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Thread: Absolute Minimum

  1. #1
    Member mybrohshi5's Avatar
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    Absolute Minimum

    Find the -coordinate of the absolute minimum for the function

    I found the derivative and set it equal to 0 and solved for x and got that x=e^(11/3)

    Isn't that where the absolute minimum occurs?

    my homework says it is wrong.

    could i get some help please.

    thank you
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  2. #2
    Super Member 11rdc11's Avatar
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    Thats the same thing I came up with
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  3. #3
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    You have probably found the LOCAL minimum. The absolute minimum doesn't exist because as x approaches 1, y approaches neg infinity (ie just keeps getting small). Have a look at the graph.
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  4. #4
    Member mybrohshi5's Avatar
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    i did look at the graph and as x increases y decreases, so there is never a minimum

    but my webwork says its a number because i tried entering DNE and -infinity but neither of those work.

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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Debsta View Post
    You have probably found the LOCAL minimum. The absolute minimum doesn't exist because as x approaches 1, y approaches neg infinity (ie just keeps getting small). Have a look at the graph.
    Actually if you look at the graph past e^\frac{11}{3} it starts increasing again
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  6. #6
    Member mybrohshi5's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    Actually if you look at the graph past e^\frac{11}{3} it starts increasing again

    I must be graphing it wrong cause i see a decreasing curve (looks like the left half of a U) that continues to decrease as x goes to infinity.

    here is how i enter it into my calculator

    (8-3ln(x))/(x)
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  7. #7
    Member mybrohshi5's Avatar
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    i just entered e^11/3 in again and it worked? weird lol

    thanks everyone
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  8. #8
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    Zoom in. The LOCAL min occurs when x = e^(11/3) ie approx 39.13.
    The question though is asking for the ABSOLUTE min.
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