Calculus

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Find the global maximum and minimum values of f(t)=4t/(2+t^2) if its domain is all real numbers?

  • Calculus -

    f' = [4(t^2+2) -4t (2t)]/(t^2+2)^2
    = [-4t^2+8]/(t^2+2)^2
    0 at max or min
    t = + or - sqrt 2

    f" = [ (t^2+2)^2 (-8t) + 4(t^2-2)(2)(t^2+2)(2t) ] / (t^2+2)^4

    if t = +sqrt 2
    t^2+2 = 4
    t^2-2 = 0
    f" = [ -32 sqrt 2 +4*0 etc } /4^4
    which is positive so a minimum

    if t = - sqrt 2
    t^2+2 is still 4
    t^2-2 is still zero
    so
    f" = { +32 sqrt 2 .... etc
    which is negative so a maximum

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