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January 31, 2015

January 31, 2015

Posted by **Stuck** on Monday, March 22, 2010 at 10:12pm.

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 -
**Reiny**, Monday, March 22, 2010 at 10:20pmWhy can't lnx = -1 ??

You were correct to have

ln x = -1

by definition

e^-1 = x

or

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 -
**drwls**, Monday, March 22, 2010 at 10:25pmThe 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 -
**Stuck**, Monday, March 22, 2010 at 10:28pmOh! So ln(x) can equal a negative number! I just didn't realize that. Thank you!

- Calculus - another question -
**Reiny**, Monday, March 22, 2010 at 10:35pmlook at the graph of y = lnx

the range is any real number, it is in the domain where x > 0

take ln (.5)

or

ln (.367879441)

- Calculus - another question -
**Stuck**, Monday, March 22, 2010 at 10:43pmThanks!

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