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

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^22)(2)(t^2+2)(2t) ] / (t^2+2)^4
if t = +sqrt 2
t^2+2 = 4
t^22 = 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^22 is still zero
so
f" = { +32 sqrt 2 .... etc
which is negative so a maximum