At x = 1/e, where x^x = 1/e[sup]1/e[/sup].
It’s minimum occurs at x = 1/e =(apprx) 0.367879. The value at the minimum is approximately 0.692201.
It’s simplest to find by noting that the minimum of f(x) and log(f(x)) occur at the same value of x provided f(x) > 0 so for f(x) = x^x, log(f(x)) = xlog x. The derivative of xlog(x) is log(x) + x/x= log(x)+1 . so we want log(x) = -1 which is 1/e.
Excellent. Another way to do it is to use the identity x^x=exp(x ln x), then use the elementary rules of differentiation.
I’m a “high-and-mighty” academic, and my lesson plan includes teaching my students Newton’s Method this coming Tuesday. Come on by, maybe you’ll learn something.
Can this old rusty tool show up to your class? What time is the class, and can I just buy a parking pass and just sit in the back of the class? I’m 50 miles away from UC Riverside. I’m hoping it’s an evening class.
blinks
I must confess: despite my flippant response to Jinx’s ill-informed tirade, I don’t really have the authority to issue invitations. The class is for the benefit of the paying students.
That said, if someone sits quietly in the back, doesn’t annoy or harass the students, and doesn’t take a seat from a registered student, it’s no skin off my nose. If you’re serious, send me an e-mail.
(And please understand, I’m not trying to be hostile to the idea. It’s just that my position obligates me to be very, very careful that my responbilities to my students are adequately carried out. Working with students is a bit of a minefield, even when the students are technically adults.)
Oooh, there I have to disappoint you. I teach at 8:10 in the morning.