# What is the range of the function f(x)=1/(1+e^-x)?

Aug 10, 2017

The range is $\left(0 , 1\right)$

#### Explanation:

The function is $f \left(x\right) = \frac{1}{1 + {e}^{-} x}$

The domain of the functions ${e}^{x}$ and ${e}^{-} x$ are $\mathbb{R}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R}$

We calculate the limits to $\pm \infty$ of $f \left(x\right)$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{1}{1 + {e}^{-} x} = \frac{1}{1 + 0} = 1$

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{1}{1 + {e}^{-} x} = \frac{1}{1 + \infty} = 0$

Therefore,

The range is $\left(0 , 1\right)$

graph{1/(1+e^-x) [-7.9, 7.904, -3.95, 3.95]}