Does $f''(0^+)=f''(0^-)$? Consider the function, which is the join of two semicircles

$$ f(x) = \left\{ \begin{array}{cc} \sqrt{1 - x^2} &amp; x &gt; 0 \\ 
1 &amp; x = 0 \\
\sqrt{ 2 - (x-1)^2} &amp; x &lt; 0\end{array} \right.$$

Near $x = 0$, $f(x)$ is continuous and $f'(0)$ makes sense.  Does $f''(0^+)=f''(0^-)$?
$$f''(x) =
\begin{cases}
-(1-x^2)^{-3/2}      &amp; x &gt; 0 \\
0,                   &amp; x = 0 \\
-2(2-(x-1)^2)^{-3/2} &amp; x &lt; 0
\end{cases}
$$

So what are
$$
\lim_{x \to 0^-} f''(x) \text{ and } \lim_{x \to 0^+} f''(x)?
$$
$$f''(x) =
\begin{cases}
-(1-x^2)^{-3/2}      &amp; x &gt; 0 \\
0,                   &amp; x = 0 \\
-2(2-(x-1)^2)^{-3/2} &amp; x &lt; 0
\end{cases}
$$

So what are
$$
\lim_{x \to 0^-} f''(x) \text{ and } \lim_{x \to 0^+} f''(x)?
$$