Question

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

Answer 1

kush^420

Answer 2

The range of the function consists of all possible values f(t) can assume, for values of t from 0 to inf.
Since we now know that f(t) is strictly decreasing, the maximum value (a) of the range is at t=0, or f(0). The minimum value of f(t) is when t=inf (b).
Can you find these values a and b? The range is therefore [a,b).

Answer 3

would the range be [1.9, 1) because (1+.9(1)) and (1+.9(0))

Answer 4

Almost, the numbers are correct, but not in the right place (my bad).
The range is expressed in smaller number first, so it is
(1, 1.9].
The parentheses are reversed because 1.9 corresponds to f(0), so it is included.
1.0 corresponds to f(x)->1 as x-<>infinity, which is excluded (so use "(" ).

Answer 5

* x-> infinity

Answer 6

wonderful thank you so much!

Answer 7

yw! :)