Question

f(t)= 1+.9e^(-.02t) with t ≥ 0
for what value of t is the function decreasing most rapidly? justify your answer

Answer 1

Normally, you would say, try looking for the minimum of the derivative f', because f' describes the way f is changing: A negative value of f' means f is decreasing. So you would go looking for the value of t where f' has its most negative value, or its minimum.
To find it you would have to solve f''(t) =0, so you would have to differentiate f' again! A lot of work :(

Why is all this not needed here?
Well, f is essentially a decreasing exponential function. Adding 1 doesn't change that, nor multiplying it with 0.9. A decreasing exponential function decreases less and less as t grows. In other words, at t =0, at the start of the domain, it decreases most rapidly.
As you see, there are no calculations needed here...