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

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New post 17 Jun 2019, 22:45
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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
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Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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New post 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



Can someone please help me with approach of this problem.
I guess it involves F inverse.
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Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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New post 20 Jun 2019, 08:49
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\(f_2 (J(f_1)) = f_3\)

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

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

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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Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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New post 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.
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If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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New post 20 Jun 2019, 11:23
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\(f_2(J(f_1(x)))= f_3(x)\)

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

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

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



Can someone please help me with approach of this problem.
I guess it involves F inverse.
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Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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New post 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.
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Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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New post 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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Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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New post 23 Jun 2019, 23:34
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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)

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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Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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New post 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
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Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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New post 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
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Re: If f1(x) = 1/x, f2(x) = 1 - x, f3(x) = 1/(1-x) then find J(x)  [#permalink]

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New post 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



Can someone please help me with approach of this problem.
I guess it involves F inverse.
[/quote]


Thanks Nick!!
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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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