Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $\frac{1}{4}$
There are no x-intercepts.
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} (1+e^x)^{-2} = \lim\limits_{x \to -\infty} \frac{1}{(1+e^x)^2} = 1$
$y = 1$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} (1+e^x)^{-2} = \lim\limits_{x \to \infty} \frac{1}{(1+e^x)^2} = 0$
$y = 0$ is a horizontal asymptote.
E. The function is decreasing on the interval $(-\infty,\infty)$
F. There is no local minimum or local maximum.
G. The graph is concave up on the interval $(-0.693, \infty)$
The graph is concave down on the interval $(-\infty, -0.693)$
The point of inflection is $(-0.693, 0.44)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = (1+e^x)^{-2}$
A. The function is defined for all real numbers.
The domain is $(-\infty, \infty)$
B. When $x=0$, then $y = (1+e^0)^{-2} = \frac{1}{(1+1)^2} = \frac{1}{4}$
The y-intercept is $\frac{1}{4}$
When $y = 0$:
$(1+e^x)^{-2} = 0$
$\frac{1}{(1+e^x)^2} = 0$
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. $\lim\limits_{x \to -\infty} (1+e^x)^{-2} = \lim\limits_{x \to -\infty} \frac{1}{(1+e^x)^2} = 1$
$y = 1$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} (1+e^x)^{-2} = \lim\limits_{x \to \infty} \frac{1}{(1+e^x)^2} = 0$
$y = 0$ is a horizontal asymptote.
E. We can try to find the values of $x$ such that $y' = 0$:
$y' = -2(1+e^x)^{-3}~e^x = -\frac{2e^x}{(1+e^x)^3}$
There are no values of $x$ such that $y' = 0$
For all values of $x,~~~~$ $y' \lt 0$
The function is decreasing on the interval $(-\infty,\infty)$
F. There is no local minimum or local maximum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = -\frac{(2e^x)(1+e^x)^3 - (2e^x)(3)(1+e^x)^2(e^x)}{(1+e^x)^6}$
$y'' = -\frac{(2e^x)(1+e^x) - (2e^x)(3)(e^x)}{(1+e^x)^4}$
$y'' = -\frac{2e^x+2e^{2x} - 6e^{2x}}{(1+e^x)^4}$
$y'' = \frac{4e^{2x} - 2e^x}{(1+e^x)^4} = 0$
$4e^{2x}-2e^x = 0$
$4e^{2x}=2e^x$
$2e^x=1$
$e^x = \frac{1}{2}$
$x = ln(\frac{1}{2})$
$x = -0.693$
When $-0.693 \lt x$, then $y'' \gt 0$
The graph is concave up on the interval $(-0.693, \infty)$
When $x \lt -0.693$, then $y'' \lt 0$
The graph is concave down on the interval $(-\infty, -0.693)$
When $x=-0.693$, then $y = (1+e^{-0.693})^{-2} = 0.44$
The point of inflection is $(-0.693, 0.44)$
H. We can see a sketch of the curve below.