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Math Help - Function domain

  1. #1
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    Function domain

    The function is f(x)=-ln(1/x).
    1)Find domain.
    Denominator must not be zero and the whole value as well. If the domain is [1;infinity) I reckon it is wrong because when the denominator is for example 9 milliard the answer is not calculated becuase it's value is negative.
    Last edited by totalnewbie; July 27th 2005 at 05:52 AM.
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  2. #2
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    Domain is the set of x-values that will define or validate f(x), or make f(x) true.

    ln is natural log.
    (-infinity,0] don't define logs, so x cannot be from -infinity up to zero.
    (0,infinity) will define logs, so x can be from, but not including, 0 to infinity.

    f(x) = -ln(1/x)
    f(x) = (-1)ln(1/x)
    f(x) = ln[(1/x)^(-1)]
    f(x) = ln(1 / 1/x)
    f(x) = ln(x)
    That means x can be from right after zero up to right before infinity.

    Therefore, domain of f(x) is (0,infinity). ----answer.
    Last edited by ticbol; July 27th 2005 at 02:10 AM.
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  3. #3
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    I encountered an another problem on my function.
    It is said to make the function y=f(x) tangent of the graph where x-coorinate is 1.
    I did it and and the equation of tangent is: y=x-1
    But the following point said to make square function g(x)=ax^2+c according to the circumstance that tangent y=x-1 would be tangent of the graph y=g(x) whose x-coordinate is 1.

    How ?
    Last edited by totalnewbie; July 27th 2005 at 03:38 AM.
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  4. #4
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    Don't understand what you mean.

    Why don't you post the original question as you see it from the source. I mean, do not interpret the question. Copy it as it is.

    "I did it and and the equation of tangent is: y=x-1"
    That alone will confuse anyone. Because, how did you get that?
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  5. #5
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    I have to interpret the question because the questions were not in english.
    The function was f(x) = ln(x) and it's derivative is y=1/x. According to the x-coordinate which was 1 I found the slope which was 1 too, because f'(x-coordinate)=slope.
    General equation of tangent is the following: y-y0=k(x-x0) where k is the slope. So I replaced the numbers: y+y0=1*(x-1) where y0=y(x0)=ln(1)=0
    And then y+0=1*(x-1) which is actually y=x-1. And this way I got this equation of tangent

    The equation of tangent was easy but the next is complicated.
    How to determine letters a and c on the square function g(x)=ax^2+c according to the circumstance that tangent y=x-1 would be the tangent of graph y=g(x) whose x-coordinate is 1.

    Did you understand my english ?
    Last edited by totalnewbie; July 27th 2005 at 12:24 PM. Reason: corrections
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  6. #6
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    I understand your English. That is not the problem. The problem is I don't understand very well what you mean.

    So you have a function
    f(x) = ln(x) ----(i)

    You were told to find the tangent line to (i) at the point where x=1.
    And you found:
    >>>slope of the tangent line there is 1.
    >>>equation of tangent line is y = x-1
    >>>the point on f(x) where x=1 is (1,0)

    Very good, those are all correct.
    Although "General equation of tangent is the following: y+y0=k(x-x0)" should be
    y-y0=k(x-x0) -----not y+y0.

    Now you are told to find a new function such that at point (1,0) of the graph of this new function the tangent line is the same y = x-1?
    And this new function is
    g(x) = ax^2 +c ? ---------(ii)

    If my understanding then is correct,
    We get the slope of the tangent line (the slope is 1, allright),
    Differentiate both sides of (ii) with respect to x,
    g'(x) = 2ax
    Since that should be equal to 1, and x=1, then,
    1 = 2a(1)
    a = 1/2 ----***

    Substitute those into (ii), knowing g(x) = 0 when x=1, point (1,0) remember?,
    0 = (1/2)(1^2) +c
    0 = 1/2 +c
    c = -1/2 ----***

    Therefore, the new function is
    g(x) = (1/2)x^2 -1/2 ------------answer.
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  7. #7
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    Yes, your understanding was really correct. g(x) = (1/2)x^2 -1/2 is the right answer. Thank you.
    Last edited by totalnewbie; July 28th 2005 at 12:34 AM.
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  8. #8
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    You post it separately and I will answer it.

    One posting, one question. Or, new question, new posting.
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