Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $\frac{1}{2}$
There are no x-intercepts.
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \frac{1}{1+e^{-x}} = 0$
$y = 0$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} \frac{1}{1+e^{-x}} = 1$
$y = 1$ is a horizontal asymptote.
E. The function is increasing on the interval $(-\infty, \infty)$
F. There is no local maximum or local minimum.
G. The graph is concave down on the interval $(0, \infty)$
The graph is concave up on the interval $(-\infty, 0)$
The point of inflection is $(0, \frac{1}{2})$
H. We can see a sketch of the curve below.
Work Step by Step
$y = \frac{1}{1+e^{-x}}$
A. The function is defined for all real numbers.
The domain is $(-\infty, \infty)$
B. When $x=0$, then $y = \frac{1}{1+e^{-0}} = \frac{1}{1+1} = \frac{1}{2}$
The y-intercept is $\frac{1}{2}$
When $y = 0$:
$\frac{1}{1+e^{-x}} = 0$
There are no values of $x$ such that $y = 0$
There are no x-intercepts.
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \frac{1}{1+e^{-x}} = 0$
$y = 0$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} \frac{1}{1+e^{-x}} = 1$
$y = 1$ is a horizontal asymptote.
E. We can find the values of $x$ such that $y' = 0$:
$y' = \frac{-(-e^{-x})}{(1+e^{-x})^2} = \frac{e^{-x}}{(1+e^{-x})^2} = 0$
There are no values of $x$ such that $y' = 0$
For all values of $x,~~~$ $y' \gt 0$
The function is increasing on the interval $(-\infty, \infty)$
F. There is no local maximum or local minimum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{(-e^{-x})(1+e^{-x})^2-(e^{-x})(2)(1+e^{-x})(-e^{-x})}{(1+e^{-x})^4}$
$y'' = \frac{(-e^{-x})(1+e^{-x})-(e^{-x})(2)(-e^{-x})}{(1+e^{-x})^3}$
$y'' = \frac{e^{-2x}-e^{-x}}{(1+e^{-x})^3} = 0$
$e^{-2x}-e^{-x} = 0$
$e^{-2x} = e^{-x}$
$e^{-x} = 1$
$\frac{1}{e^x} = 1$
$e^x = 1$
$x = 0$
When $x \gt 0$, then $y'' \lt 0$
The graph is concave down on the interval $(0, \infty)$
When $x \lt 0$, then $y'' \gt 0$
The graph is concave up on the interval $(-\infty, 0)$
When $x = 0$, then $y = \frac{1}{1+e^{-0}} = \frac{1}{1+1} = \frac{1}{2}$
The point of inflection is $(0, \frac{1}{2})$
H. We can see a sketch of the curve below.