The discussion focuses on solving the derivative of the function g(T) = 5^(2/3) + t^(5/3) and finding its critical points. Participants clarify that to solve g'(t) = 0, one should factor out t^(-1/3) and ensure proper use of parentheses to avoid confusion with exponent notation. The derivative is expressed as g'(T) = (10/3)T^(-1/3) + (5/3)T^(2/3) = 0, and it is noted that the derivative does not exist at T = 0, marking it as a critical point. Clear differentiation between variables T and t is emphasized for accuracy.
PREREQUISITESStudents and educators in mathematics, particularly those focusing on calculus, as well as anyone needing to clarify derivative calculations and critical point analysis.
I assume you meant g(T)= 6T^(2/3)+ T^(5/3).afcwestwarrior said: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