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# If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)

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Manager
Joined: 27 May 2010
Posts: 200
If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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17 Jun 2019, 22:45
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85% (hard)

Question Stats:

49% (02:20) correct 51% (03:03) wrong based on 74 sessions

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If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x) such that f2(J(f1(x))) = f3(x)

A) f1(x)
B) f3(x)/x
C) f3(x)
D) f2(x)
E) f2(x)/x
Senior Manager
Joined: 09 Jun 2014
Posts: 343
Location: India
Concentration: General Management, Operations
Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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20 Jun 2019, 08:22
prashanths wrote:
If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x) such that f2(J(f1(x))) = f3(x)

A) f1(x)
B) f3(x)/x
C) f3(x)
D) f2(x)
E) f2(x)/x

I guess it involves F inverse.
Manager
Joined: 21 Feb 2019
Posts: 125
Location: Italy
Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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20 Jun 2019, 08:49
3
$$f_2 (J(f_1)) = f_3$$

$$1 - J(\frac{1}{x}) = \frac{1}{1-x}$$

$$J(\frac{1}{x}) = - \frac{x}{1-x}$$

To get $$J(x)$$, you need to reverse $$x$$ to the power of negative 1. So you derive $$J(x) = \frac{1}{1- x}$$. Correct answer is C.
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MEMENTO AUDERE SEMPER
Senior Manager
Joined: 09 Jun 2014
Posts: 343
Location: India
Concentration: General Management, Operations
Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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20 Jun 2019, 09:09
lucajava wrote:
$$f_2 (J(f_1)) = f_3$$

$$1 - J(\frac{1}{x}) = \frac{1}{1-x}$$

$$J(\frac{1}{x}) = - \frac{x}{1-x}$$

To get $$J(x)$$, you need to reverse $$x$$ to the power of negative 1. So you derive $$J(x) = \frac{1}{1- x}$$. Correct answer is C.

Thanks Lucajava. I was missing this highligted text.
Director
Joined: 19 Oct 2018
Posts: 790
Location: India
If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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20 Jun 2019, 11:23
2
$$f_2(J(f_1(x)))= f_3(x)$$

$$1-J(\frac{1}{x})$$=$$\frac{1}{(1-x)}$$

$$J(\frac{1}{x})$$=$$\frac{-x}{(1-x)}$$

$$J(\frac{1}{x})$$= (1/x)/[1-(1/x)]

$$J(x)=\frac{x}{(1-x)}$$

prabsahi wrote:
prashanths wrote:
If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x) such that f2(J(f1(x))) = f3(x)

A) f1(x)
B) f3(x)/x
C) f3(x)
D) f2(x)
E) f2(x)/x

I guess it involves F inverse.
Manager
Joined: 11 Feb 2018
Posts: 74
Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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22 Jun 2019, 08:09
lucajava wrote:
$$f_2 (J(f_1)) = f_3$$

$$1 - J(\frac{1}{x}) = \frac{1}{1-x}$$

$$J(\frac{1}{x}) = - \frac{x}{1-x}$$

To get $$J(x)$$, you need to reverse $$x$$ to the power of negative 1. So you derive $$J(x) = \frac{1}{1- x}$$. Correct answer is C.

Hi could you please elaborate on the solution. Unfortunately I am not following. I got the right answer by substitution functions one at a time but took 4+ minutes.
Manager
Joined: 21 Feb 2019
Posts: 125
Location: Italy
Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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23 Jun 2019, 08:25
1
aliakberza wrote:
lucajava wrote:
$$f_2 (J(f_1)) = f_3$$

$$1 - J(\frac{1}{x}) = \frac{1}{1-x}$$

$$J(\frac{1}{x}) = - \frac{x}{1-x}$$

To get $$J(x)$$, you need to reverse $$x$$ to the power of negative 1. So you derive $$J(x) = \frac{1}{1- x}$$. Correct answer is C.

Hi, could you please elaborate on the solution. Unfortunately, I am not following it. I got the right answer by substitution functions one at a time but took 4+ minutes.

So, we are demanded to find out what is $$J(x)$$ such that $$f_2 (J(f_1(x))) = f_3(x)$$.

We know that $$f_2(x) = 1 - x$$. Thus, $$f_2(J(f_1(x)) = 1 - J(f_1(x)$$.

Now, knowing that $$f_1(x) = \frac{1}{x}$$, we can rewrite it as $$f_2(J(f_1(x)) = 1 - J(\frac{1}{x})$$.

Turn back to the original equation, and substitute what we've found into it: $$1 - J(\frac{1}{x}) = \frac{1}{1-x}$$.

After some computations, we derive $$J(\frac{1}{x}) = - \frac{x}{1-x}$$.

To get $$J(x)$$, you need to raise $$x$$ to the -1 power. In fact $$\frac{1}{x}^{-1} = x$$. Elaborating on this, what you find out is:

$$- \frac{x^{-1}}{1 - x^{-1}} = \frac{1}{x} * \frac{x}{x - 1} = \frac{1}{1 - x}$$.

I hope it's clear!
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Manager
Joined: 27 May 2010
Posts: 200
Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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23 Jun 2019, 23:34
2
If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x) such that f2(J(f1(x))) = f3(x)

f2(J(f1(x))) = f3(x)

Since f2(x) = 1 - x and f3(x) = 1/(1-x)

f2(J(f1(x))) = 1 - J(f1(x)) = 1/(1-x)

Since f1(x) = 1/x,

J(1/x) = 1 - 1/(1-x) = -x/(1-x)

Substitute 1/x as x,

J(x) = - (1/x)/(1 - 1/x) = -(1/x)/((x-1)/x)

= 1/(1-x)

which equals f3(x).

Option C
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Manager
Joined: 03 Mar 2017
Posts: 112
Location: India
GMAT 1: 650 Q45 V34
Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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24 Jun 2019, 03:08
when you simplify LHS you get
f2(j(1/x))=1-j(1/x)
1-j(1/x)=1/1-x
1-1/1-x=j(1/x)
j(1/x)=-1/(1/x)-1=1/1-(1/x)=f3
Manager
Joined: 07 May 2018
Posts: 66
Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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30 Jun 2019, 10:55
lucajava wrote:
$$f_2 (J(f_1)) = f_3$$

$$1 - J(\frac{1}{x}) = \frac{1}{1-x}$$

$$J(\frac{1}{x}) = - \frac{x}{1-x}$$

To get $$J(x)$$, you need to reverse $$x$$ to the power of negative 1. So you derive $$J(x) = \frac{1}{1- x}$$. Correct answer is C.

How do you reverse x
Senior Manager
Joined: 09 Jun 2014
Posts: 343
Location: India
Concentration: General Management, Operations
Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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20 Jul 2019, 21:20
$$f_2(J(f_1(x)))= f_3(x)$$

$$1-J(\frac{1}{x})$$=$$\frac{1}{(1-x)}$$

$$J(\frac{1}{x})$$=$$\frac{-x}{(1-x)}$$

$$J(\frac{1}{x})$$= (1/x)/[1-(1/x)]

$$J(x)=\frac{x}{(1-x)}$$

prabsahi wrote:
prashanths wrote:
If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x) such that f2(J(f1(x))) = f3(x)

A) f1(x)
B) f3(x)/x
C) f3(x)
D) f2(x)
E) f2(x)/x

I guess it involves F inverse.
[/quote]

Thanks Nick!!
Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)   [#permalink] 20 Jul 2019, 21:20
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