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Calculus - another question

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"Find the absolute minimum value of the function f(x) = x ln(x)".

I know that one of the critical values is 1 (domain restriction). Am I right?

So I did the derivative, and got 1+ ln(x), but I cannot figure out the critical values, because ln(x) can't equal -1...

Am I doing something wrong??

(Answer is -1/e)

  • Calculus - another question - ,

    Why can't lnx = -1 ??

    You were correct to have
    ln x = -1
    by definition
    e^-1 = x
    x = 1/e

    now put that back ...
    f(1/e) = (1/e)ln(1/e)
    = (1/e)(-1) = -1/e

    so the minimum value of the function is -1/e and it occurs when x = 1/e

  • Calculus - another question - ,

    The derivative of
    The log of a quantity CAN be negative, and is if the number is between 0 1nd 1. You were thinking of the rule that you cannot take the log of a negative number.

    f(x) = x lnx leads to

    f'(x) = lnx + 1

    That equals zero when
    lnx = -1

    Make both sides the same power of e and the equation will still be valid.
    x = e^-1

    That is where the function is a minimum. The value of that minimum is
    ln(e^-1)*e^-1 = -e^-1

  • Calculus - another question - ,

    Oh! So ln(x) can equal a negative number! I just didn't realize that. Thank you!

  • Calculus - another question - ,

    look at the graph of y = lnx
    the range is any real number, it is in the domain where x > 0

    take ln (.5)
    ln (.367879441)

  • Calculus - another question - ,


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