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    May i have some help please in finding the inverse of this equation, also please show step by step how you did it its been bugging me for ages.  f:x \rightarrow \frac{2x+3}{x-1} thanks for your help.
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    (Original post by Core)
    May i have some help please in finding the inverse of this equation, also please show step by step how you did it its been bugging me for ages.  f:x \rightarrow \frac{2x+3}{x-1} thanks for your help.
    Note that, if g(x) is the inverse of f, then f(x)=\frac{2x+3}{x-1} \implies x=\frac{2g(x)+3}{g(x)-1}.

    Rearrange for g(x) and you have your inverse. If you can't do that, post your attempt at it.
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    set f(x)=y. So y=\frac{2x+3}{x-1}. Swap your x's and y's around:

    x=\frac{2y+3}{y-1}

    Multiply through by y-1

    x(y-1)=2y+3
    xy-x=2y+3

    Move all the y's to one side, and all the x's to the other:

    xy-2y=3+x
    y(x-2)=3+x

    And divide through by x-2.

    y=\frac{3+x}{x-2}

    f^{-1}(x)=\frac{3+x}{x-2}
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    I can just swap them like that? Anyway thanks for that, il use what i learned from your post in the future.
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    Pretty much. It's not entirely mathematically sound, notation wise, but it gets the answer. The notation used by Farhan.Hanif93 is the more accurate way of going.

    You'll also often need to ensure the initial function is one-one, that there is only one value of y for a given x and vice versa - the function you gave is, but not all functions necessarily are, sometimes the domain needs to be restricted.
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    (Original post by TheDavibob)
    Pretty much. It's not entirely mathematically sound, notation wise, but it gets the answer. The notation used by Farhan.Hanif93 is the more accurate way of going.

    You'll also often need to ensure the initial function is one-one, that there is only one value of y for a given x and vice versa - the function you gave is, but not all functions necessarily are, sometimes the domain needs to be restricted.
    Oh i see how you swaped, you multiplied through by one of the terms then divided by the other, also i am going to try the other method, yours was the first i saw because i clicked on "last post". My issue was the rearranging becuase i had forgotten the multiply through method that you used.
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    (Original post by TheDavibob)
    ...
    If you haven't already, you should take a read of this. Since you're new, you probably aren't aware of how things are run on the maths forum.

    Welcome to TSR.
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    Welcome and ty for your help once again.
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    (Original post by Farhan.Hanif93)
    If you haven't already, you should take a read of this. Since you're new, you probably aren't aware of how things are run on the maths forum.

    Welcome to TSR.
    I realised after I posted that I'd probably answered too much, as it were. I think I got a bit carried away.

    Thanks for the welcome, and it won't happen again.
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    (Original post by TheDavibob)
    I realised after I posted that I'd probably answered too much, as it were. I think I got a bit carried away.

    Thanks for the welcome, and it won't happen again.
    No worries.
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    All i do is, for example, if the equation is y=, then you will rearrange it for x=, then once you've done that you change all the x's into y's and vice versa. For example, y = sin x , rearrange for x ---> x = sin-1 y , swap the x and y around --> y = sin-1 x and thats you're inverse function
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    (Original post by bonana567)
    All i do is, for example, if the equation is y=, then you will rearrange it for x=, then once you've done that you change all the x's into y's and vice versa. For example, y = sin x , rearrange for x ---> x = sin-1 y , swap the x and y around --> y = sin-1 x and thats you're inverse function
    I knew what i had to do, it was the actualy rearrange that was my issue.
 
 
 
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