Show that for x>0, x1/x has its greatest value when x=e.
so dydx=y(1−lnx)x2=x1/x(1−lnx)x2. Turning points occur when dydx=0, which is precisely when 1−lnx=0, as x1/x is never zero for x>0. So lnx=1, and the only stationary point is at x=e.
Alternatively we can say, let x1/x=ef(x). Taking natural logs shows that f(x)=lnxx.
We can now differentiate y=ef(x) using the chain and quotient rules.
To determine the nature of the stationary point, we’ll look at the value of y or dydx either side of x=e.
We have dydx=x1/x(1−lnx)x2, and since x1/x and x2 are always positive for x>0, the sign of the derivative is determined by the sign of 1−lnx.
So when x<e, dydx>0 and the function is increasing; when x>e, dydx<0 and the function is decreasing.
Therefore the function has a maximum at x=e, and this is the global maximum of the function.
Hence, or otherwise, but without use of a calculator, determine which is the greater of eπ and πe.
From our argument above, we must have π1/π<e1/e.
Raising both sides to the power eπ gives πe<eπ (this step is valid, since the function f(x)=xeπ is strictly increasing for x>0).
So eπ is larger.