D. \(1 - f(x)\)
E. none of the above


\(f(\frac{1}{x})= \frac{\frac{1}{x} }{\frac{1}{x} + 1} =\)

\(= \frac{\frac{1}{x} }{(\frac{1+x}{x})} =\)

\(=\frac{1}{x} *\frac{x}{x+1} =\)

\(=\frac{1}{1 + x} =\)

\(=\frac{(1 + x) - x}{1 + x} =\)

\(=\frac{1 + x}{1 + x}-\frac{x}{1+x}=\)

\(=1 - \frac{x}{1 + x} =\)

\(=1 - f(x)\)


Answer: D


------------------

Hi Bunuel,
It's not given to us that 'x' is not equal to 0 so how can we cancel 'x' when solving for f(1/x)??


If x were 0, 1/x would be undefined and f(1/x) wouldn't make any sense.
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Re: M19-14 [#permalink]
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