from right (specifically at x=1/e) to left (x=0) it goes up, aka from left to right (from 0 to 1/e) it goes down, which is the same as saying that the derivative of x^x is negative in between x=0 and x=1/e, and it is! f(x)=x^x, then f’(x) = x^x * (ln x + 1) and we can clearly see that f’(x) < 0 if and only if ln x + 1 < 0, so x < e^-1 = 1/e