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    Attachment 616050616056 hi I am a bit stuck on finding the domain for the inverse function of part c. I worked out that the range for fx is fx<1 . So why isn't the domain of the inverse function x<1?

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    (Original post by coconut64)
    hi I am a bit stuck on finding the domain for the inverse function of part c. I worked out that the range for fx is fx<1 . So why isn't the domain of the inverse function x<1?

    Thanks
    Domain of f(x) is x&lt;0 not x&gt;0 as from your sketch... Also yes the range of f(x) is less than one but it's not everything below 1. It's not -1 for example.
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    (Original post by RDKGames)
    Domain of f(x) is x&lt;0 not x&gt;0 as from your sketch... Also yes the range of f(x) is less than one but it's not everything below 1. It's not -1 for example.
    Why is that? Because the graph y=1-e^2x does continue all the way down as x is greater than 0. I don't get what you mean.

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    (Original post by coconut64)
    Why is that? Because the graph y=1-e^2x does continue all the way down as x is greater than 0. I don't get what you mean.

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    In the question the domain of x is given as x is real and x<0. So for x<0, 0<e^2x<1, so what is the range for f? How does this related to the domain for the inverse of f?


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    (Original post by coconut64)
    Why is that? Because the graph y=1-e^2x does continue all the way down as x is greater than 0. (<---- ????) I don't get what you mean.

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    f(x)=1-e^{2x} is defined for x&lt;0 (read the question for the proper domain... only if this was x>0 would the graph continue all the way down). Take e^x defined for x&lt;0 then you have 0&lt;e^x&lt;1 also 0&lt;e^{2x}&lt;1 then 0&gt;-e^{2x}&gt;-1 and so 1&gt;1-e^{2x}&gt;0

    That's the range. So that's the domain of the inverse.
 
 
 
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