Answer
Concave upward on
$$
(-\infty, -\frac{1}{\sqrt 2} ), \quad (\frac{1}{\sqrt 2}, \infty ),
$$
concave downward on
$$
( -\frac{1}{\sqrt 2} ,\frac{1}{\sqrt 2} )
$$
inflection points at
$$(-\frac{1}{\sqrt 2} , \frac{2}{\sqrt{e}}),\quad\quad (\frac{1}{\sqrt 2} , \frac{2}{\sqrt{e}}).$$
Work Step by Step
$$
f(x)=2e^{-x^{2}}
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=2(-2x)e^{-x^{2}}\\
&=-4xe^{-x^{2}} ,
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=-4x((-2x)e^{-x^{2}})-4e^{-x^{2}}(1) ,\\
&=8x^{2}e^{-x^{2}}-4e^{-x^{2}}\\
&=[2x^{2}-1]4e^{-x^{2}}
\end{aligned}
$$
We factor $f^{\prime\prime}(x)$ as
$$
\begin{aligned}
f^{\prime\prime}(x) &=[(\sqrt {2}x-1)(\sqrt {2}x+1)]4e^{-x^{2}}
\end{aligned}
$$
look for an $x$-value that makes $f^{\prime\prime}(x)=0$.
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime\prime}(x) =[(\sqrt {2}x-1)(\sqrt {2}x+1)]4e^{-x^{2}}&=0 \\
x &=-\frac{1}{\sqrt 2}\quad\quad \text {or}\quad\quad x =\frac{1}{\sqrt 2}
\end{aligned}
$$
(1)
Test a number in the interval $(-\infty, -\frac{1}{\sqrt 2} )$ say $ -2$
$$
\begin{aligned}
f^{\prime\prime}(-2) & =[2(-2)^{2}-1]4e^{-(-2)^{2}}\\
&=\frac{28}{e^4} \\
&\gt0
\end{aligned}
$$
to see that $ f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $(-\infty, -\frac{1}{\sqrt 2} )$.
(2)
Test a number in the interval $( -\frac{1}{\sqrt 2} ,\frac{1}{\sqrt 2} )$ say $ 0$
$$
\begin{aligned}
f^{\prime\prime}(0) & =[2(0)^{2}-1]4e^{-(0)^{2}}\\
&=-4 \\
&\lt 0
\end{aligned}
$$
to see that $ f^{\prime\prime}(x)$ is negative in that interval,, so $f(x)$ is concave downward on $( -\frac{1}{\sqrt 2} ,\frac{1}{\sqrt 2} )$
(3)
Test a number in the interval $ (\frac{1}{\sqrt 2}, \infty )$ say $ 2$
$$
\begin{aligned}
f^{\prime\prime}(2) & =[2(2)^{2}-1]4e^{-(2)^{2}}\\
&=\frac{28}{e^4} \\
&\gt0
\end{aligned}
$$
to see that $ f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $ (\frac{1}{\sqrt 2}, \infty )$.
Therefore,
Concave upward on
$$
(-\infty, -\frac{1}{\sqrt 2} ), (\frac{1}{\sqrt 2}, \infty ),
$$
concave downward on
$$
( -\frac{1}{\sqrt 2} ,\frac{1}{\sqrt 2} )
$$
Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $ -\frac{1}{\sqrt 2} $, the graph changes from concave upward to concave downward at that point, so $(-\frac{1}{\sqrt 2} , f(-\frac{1}{\sqrt 2})$ is inflection point.
Also, the sign of $f^{\prime\prime}(x)$ changes from negative to positive
at $ \frac{1}{\sqrt 2} $, the graph changes from concave downward to concave upward at that point, so $(\frac{1}{\sqrt 2} , f(\frac{1}{\sqrt 2})$ is inflection point.
Now ,we find $f( -\frac{1}{\sqrt 2}), f(\frac{1}{\sqrt 2})$as follows:
$$
f(-\frac{1}{\sqrt 2})=2e^{-(-\frac{1}{\sqrt 2})^{2}}=\frac{2}{\sqrt{e}},
$$
$$
f(\frac{1}{\sqrt 2})=2e^{-(\frac{1}{\sqrt 2})^{2}}=\frac{2}{\sqrt{e}}
$$
Finally, we have inflection points at $(-\frac{1}{\sqrt 2} , \frac{2}{\sqrt{e}}), (\frac{1}{\sqrt 2} , \frac{2}{\sqrt{e}})$.
