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Math Help - Help with calculus problem

  1. #1
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    Help with calculus problem

    Hi. Someone can help me with this.

    f(x)=x-ln x

    a) Discover in the grafic of the funcion the point with abscissa c knowing that the line that is tangent with c is parallel with the secant line that passes on the points with abscissa 1 and e.

    b) Show that x \ge 1+ ln x
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by usual_suspect View Post
    Hi. Someone can help me with this.

    f(x)=x-ln x

    a) Discover in the grafic of the funcion the point with abscissa c knowing that the line that is tangent with c is parallel with the secant line that passes on the points with abscissa 1 and e.
    what does "abscissa" mean?

    b) Show that x \ge 1+ ln x
    consider the function f(x) = x - \ln x - 1 defined on (0, \infty)

    show that this function is always nonnegative. u can do this by showing it is nonnegative on (0,1] then show it is strictly increasing for x > 1
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    Quote Originally Posted by Jhevon View Post
    what does "abscissa" mean?
    abcissa is the x-coordinate or the coordinate in the x-axis
    Cartesian coordinate system - Wikipedia, the free encyclopedia
    Thanks
    Last edited by usual_suspect; October 28th 2007 at 02:25 PM.
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    Quote Originally Posted by Jhevon View Post
    consider the function f(x) = x - \ln x - 1 defined on (0, \infty)

    show that this function is always nonnegative. u can do this by showing it is nonnegative on (0,1] then show it is strictly increasing for x > 1
    I really don't understant what you want to say with that. Can you explain.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by usual_suspect View Post
    I really don't understant what you want to say with that. Can you explain.
    since \lim_{x \to 0^+} - \ln x = \infty we have that f(x) = x - \ln x - 1 > 0 for x \in (0,1)

    when x = 1 we have f(1) = 1 - 0 - 1 = 0

    thus, f(x) \ge 0 for x \in (0,1]

    now, f'(x) = 1 - \frac 1x. obviously f'(x) > 0 for all x > 1

    thus, f(x) is increasing on (1, \infty). which means f(x) \ge 0 for x \in (1, \infty)

    thus, for x \in (0, \infty) we have that f(x) \ge 0

    but f(x) = x - \ln x - 1, so this means x - \ln x - 1 \ge 0

    adding \ln x + 1 to both sides, we get x \ge \ln x + 1 for all x > 0 as desired
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  6. #6
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    Abscissa?. Why not just say x=c. Wonder why some texts insist on showing off with Rube Goldbergian jargon?.

    If I am reading this correctly, it appears to be a problem related to the Mean Value Theroem.

    Use

    \frac{f(b)-f(a)}{b-a}=\frac{(e-ln(e))-(1-ln(1))}{e-1} = \frac{e-2}{e-1}

    \frac{d}{dx}[x-ln(x)]=1-\frac{1}{x}

    To find the point c, which by the MVT should be between 1 and e:

    1-\frac{1}{c}=\frac{e-2}{e-1}

    Solve for c.

    Here's the graph with the tangent line and the secant line. Note they are parallel as per the MVT.

    You can now find the equation of your tangent line, if needed.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  7. #7
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    Quote Originally Posted by galactus View Post
    Abscissa?. Why not just say x=c. Wonder why some texts insist on showing off with Rube Goldbergian jargon?.
    Here in portugal abscissa (or how we say here abcissa) is a very a term used much times. I didn't noticed that you guys don't use it. English is not my native English, so sometimes I do some mistakes.

    Thanks for your help
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  8. #8
    Eater of Worlds
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    That's OK. I wasn't criticizing you. I have heard it used, but not much.

    Isn't there another wacky word for the y coordinate?.
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  9. #9
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    Quote Originally Posted by usual_suspect View Post
    I did noticed that you guys don't use it.
    That happens to not be true. Classically written mathematics, even in English, does use the term abscissa. While it does mainly appear in older texts it is by no means obsolete. The difference is between those classically train and those who are not.
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  10. #10
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    I remember now. The y-coordinate is called the 'ordinate'. As opposed to the x-coordinate is the abscissa. I reckon I am not a classic then. I certainly knew what it meant, but rarely hear it used except for seeing it in old books.
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