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    (Original post by Remarqable M)
    what about f(X)=-2^-x
    This
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    can someone explain why the ln(3-x) graph translates 3 to the right?

    I can calculate that it would cross the x axis at 2, therefore the asymptote is at 3, but I don't know why
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    (Original post by JTeighty)
    The person who sketched their graph above (Now they deleted it) is wrong. lnx is a reflection of y=e^x in y=x not y=0.

    So the graph would be a translation of 3 to the left and then a reflection in x y axis.
    This is right.
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    (Original post by unamed)
    can someone explain why the ln(3-x) graph translates 3 to the right?
    Just checking why as well.
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    (Original post by Clarity Incognito)
    It doesn't, it goes left 3 units.
    but that doesn't work out when you plug the values in!
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    (Original post by Clarity Incognito)
    Unfortunately, you are wrong as well.

    The graph ln(3-x) is a reflection of ln(x) in the y axis AND THEN a translation 3 units to the left.
    So, would the x coordinate be -3?
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    (Original post by unamed)
    but that doesn't work out when you plug the values in!
    Well bugger me sideways, you're right. It must be translation of 3 to the left and then the reflection in y axis. I always thought it was reflections and stretches first, and translations last...
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    (Original post by I Have No Imagination)
    So, would the x coordinate be -3?
    No, no, I got it wrong, asymptote is at x = 3, crosses x axis when y = 0 and that's when x = 2. I thought it was reflections/stretches first, translations last but in this case, that's wrong.
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    how do you find the range when you have just solved the inverse to a function


    ps. i need HELP
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    (Original post by morebritishthanindian)
    how do you find the range when you have just solved the inverse to a function


    ps. i need HELP
    Depends on the question/function, whip it on here. Essentially for functions and their inverses, their domains and ranges are swapped so the domain of the function is the range of the inverse, and the range of the function is the domain of the inverse.
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    ln3-X

    Yes or no?
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    (Original post by unamed)
    can someone explain why the ln(3-x) graph translates 3 to the right?

    I can calculate that it would cross the x axis at 2, therefore the asymptote is at 3, but I don't know why
    think it of this way, when x=0, the y=ln(3) so u mark a point y=1 in the y-axis and then find when it crosses the x-axis this is when the y=0, so u get 0=ln(3-x) >> e^0=3-x >>> x =3-e^0=2 so u plot the points y=1 and x=2 then ask urself what is the graph of lnx draw it out and then hold a ruler in the y-axis reflect the lnx and mark your aymptote which is 3. If you move it 3 to the left what happens is that the on your calculator you get error. It is always good to work through using your calculator very helpful(you can't afford to lose 3 marks!)
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    page 37 Edexcel C3 book
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    f(x) = ln(4-2x) x<2

    inverse = 2-1/2.e^x
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    (Original post by I Have No Imagination)
    ln3-X

    Yes or no?
    yes! goes through y axis at Ln 3
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    (Original post by I Have No Imagination)
    ln3-X

    Yes or no?
    Yes, that's what I've got now from plugging in values but do you know why the translation is first and the reflection is second?
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    (Original post by morebritishthanindian)
    f(x) = ln(4-2x) x<2

    inverse = 2-1/2.e^x
    The range of the inverse is y<2. Also the inverse is (4 - e^x)/2. If you split the inverse into partial fractions to get 2 - (e^x)/2 then as x tends to negative infinity, f'(x) tends to 2. f'(x) never gets larger than 2 because e^x never becomes negative.
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    (Original post by Clarity Incognito)
    Yes, that's what I've got now from plugging in values but do you know why the translation is first and the reflection is second?
    is only for equation involving e^x and lnx that you need to be careful, and use calculator but for all the other graphs like cubic and quadratice the normal order applies.

    when u see any cubic function then apply this rule first any stretches, then translation and then finally reflection.

    if you have something like sketch y=(x-2)^2+3 start with sketching y=x^3 and then move it +2 to the right and then a vertical translation of +3 in the y direction.
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    (Original post by Remarqable M)
    is only for equation involving e^x and lnx that you need to be careful, and use calculator but for all the other graphs like cubic and quadratice the normal order applies.
    There's probably a reason for that somewhere down the line, which I'll look for later, for now, I'll just accept that, thanks!
 
 
 
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