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    http://www.examsolutions.net/a-level...e/paper.php#Q4
    part c
    i got for the inverse \frac{1}{x} -1 which told me that it's a graph of \dfrac{1}{x} translated on unit down

    so i thought the range was f(x)< -1

    but it's not why is that?
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    (Original post by thegreatwhale;[url="tel:66469010")
    66469010[/url]]http://www.examsolutions.net/a-level...e/paper.php#Q4
    part c
    i got for the inverse \frac{1}{x} -1 which told me that it's a graph of \dfrac{1}{x} translated on unit down

    so i thought the range was f(x)< -1

    but it's not why is that?
    Because it links into the domain of the original function. Do you agree that the range of f(x) is less than 1/4?
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    (Original post by RDKGames)
    Because it links into the domain of the original function. Do you agree that the range of f(x) is less than 1/4?
    i don't quite understand...
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    (Original post by thegreatwhale;[url="tel:66469846")
    66469846[/url]]i don't quite understand...
    Sketch f(x) for X>3 and tell me the range that you find
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    (Original post by RDKGames)
    Sketch f(x) for X>3 and tell me the range that you find
    i've sketched it but i don't know how i can get the range from it....
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    (Original post by thegreatwhale;[url="tel:66470300")
    66470300[/url]]i've sketched it but i don't know how i can get the range from it....
    Plug in X=3 and the range is everything less than that answer. Hence why you exclude the equal bit, JUST less than. So range is everything less than f(3)
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    (Original post by RDKGames)
    Plug in X=3 and the range is everything less than that answer. Hence why you exclude the equal bit, JUST less than. So range is everything less than f(3)
    so x<3 ???
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    (Original post by thegreatwhale;[url="tel:66470388")
    66470388[/url]]so x<3 ???
    Are you getting confused between range and domain? Domain is what you are allowed to PLUG IN, range is what you get out as a result of plugging in the domain. So when you plug in X=3, you get 1/4. From your sketch you should see that as X increases from 3, the y value gets smaller and smaller thus the range is less than 1/4.

    For the inverse, the domain the range of the original therefore the domain would be X<1/4 for the inverse.
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    (Original post by RDKGames)
    Are you getting confused between range and domain? Domain is what you are allowed to PLUG IN, range is what you get out as a result of plugging in the domain. So when you plug in X=3, you get 1/4. From your sketch you should see that as X increases from 3, the y value gets smaller and smaller thus the range is less than 1/4.

    For the inverse, the domain the range of the original therefore the domain would be X<1/4 for the inverse.
    if i plug in 3 into  \dfrac{1}{x} -1 i get -\dfrac{2}{3} not 0.25 >.>
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    (Original post by thegreatwhale;[url="tel:66470630")
    66470630[/url]]if i plug in 3 into  \dfrac{1}{x} -1 i get -\dfrac{2}{3} not 0.25 >.>
    I said into f(x), you are plugging it into f-1(X)
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    (Original post by RDKGames)
    I said into f(x), you are plugging it into f-1(X)
    oh shintezewizel that's why i got a weird answer.... ah thanks so f(x)>0.25 ??
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    Name:  ImageUploadedByStudent Room1468788955.160967.jpg
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    Hence answer to part c is between 0 and 1/4 for x

    The 0 comes from the fact that the range for the original function is between 0 and 1/4; it never reaches 0.


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    (Original post by thegreatwhale;[url="tel:66470806")
    66470806[/url]]oh shintezewizel that's why i got a weird answer.... ah thanks so f(x)>0.25 ??
    No, less than that, observe your sketch for X>3 (assuming you got it right)
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    (Original post by RDKGames)
    No, less than that, observe your sketch for X>3 (assuming you got it right)
    tried looking at the graph for a while and the graph isn't very tall but goes on long for "infinity" but the range is the height so it's from 0 to 1/4 ???
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    (Original post by thegreatwhale)
    tried looking at the graph for a while and the graph isn't very tall but goes on long for "infinity" but the range is the height so it's from 0 to 1/4 ???
    Exactly. So that becomes the domain for its inverse then.
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    (Original post by RDKGames)
    Exactly. So that becomes the domain for its inverse then.
    thanks a bunch
 
 
 
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