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Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = (36 − t^2)^ 1/t, [−1, 6]Asked by TayB
Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = (e^−x)*(− e^−6x), [0, 1]
f(x) = (e^−x)*(− e^−6x), [0, 1]
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Answered by
Steve
well, (e^−x)*(−e^−6x) = -e^-7x
as you know exponentials have no local max/min, so the extrema are at the ends of the interval.
That should pose no difficulty, if you sketch the graph.
as you know exponentials have no local max/min, so the extrema are at the ends of the interval.
That should pose no difficulty, if you sketch the graph.
Answered by
TayB
That's not the derivative of the function tho Steve.
Answered by
Steve
well, the derivative is 7e^-7x
where is that zero?
where is that zero?
Answered by
TayB
no the derivative is not 7e^-7x
when the function is e^-x-e^-6x
when the function is e^-x-e^-6x
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