# How do you find the Vertical, Horizontal, and Oblique Asymptote given (e^x)/(1+e^x)?

Jul 17, 2016

The function has horizontal asymptotes at $f \left(x\right) = 1$ and $f \left(x\right) = 0$ and a point of inflection at the point $\left(0 , \frac{1}{2}\right)$

#### Explanation:

$f \left(x\right) = {e}^{x} / \left(1 + {e}^{x}\right)$

This can also be written as:
$f \left(x\right) = \frac{1}{\frac{1}{e} ^ x + 1}$

Now consider $f \left(x\right)$ as x→+∞ and x→−∞

${\lim}_{\text{x->+oo}} f \left(x\right) = \frac{1}{0 + 1} = 1$

${\lim}_{\text{x->-oo" f(x)= lim_"x->+oo}} \frac{1}{{e}^{x} + 1} = 0$

Consider the graph of f(x) below. By inspection we can see that limiting values arrived at above form horizontal asymptotes.

Evaluating $f \left(0\right)$ as a point of interest.

$f \left(0\right) = {e}^{0} / \left(1 + {e}^{0}\right) = \frac{1}{1 + 1} = \frac{1}{2}$

Notice that the curvature of $f \left(x\right)$ changes at $\left(0 , \frac{1}{2}\right)$ from concave downward to concave upward. This indicates a point of inflection.

graph{e^x/(1+e^x) [-3.08, 3.08, -1.538, 1.54]}

Hence:

The function has horizontal asymptotes at $f \left(x\right) = 1$ and $f \left(x\right) = 0$ and a point of inflection at the point $\left(0 , \frac{1}{2}\right)$