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Inverse functions 
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#1
Apr207, 01:31 AM

P: 520

how do i find
f(x)^1 of y=4x+9/2x3 the answer in the back of the text book is f(x)^1=3x+9/2x4 


#2
Apr207, 01:39 AM

P: 529

Show your attempt at a solution, or pay me 1 dollar for doing your homework for you (your choice).



#3
Apr207, 01:48 AM

P: 520

2x=4x+9/y + 3y/y
2x=4x+9+3y/y How do i move 4x to the other side? 


#4
Apr207, 02:07 AM

P: 529

Inverse functions
is 9/2x in your original post 9/(2x)? Just want to check.



#5
Apr207, 02:14 AM

P: 520

Heres the full working out:
y(2x3)=4x+9 2x3=4x+9/y 2x=4x+9/y + 3 2x=4x+9/y + 3y/y 


#6
Apr207, 02:50 AM

P: 529

You gotta learn to put brackets in your eqns. There's no way anyone could have understood your original problem without the brackets.
y(2x3)=4x+9 x(2y4)3y=9 x=(9+3y)/(2y4) 


#7
Apr207, 03:17 AM

P: 520

Sorry to bother u again but how
do u get from y(2x3)=4x+9 to x(2y4)3y=9 


#8
Apr207, 03:22 AM

P: 529

expand and collect terms in x
y(2x3)=4x+9 > y2x3y=4x+9 > x(2y)3y=x(4)+9 > x(2y4)3y=9 


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