Asked by Carly
Sorry I have been trying to figure this out all day. Find the largest and smallest values of the given function over the prescribed closed, bounded interval.
g(t)= (t^3/2)(e^-2t) for 0<t<1.
g(t)= (t^3/2)(e^-2t) for 0<t<1.
Answers
Answered by
Reiny
Using the product rule
g'(t) = t^(3/2)(-2)e^(-2t) + (3/2)t^(1/2)(e^(-2t))
common factor ...
= (1/2)e^(-2t)(t^(1/2)) [-4t + 3}
set this equal to zero
from the first factor we get t=0
and from the second one we get t = 3/4
now evaluate
g(0) = ..
g(3/4) = ..
g(1) = ...
and determine which is the largest and smallest value.
g'(t) = t^(3/2)(-2)e^(-2t) + (3/2)t^(1/2)(e^(-2t))
common factor ...
= (1/2)e^(-2t)(t^(1/2)) [-4t + 3}
set this equal to zero
from the first factor we get t=0
and from the second one we get t = 3/4
now evaluate
g(0) = ..
g(3/4) = ..
g(1) = ...
and determine which is the largest and smallest value.
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