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    im useless at these
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    a) the inverse function is 3 - e^(x/2)

    b) put -4 for x in the inverse function

    c) i) the first step is to sketch ln x
    the second step is to sketch ln ( x - 3 ) which is a translation to the right of 3 units
    the third step is to sketch ln -(x-3) which is a reflection in the y axis
    the fourth step is to apply a vertical stretch of scale factor 3

    ii) reflect any parts of the previous graph which are below the x axis so they are now above the x axis
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    a) ....the domain of the inverse function must be indentical to the range of the original function. the range of f(x) appears to be all real values
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    okie thanks for replying
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    (Original post by the bear)
    a) the inverse function is 3 - e^(x/2)

    b) put -4 for x in the inverse function

    c) i) the first step is to sketch ln x
    the second step is to sketch ln ( x - 3 ) which is a translation to the right of 3 units
    the third step is to sketch ln -(x-3) which is a reflection in the y axis
    the fourth step is to apply a vertical stretch of scale factor 3

    ii) reflect any parts of the previous graph which are below the x axis so they are now above the x axis

    could you go through how you got the inverse function as i dont quite understand
    thanks
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    (Original post by lil_kk)
    could you go through how you got the inverse function as i dont quite understand
    thanks
    f(x) = 2ln(3-x)

    Let y = f(x). Doing this, and then rearranging to find x in terms of y will give you the solution. You then jut replace y with x.

    y = 2ln(3-x)
    ½y = ln(3-x)
    e½y = 3-x
    x = 3 - e½y

    Hence f-1(x) = 3 - e½x
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    (Original post by JohnSPals)
    f(x) = 2ln(3-x)

    Let y = f(x). Doing this, and then rearranging to find x in terms of y will give you the solution. You then jut replace y with x.

    y = 2ln(3-x)
    ½y = ln(3-x)
    e½y = 3-x
    x = 3 - e½y

    Hence f-1(x) = 3 - e½x
    thanks for the explanation
    much appreciated
 
 
 
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