Max/min value of f(x) = x(e^(-3x)) on(-infinity, infinity)

[Sophie]

Fine the absolute maximum and absolute minimum values of f on the given interval.

f(x) = x(e^(-3x)) on(-infinity, infinity)

f'(x) = (e^(-3x)) (1-3x)

f'(x) =0 when x=1/3

f(1/3) = 1/3 (e^-1)
f(-infinity)=infinity
f(infinity) = - infinity

Absolute Max = f(1/3) = 1/3 (e^-1)
there is no Absolute Min as there is no difinitive end

my problem is with infinity, which is not an exact answer and therefore I am confused, but I feel relativly sure my answer is right! My text did not specify absolute values with infinity, I am a little confused. If someone could clarify I would be greatful

Thanks Sophie

 

[ChaoticLlama]

Your analysis is correct up until this point

f(-infinity) -> infinity
f(1/3) = 1/(3e)
f(infinity) -> - infinity

Check that which is highlighted in red again.

However, your final analysis

Absolute Max = f(1/3) = 1/(3e)
there is no Absolute Min as there is no difinitive end

is correct somehow. (although I would change the explanation for the absolute min)