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Math Help - Finding Critical Points

  1. #1
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    Finding Critical Points

    Hi. I was wondering if anybody could help me out with this problem.

    F(x) equals the Integration from x^2 to x^3 of (lnt)dt. We have to find all critical points of f(x).

    What I (think) I know.

    I think you have to use the fundamental theroem of calculus (1 and 2?), but I'm unsure of how to apply it. A friend helped me a bit and so far we have gotten this:

    ln(x^3)3x - ln(x^2)2x = 0

    from there though, I'm not sure what you do. I'm not sure (if that is the correct procedure) how you go about solving for those points.

    Any help is greatly appreciated.
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  2. #2
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    Quote Originally Posted by ChaosBlue View Post
    Hi. ...

    F(x) equals the Integration from x^2 to x^3 of (lnt)dt. We have to find all critical points of f(x).
    ...
    Hello,

    I've attached a diagram with the complete calculations. It looks to me as if you've made a small mistake in calculating the first derivation.

    EB
    Attached Thumbnails Attached Thumbnails Finding Critical Points-ln_vargrenztext.gif  
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  3. #3
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    Hello,

    I'm awfully sorry that I forgot to attach the diagram of the function.

    EB
    Attached Thumbnails Attached Thumbnails Finding Critical Points-ln_vargrenz.gif  
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  4. #4
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    Quote Originally Posted by ChaosBlue View Post
    Hi. I was wondering if anybody could help me out with this problem.

    F(x) equals the Integration from x^2 to x^3 of (lnt)dt. We have to find all critical points of f(x).

    What I (think) I know.

    I think you have to use the fundamental theroem of calculus (1 and 2?), but I'm unsure of how to apply it. A friend helped me a bit and so far we have gotten this:

    ln(x^3)3x - ln(x^2)2x = 0

    from there though, I'm not sure what you do. I'm not sure (if that is the correct procedure) how you go about solving for those points.

    Any help is greatly appreciated.
    If I understand what you mean....

    F(x) = INT.(x^2 --->x^3)[ln(t)]dt
    So,
    F(x) = [1/t]|(x^2 ---> x^3)
    F(x) = 1/(x^3) -1/(x^2) --------------(i)

    Now to the critical points of F(x), we get the 1st derivative of F(x) and equate that to zero.
    F(x) = x^(-3) -x^(-2) ---------rewriting (i) in another form.
    F'(x) = (-3)[x^(-4)] -(-2)[x^(-3)]
    F'(x) = -3/(x^4) +2/(x^3)
    Set that ot zero,
    0 = -3/(x^4) +2/(x^3)
    Clear the fractions, multiply both sides by (x^4)(x^3),
    0 = -3x^3 +2x^4
    0 = (-3 +2x)x^3
    x = 0 or 3/2

    Hence, the critical points of F(x) are:

    When x = 0,
    F(0) = 1/0 -1/0 = indeterminate.
    Meaning, zero is not a root of F(x). So discard x=0.

    When x = 3/2,
    F(3/2) = 1/[(3/2)^3] -1/[(3/2)^2]
    F(3/2) = 8/27 -4/9
    F(3/2) = (8 -12)/27 = -4/27
    So point (3/2,-4/27) --------------------the only critical point.

    ==================================================
    :-), Heck, I made a honest mistake. INT.[ln x] dx = 1/x. Yeah, right. Because INT.[1/u]du = ln(u) +C. :-)
    I will not edit this wrong solution, nor will i delete it. Let me be wrong once in a while.

    -----------------------
    Last edited by ticbol; February 15th 2007 at 01:32 AM.
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  5. #5
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    Quote Originally Posted by ticbol View Post
    :-), Heck, I made a honest mistake. INT.[ln x] dx = 1/x. Yeah, right. Because INT.[1/u]du = ln(u) +C. :-)
    I will not edit this wrong solution, nor will i delete it. Let me be wrong once in a while.

    -----------------------
    I have found that seeing an instructor make a mistake can be as educational as seeing an instructor doing it correctly. Both expose problem solving techniques and can point out possible errors that students can make. No shame there!

    -Dan
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