What are the Maximum and Minimum Values of the Function s(t) = 1+2t-8/(t^2+1)?

Thread starter thomasrules
Start date Jan 26, 2005
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Minimum

AI Thread Summary

To find the maximum and minimum values of the function s(t) = 1 + 2t - 8/(t^2 + 1), the derivative s'(t) is calculated as s'(t) = 2 - 16t/(t^2 + 1)^2. The discussion reveals confusion over the correct formulation of the function and its derivative. Participants clarify the equation and attempt to solve for critical points by setting the derivative equal to zero. The conversation emphasizes the importance of correctly deriving and simplifying the function to identify maximum and minimum values effectively.

Jan 26, 2005

thomasrules

I have to find the Max and Minimum in this question.
I'm stuck because I can't find t.

s(t)= 1+2t-8/(t^2+1)

I've got:

s prime(t)= 16t(t^2+1)^-2 +2

Thanks

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Jan 26, 2005

courtrigrad

ok so I assume your equation is s(t) = (2t+1) - \frac{8}{t^2+1} Is this correct? Then \frac{ds}{dt} = 2 - \frac{16t}{(t^2+1)^2} Then t = 0

Last edited: Jan 26, 2005

Jan 26, 2005

thomasrules

sorry wrong equation:

(1+2t)-(8/t^2+1)

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