>>12943944what is it you're wondering about?
can you add any detail of what else you were hoping to explain?
* you see on your graph of y = x^(1/x) that x = 2 and x = 4 give the same y.
* you see that y is higher when x is in between 2 and 4, and lower on either side.
* you see that y gets higher as x gets higher until a certain point, so the slope is positive, and that y gets lower as x gets higher after this point, so the slope is negative.
* you can find what this point is, by solving if and where the slope of your graph is 0.
* you can solve that by taking the derivative of your graph y' = (log(x)-1)(x^(1/x))/x^2, and seeing where y' = 0
* x^(1/x) and x^2 are both positive whenever x is positive, so this can only happen if log(x)-1 = 0 -> log(x) = 1 -> x = e.
* you can find the same point with a less messy derivative by taking the log of both sides, to remove the fractional exponent.
