# calculus

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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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