f' = 7-3/t^2
f'=0 when t=±√(3/7)
f" = 6/t^3
f" > 0 when t>0 so min at t=√(3/7)
f" < 0 when t<0 so max at t=-√(3/7)
Find the relative extrema, if any, of the function. Use the second derivative test, if applicable. (If an answer does not exist, enter DNE.)
f(t) = 7 t + 3/t
relative maximum (x, y) =
relative minimum (x, y) =
2 answers
Eric Johnson