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

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The discussion focuses on finding the maximum and minimum values of the function s(t) = 1 + 2t - 8/(t^2 + 1). The derivative of the function, s'(t) = 2 - 16t/(t^2 + 1)^2, is derived to locate critical points. The participants clarify the correct formulation of the function and its derivative, emphasizing the importance of accurate algebraic manipulation in calculus. Ultimately, the goal is to determine the values of t that yield the extrema of the function.

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thomasrules
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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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ok so I assume your equation is [tex]s(t) = (2t+1) - \frac{8}{t^2+1}[/tex] Is this correct? Then [tex]\frac{ds}{dt} = 2 - \frac{16t}{(t^2+1)^2}[/tex] Then [tex]t = 0[/tex]
 
Last edited:
sorry wrong equation:

(1+2t)-(8/t^2+1)
 

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