Why 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
"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)
5 answers
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
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
Oh! So ln(x) can equal a negative number! I just didn't realize that. Thank you!
look 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)
the range is any real number, it is in the domain where x > 0
take ln (.5)
or
ln (.367879441)
Thanks!
1 answer