Posted by **George** on Thursday, October 16, 2008 at 6:21pm.

Use analytical methods to find the exact global maximum and minimum values of the function f(x)=8x-ln(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.

- Calculus -
**Reiny**, Thursday, October 16, 2008 at 7:15pm
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.

- Calculus -
**Damon**, Thursday, October 16, 2008 at 7:20pm
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

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