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Calculus:Maximum and Minimum

  1. Jan 26, 2005 #1
    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

  2. jcsd
  3. Jan 26, 2005 #2
    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: Jan 26, 2005
  4. Jan 26, 2005 #3
    sorry wrong equation:

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