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.

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.

To find the largest and smallest values of the given function g(t) = (t^(3/2))(e^(-2t)) over the closed and bounded interval 0 < t < 1, we can use the concepts of calculus.

First, we will find the critical points of the function within the given interval. Critical points occur where the derivative of the function is either zero or undefined.

1. Find the derivative of g(t) with respect to t:
g'(t) = [(3/2)t^(1/2)e^(-2t)] + [(t^(3/2))(-2e^(-2t))]

2. Set g'(t) equal to zero and solve for t:
[(3/2)t^(1/2)e^(-2t)] + [(t^(3/2))(-2e^(-2t))] = 0

Unfortunately, this equation is difficult to solve analytically. However, we can use numerical methods like the Newton-Raphson method to approximate the solution. Alternatively, you can use graphing software or calculators that can find the zeros of a function.

Once you find the critical points within the interval 0 < t < 1, evaluate g(t) at these critical points, as well as at the endpoints t = 0 and t = 1. The largest and smallest values of g(t) will be the maximum and minimum values among these points.

It's worth mentioning that finding the exact maximum and minimum values of a function over a closed, bounded interval can sometimes be challenging. In such cases, numerical approximation methods prove useful.