Another critical numbers problem

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Homework Help Overview

The discussion revolves around finding critical numbers for the function g(t) = 5^(2/3) + t^(5/3) and its derivative g'(t). Participants are exploring the implications of setting the derivative equal to zero and the proper handling of exponents in the expressions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of solving g'(t) = 0 and the potential need to factor out terms. There is also a focus on the interpretation of exponents, particularly negative exponents, and the importance of using parentheses for clarity.

Discussion Status

The discussion is active, with participants providing guidance on factoring and clarifying notation. There is an acknowledgment of the critical point at T = 0, although no consensus on the next steps has been reached.

Contextual Notes

There is a noted confusion regarding the use of "T" and "t" in the expressions, which may affect the clarity of the discussion. Additionally, the derivative's behavior at T = 0 is highlighted as a critical point, but the implications of this are still being explored.

afcwestwarrior
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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
 
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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.
 
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
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.
 

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