Find the interval on which the curve y= integral x to 0 of 6/(1+2t+t^2) dt is concave upward
2 answers
your limits make no sense.
y is concave up if y" > 0
By the 2nd FT of Calculus,
y' = -f(x) = -6/(x+1)^2
y" = 12(x+1)^3
so, where is y" positive?
By the 2nd FT of Calculus,
y' = -f(x) = -6/(x+1)^2
y" = 12(x+1)^3
so, where is y" positive?