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

Jul 24, 2018

The range is $\left(1 , + \infty\right)$

#### Explanation:

The function is

$f \left(t\right) = 1 + 0.9 {e}^{- 0.02 t}$

The domain of this function is $t \in \mathbb{R}$

That is

$t \in \left(- \infty , + \infty\right)$

The limits are

${\lim}_{t \to - \infty} f \left(t\right) = {\lim}_{t \to - \infty} \left(1 + 0.9 {e}^{- 0.02 t}\right)$

$= 1 + 0.9 {e}^{+ \infty}$

$= + \infty$

${\lim}_{t \to + \infty} f \left(t\right) = {\lim}_{t \to + \infty} \left(1 + 0.9 {e}^{- 0.02 t}\right)$

$= 1 + 0.9 {e}^{- \infty}$

$= 1 + 0$

$= 1$

The range is $\left(1 , + \infty\right)$

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