g't= 10/3T^-1/3+ 5/3t^2/3
ok do i factor this one out, this one looks confusing
Are you trying to solve g'(t)=0? Then sure, factor out a t^(-1/3).
remember that when it says ^-1/3
its not saying (5t^-1)/3
That's why I put parentheses around the (-1/3). You should too.
I assume you meant g(T)= 6T^(2/3)+ T^(5/3).
After you got g'(T)= (10/3)T^(-1/3)+ (5/3)T^(2/3)= 0 you can multiply the entire equation by T^(1/3). (Obviously, the derivative does not exist at T= 0 so that is also a critical point.)
It's a lot easier to read with parentheses! Also, please do not use "T" and "t" to mean the same thing.
Separate names with a comma.