Calculus

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

• Calculus -

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^2-2)(2)(t^2+2)(2t) ] / (t^2+2)^4

if t = +sqrt 2
t^2+2 = 4
t^2-2 = 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^2-2 is still zero
so
f" = { +32 sqrt 2 .... etc
which is negative so a maximum

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