Another critical numbers problem

  • #1
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
 

Answers and Replies

  • #2
Dick
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Are you trying to solve g'(t)=0? Then sure, factor out a t^(-1/3).
 
  • #3
remember that when it says ^-1/3
its not saying (5t^-1)/3
 
  • #4
Dick
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That's why I put parentheses around the (-1/3). You should too.
 
  • #5
HallsofIvy
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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
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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