# Another critical numbers problem

g(T)=5^2/3+ t^5/3

g't= 10/3T^-1/3+ 5/3t^2/3

ok do i factor this one out, this one looks confusing

Dick
Homework Helper
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

Dick
Homework Helper
That's why I put parentheses around the (-1/3). You should too.

HallsofIvy
Homework Helper
g(T)=5^2/3+ t^5/3

g't= 10/3T^-1/3+ 5/3t^2/3

ok do i factor this one out, this one looks confusing
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.