Calculus
posted by George .
Use analytical methods to find the exact global maximum and minimum values of the function f(x)=8xln(4x) for x >0. If a global maximum or minimum does not exist, enter the word NONE.
For the global maximum at x=none, But for the Global minimum at x=?
Is the Global minimum 0 or nonw? Any help would be greatly appreciated.

I always encouraged my students to make a rough sketch of the graph.
Pick any 5 or 6 positive x's, then find their y values.
Pick an x close to zero (e.g. x = .00001) and see what y you get. This gives you an indication of the graph does close to zero. (remember x cannot be zero for this function)
In general, the max/min is found by setting the derivative equal to zero, solving for the variable, and then subbing that back into the original equation.
so
f(x) = 8x  ln(4x)
f'(x) = 8  1/x
set this equal to zero, you will get x = 1/8
I will let you find the minimum. 
y = 8 x  ln (4x)
y' = 8 4/x
when is that zero?
4/x = 8
x = 1/2
is that a maximum or a minimum?
y" = 0 +4/x^2
for x>0, y" is positive so it is a minimum
so minimum at x = 1/2
what is the value of the function at that minimum?
y = 8(1/2)  ln 2
4  ln 2 = about 3.3