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    Something that really bothers me...I've grasped all the hard concepts yet I cant quite get this.... Is there some sort of a direct formula..

    Example

    Find the range of f(x)= ln(4-2x) x<2 and allreal numbers( X E R)


    Thanks
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    Draw the graph. See what values Y can take.
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    I really do agree with you. Range and domain are beyond my understanding. Some graphs are too complicated. Thats one mark I might just let fly tommorrow.
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    The range is what y can be. A normal ln graph can't be less than 0, but since we have -2x, it can't be less than 4. Do you see why?
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    (Original post by face05)
    Example

    Find the range of f(x)= ln(4-2x) x<2 and allreal numbers( X E R)


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    I still get the range as all real numbers so (X E R). But I could be wrong.
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    Well ln 0 doesn't exist. And you can't have ln of any negative number. Therefore you have to find all values of x where (4-2X) is greater than 0, so you would say X>2.
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    (Original post by mirah12)
    I still get the range as all real numbers so (X E R). But I could be wrong.
    What it tells you in the question is that the values of x are all real numbers greater than 2. The range is what y can be, the domain is what x can be - which is what the question tells you.
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    (Original post by Jimlon0)
    Well ln 0 doesn't exist. And you can't have ln of any negative number. Therefore you have to find all values of x where (4-2X) is greater than 0, so you would say X>2.
    that's the domain though, right?
    The range for this would be all real numbers, right?
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    The easiest way is to put in some numbers in and make a judgement.
    So for example, if the equation was, y=1/(X+1) , and the domain was X>3, Given X+1 will therefore always be positive so if you put it 3, then try 4, you'll see that it will get smaller. So the range would be y<1/4.
    If you just look at the equation and just think about what numbers can go in, and therefore what answers you will get out.
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    (Original post by unamed)
    that's the domain though, right?
    The range for this would be all real numbers, right?
    Oh yes sorry, I misread the top part. The range would be greater than or equal to 0, because logs don't result in negative numbers.
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    (Original post by Jimlon0)
    Oh yes sorry, I misread the top part. The range would be greater than or equal to 0, because logs don't result in negative numbers.
    right. Thank you!

    I just properly did this, and the poster below me is right. The range would be greater than or equal to for e^x, not lnx
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    This is definitely a mark I'll let fly in the exam.:sigh:

    (Original post by Jimlon0)
    Oh yes sorry, I misread the top part. The range would be greater than or equal to 0, because logs don't result in negative numbers.
    Assuming that I drew it right...if I draw the graph it passes through the y-axis at ln4 and the x-axis at 1.5 with the asymptote at 2.

    The graph then does continue down below 0... :confused:

    If x=1.9, then f(x)= ln(4-2(1.9)) = - 1.7 (a negative)
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    Range is an element of reals (X E R) for a ln function. In the graph you have given , as x -> 2 , y -> - infinity. and as x -> - infinity , y -> infinity. Therefore y can take any value between negative infinity and infinity, hence all real numbers. Hope this makes sense, and I hope I am not wrong.... else I am in trouble tommorow...
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    (Original post by paper-wings)
    Range is an element of reals (X E R) for a ln function. In the graph you have given , as x -> 2 , y -> - infinity. and as x -> - infinity , y -> infinity. Therefore y can take any value between negative infinity and infinity, hence all real numbers. Hope this makes sense, and I hope I am not wrong.... else I am in trouble tommorow...
    Yeah that made sense. Major thanks!
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    IMO the easiest way to find it is just to imagine the graph. If it's a transformation of...say...x^2 then I imagine an X^2 graph and shift it along in my head as per the equation. Same with e^x, Ln(X), etc. I know this only works for some people and some equations but...at the end of the day it's usually a 1 mark question. Not worth getting stressed out over.

    By the way good luck everyone sitting it tomorrow.
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    I thought the answer to this was f(x) < ln4 ?
 
 
 
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