What is the absolute minimum of f(x) = x!?

Question

I am not after a decimal readout from a graphing calculator (I could do that myself). I am wondering if it is possible to analitically (solve on paper) figure this out. Note : I know it is some value inbetween 0 and 1. My BC Calc. teacher said it is possible. Could someone show me how this is solved?

Answer 1

You can define the factorial continuously with the Gamma function. This is pretty complicated though.

http://en.wikipedia.org/wiki/Gamma_function

The person answering above me may have been thinking of the Gamma function, or possibly Stirling's formula:

http://en.wikipedia.org/wiki/Stirling's_approximation


Edit: oops, now it's the person above the person answering above me :)

Answer 2

Actually the factorial function is part of a function called the gamme function--it is defined for any value, not just the integers. There are values, the negative intergers for example, such that the factorial is negative infinity. Go to this site and you will see what I am talking about:http://en.wikipedia.org/wiki/Gamma_function

Answer 3

n! is only defined for non-negative integers, hence we can not talk about x! if x is continuous.

The minimum is definitely 1.
0! = 1
1! = 1
2! = 2
3! = 6
..... increasing function, so then we know the minimum is at n=0 or n=1.

Absolute min is 1

=)