Answer
See proof
Work Step by Step
Rewrite the given function:
$$\begin{align*}
g(x)=\begin{cases}
-x^2,x<0\\
x^2,x\geq 0
\end{cases}
\end{align*}$$
The first derivative $g'(x)$ is:
$$\begin{align*}
g'(x)=\begin{cases}
-2x,x<0\\
2x,x>0
\end{cases}
\end{align*}$$
The second derivative $g''(x)$ is:
$$\begin{align*}
g''(x)=\begin{cases}
-2,x<0\\
2,x\geq 0
\end{cases}
\end{align*}$$
Since $g''(x)<0$ for $x<0$ and $g''(x)>0$ for $x\gt0$, it follows that
$(0,0)$ is an inflection point even though $g''(0)$ does not exist.
