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Find the critical number of g(t)=(t^2)/(5t+4)

  • calculus -

    vertical asymptotes where denominator is zero

    By now you should know all about how 1st and 2nd derivatives tell you about critical points:

    g' = t(5t+8)/(5t+4)^2
    g'' = 32/(5t+4)^3

    g has max/min where g'=0 and g'' not 0
    g has inflection where g'' = 0
    g increasing where g' > 0
    g concave up where g'' > 0

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