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Find `y=f(f^(-1)(3x))` in the form of y=ax/(bx + c), where a,b and c are real...

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karengervasi | Student | eNoter

Posted April 17, 2013 at 6:43 AM via iOS

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Find `y=f(f^(-1)(3x))` in the form of y=ax/(bx + c), where a,b and c are real constants.

 `f(x)=2log_e(x+1)`   and  ` f^(-1)(x)=e^(x/2) - 1` .

Tagged with algebra, function, inverse, math, methods

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mathsworkmusic | (Level 3) Associate Educator

Posted April 17, 2013 at 11:40 AM (Answer #2)

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We have

`f(x) = 2log_e(x+1)`  and

`f^(-1)(x) = e^(x/2)-1`

To work out

`y = f(f^(-1)(3x))`

however, we don't need either of these as, since `f^(-1)(x)`is the inverse of `f(x)`,

`f(f^(-1)(3x))`  is simply  `3x`.

We can write `y = f(f^(-1)(3x))` in the form  `(ax)/(bx+c)`  where

`a = 3`, `b = 0`  and `c=1`

y = 3x/(0x + 1)


 

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pramodpandey | College Teacher | Valedictorian

Posted April 17, 2013 at 10:42 AM (Answer #1)

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`f(x)=2log_e(x+1)`

`f^(-1)(x)=e^(x/2)-1`

`f(f^(-1)(3x))=f(e^((3x)/2)-1)`

`=2log_e(e^((3x)/2)-1+1)`

`f(f^(-1)(3x))=2xx((3x))/2`

`f(f^(-1)(3x))=3x`

`y=f(f^(-1)(3x))=3x`

a=1,b=0,c=1

`y=(1.x)/(0.x+1)`

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