g'(t) = 8*e^(-2t) - 16t*e^(-2t)
Extreme at g'(t) = 0 = (8-16t)e^(-2t)
e^(-2t) > 0 for all t, and goes to 0 as t goes to infinity
g'(1/2) = 0
g''(t) = [-16 + (-16+32t)]*e^(-2t) = (32t - 32)*e^(-2t)
g''(1/2) < 0, so g(1/2) is the maximum.
So, global max at t=1/2 is g(1/2) = 8*(1/2)/e = 4/e
Global min is approached only in the limit ...
g(t) -> 0 as t -> 0 and as t -> infinity
Global Maximum And Minimum
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Verified answer
g'(t) = 8*e^(-2t) - 16t*e^(-2t)
Extreme at g'(t) = 0 = (8-16t)e^(-2t)
e^(-2t) > 0 for all t, and goes to 0 as t goes to infinity
g'(1/2) = 0
g''(t) = [-16 + (-16+32t)]*e^(-2t) = (32t - 32)*e^(-2t)
g''(1/2) < 0, so g(1/2) is the maximum.
So, global max at t=1/2 is g(1/2) = 8*(1/2)/e = 4/e
Global min is approached only in the limit ...
g(t) -> 0 as t -> 0 and as t -> infinity
Global Maximum And Minimum