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Math Help - Weird natural log question

  1. #1
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    Weird natural log question

    You are given the function

    f(x) = x*lnx

    (x > 0)

    a) Find the solution(s) of the equation f(x) = 0
    b) Find any minima/maxima that f(x) has.

    c) Find the value(s) of x0 for which

    INTEGRAL ( between limits x0 and 0) f(x) dx = 0

    and describe graphically what this means.

    Ok first of all, for part a) all I can think of for a solution is x = 1 because lnx = 0 only when x = 1, right? Thing is this question is worth 4 marks so I feel like im missing something.

    Secondly for part b) I didnt think there were any maxima/minima because of what I said above. However, differentiating the function I get

    lnx + x(1/x)
    equating to zero gives

    lnx + 1 = 0
    lnx = -1
    lnx = -lne
    lnx = lne^-1

    x = 1/e

    Plugging this into the second derivative (which is 1/x) gets me e as a minima.
    Is this correct?

    And for the last part it tells me to use integration by parts, but it doesnt get me anywhere...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by BogStandard View Post
    You are given the function

    f(x) = x*lnx

    (x > 0)

    a) Find the solution(s) of the equation f(x) = 0
    b) Find any minima/maxima that f(x) has.

    c) Find the value(s) of x0 for which

    INTEGRAL ( between limits x0 and 0) f(x) dx = 0

    and describe graphically what this means.

    Ok first of all, for part a) all I can think of for a solution is x = 1 because lnx = 0 only when x = 1, right? Thing is this question is worth 4 marks so I feel like im missing something.
    if x*ln(x) = 0 then either x = 0 or ln(x) = 0 because if you have a product of two terms that yield zero, one or the other must be zero. however, we have that x > 0 (we need this for the log to be defined) so, you are correct, you are left with only one solution, x = 1.

    Secondly for part b) I didnt think there were any maxima/minima because of what I said above. However, differentiating the function I get

    lnx + x(1/x)
    equating to zero gives

    lnx + 1 = 0
    lnx = -1
    lnx = -lne
    lnx = lne^-1

    x = 1/e
    good!

    Plugging this into the second derivative (which is 1/x) gets me e as a minima.
    Is this correct?
    yup. you're a genius!

    And for the last part it tells me to use integration by parts, but it doesnt get me anywhere...
    use integration by parts with u = \ln x and dv = x

    you know what u and dv represent, right?
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