Answer
(a) $y=0$ is a horizontal asymptote.
The maximum value is $f(0) = 1$
The inflection points are $(-\sigma, \frac{1}{\sqrt{e}})$ and $(\sigma, \frac{1}{\sqrt{e}})$
(b) The locations of the inflection points depend on $\sigma$
(c) We can see a sketch of the graphs for $\sigma = 1,2,3,4$
Work Step by Step
(a) $f(x) = e^{-x^2/(2\sigma^2)}$
$\lim\limits_{x \to -\infty}e^{-x^2/(2\sigma^2)} = \lim\limits_{x \to -\infty}\frac{1}{e^{x^2/(2\sigma^2)}} = 0$
$\lim\limits_{x \to \infty}e^{-x^2/(2\sigma^2)} = \lim\limits_{x \to \infty}\frac{1}{e^{x^2/(2\sigma^2)}} = 0$
$y=0$ is a horizontal asymptote.
We can find the value of $x$ such that $f'(x) = 0$:
$f'(x) = -\frac{2x}{2\sigma^2}~e^{-x^2/(2\sigma^2)}$
$f'(x) = -\frac{x}{\sigma^2}~e^{-x^2/(2\sigma^2)} = 0$
$x = 0$
We can find the maximum value:
$f(0) = e^{-0^2/(2\sigma^2)} = 1$
The maximum value is $f(0) = 1$
We can find the values of $x$ such that $f''(x) = 0$
$f''(x) = -\frac{1}{\sigma^2}~e^{-x^2/(2\sigma^2)}-(\frac{x}{\sigma^2})(-\frac{2x}{2\sigma^2})~e^{-x^2/(2\sigma^2)}$
$f''(x) = -\frac{1}{\sigma^2}~e^{-x^2/(2\sigma^2)}+\frac{x^2}{\sigma^4}~e^{-x^2/(2\sigma^2)} = 0$
$\frac{x^2}{\sigma^4}~e^{-x^2/(2\sigma^2)} = \frac{1}{\sigma^2}~e^{-x^2/(2\sigma^2)}$
$x^2 = \sigma^2$
$x = \pm \sigma$
$f(-\sigma) = e^{-(-\sigma)^2/(2\sigma^2)} = \frac{1}{\sqrt{e}}$
The inflection points are $(-\sigma, \frac{1}{\sqrt{e}})$ and $(\sigma, \frac{1}{\sqrt{e}})$
(b) The locations of the inflection points depend on $\sigma$
(c) We can see a sketch of the graphs for $\sigma = 1,2,3,4$
