Question

How do you find the range of the function: `f(t) = 1 + 0.9 e ^(-0.02t)`?

Answer 1

The range is `(1,+oo)`

Explanation:

The function is

`f(t)=1+0.9e^(-0.02t)`

The domain of this function is `t in RR`

That is

`t in (-oo, +oo)`

The limits are

`lim_(t->-oo)f(t)=lim_(t->-oo)(1+0.9e^(-0.02t))`

`=1+0.9e^(+oo)`

`=+oo`

`lim_(t->+oo)f(t)=lim_(t->+oo)(1+0.9e^(-0.02t))`

`=1+0.9e^(-oo)`

`=1+0`

`=1`

The range is `(1,+oo)`

graph{1+0.9e^(-0.02x) [-213.7, 213.6, -107, 106.9]}