The critical numbers of the function

**Asuhuman18** · Nov 8th 2009, 12:34 PM

f(t)=6^(2/3)+t^(5/3)

I understand I take the derivative of the function and set it equal to zero to find the critical numbers, however I get stuck after I take the derivative

f'(t)=4t^(-1/3)+5/3t^(2)=0

**skeeter** · Nov 8th 2009, 12:53 PM

Originally Posted by Asuhuman18

f(t)=6^(2/3)+t^(5/3)

I understand I take the derivative of the function and set it equal to zero to find the critical numbers, however I get stuck after I take the derivative

f'(t)=4t^(-1/3)+5/3t^(2)=0

I guess you meant f(t) to be ...

$f(t) = 6t^{2/3} + t^{5/3}$

$f'(t) = 4t^{-1/3} + \frac{5}{3}t^{2/3} = t^{-1/3}\left(4 + \frac{5}{3}t \right) = 0$

can you finish?

**Asuhuman18** · Nov 8th 2009, 01:16 PM

Originally Posted by skeeter

I guess you meant f(t) to be ...

$f(t) = 6t^{2/3} + t^{5/3}$

$f'(t) = 4t^{-1/3} + \frac{5}{3}t^{2/3} = t^{-1/3}\left(4 + \frac{5}{3}t \right) = 0$

can you finish?

yeah sorry about that...Thanks for all the help

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