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    I'm a little confused about functions and their inverses. I'd just like someone to explain whether my understanding is correct, patchy or just plain crap. And this is the first time I'm using latex so sorry for any errors!

    f(x) = 2x - 1 (x \in \mathbb{R} , x \geq 0)
     f^{-1}(x)= \frac{x+1}{2} (x \in \mathbb{R} , x \geq -1)

    Is my domain right? I worked it out by putting 0 into f(x) = 2x - 1...?
    And when you draw a graph of f(x) and f^-1(x), would you have to draw f(x) so that you only use values of x greater than 0 and you draw f-1(x) so you only use values of x greater than -1, starting from -1 and onwards?

    Thanks - my book isn't very good at explaining this well!
    I'll + rep too!
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    Yes that all looks right to me. As the inverse graph is just a reflection in the line x=y, the domain of the inverse function = the range of the function (and vice versa).

    So, as the graph is fairly simple I would just draw y= 2x-1, where 'y' clearly has to be any number greater than '-1', or which you've done by substituting 0 into 2x-1. So this is also your domain of the inverse. Your working all sounds fine
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    (Original post by Narik)
    I'm a little confused about functions and their inverses. I'd just like someone to explain whether my understanding is correct, patchy or just plain crap. And this is the first time I'm using latex so sorry for any errors!

    f(x) = 2x - 1 (x \in \mathbb{R} , x \geq 0)
     f^{-1}(x)= \frac{x+1}{2} (x \in \mathbb{R} , x \geq -1)

    Is my domain right? I worked it out by putting 0 into f(x) = 2x - 1...?
    And when you draw a graph of f(x) and f^-1(x), would you have to draw f(x) so that you only use values of x greater than 0 and you draw f-1(x) so you only use values of x greater than -1, starting from -1 and onwards?

    Thanks - my book isn't very good at explaining this well!
    I'll + rep too!
    sorry I think the domain should be just x E R since it is a one to one function for all the real no's..On a general note the domain of all linear functions is just x E R
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    (Original post by rbnphlp)
    sorry I think the domain should be just x E R since it is a one to one function for all the real no's..On a general note the domain of all linear functions is just x E R
    But f(x) is also linear, meaning that it should also be x E R, shouldn't it? But it isn't [example taken directly from C3 edexcel book] so... je suis tres confused! :confused:
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    (Original post by Narik)
    But f(x) is also linear, meaning that it should also be x E R, shouldn't it? But it isn't [example taken directly from C3 edexcel book] so... je suis tres confused! :confused:
    did the example gave you the domain ( you mentioned in the first post) and asked you to find the range ?

    or did the example ask you to find only the domain?
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    (Original post by rbnphlp)
    did the example gave you the domain ( you mentioned in the first post) and asked you to find the range ?

    or did the example ask you to find only the domain?
    It gives you f(x), then tells you to sketch f(x) and f-1(x) on the same axes, and to determine the equation of f-1(x), "taking care with its domain."
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    (Original post by Narik)
    But f(x) is also linear, meaning that it should also be x E R, shouldn't it? But it isn't [example taken directly from C3 edexcel book] so... je suis tres confused! :confused:
    what does your book give the range of this function as ? I simply think it should be f(x) E R
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    (Original post by rbnphlp)
    what does your book give the range of this function as ? I simply think it should be f(x) E R
    Yeah I agree with you actually. I think the question is saying if the domain was a different number, which is plain stupid and confusing, so I refuse to do it.

    Thanks for the help though!
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    (Original post by Narik)
    Yeah I agree with you actually. I think the question is saying if the domain was a different number, which is plain stupid and confusing, so I refuse to do it.

    Thanks for the help though!
    no worries
 
 
 
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