
More max/min (sort of)
We need to find:
Find the interval where the function is decreasing.
Find the inflection point.
Find the interval where the function is concave up/down
$\displaystyle F(x) = ln(4ln(x))$
$\displaystyle F'(x) = \frac{1}{4ln(x)} * \frac{1}{x} = \frac{1}{4xxln(x)}$
I thought this was right, but that gave me a decreasing interval of $\displaystyle (e^3,\infty)$
$\displaystyle F'(x) = \frac{1}{4xxln(x)}$
$\displaystyle F''(x) = \frac{0(5ln(x))(1)}{(4xxln(x))^2} = \frac{5ln(x)}{(4xxln(x))^2}$
Which is pretty much where I've gotten stuck.
Is it one of my derivatives that's gone askew? Or did I do those right and I'm just getting the wrong intervals?

in F'(x) didn't you forget the negative . I believe you timed it by 1/x instead of 1/x.
right?

Sorry. I was doing that by memory.

Still don't have an answer on this on.
Does that mean I did it correct? (Wondering)

check your work on the second derivative.
remember that (4xxln(x))' = 3ln(x)