ابتدا پیوستگی تابع $f$ را در $x=2$ بررسی می‌کنیم: $f(x)=x\left| x-2 \right|={{x}^{2}}-1\Rightarrow f(x)=\left\{ \begin{matrix}    {{x}^{2}}-2x+{{x}^{2}}-1\,\,\,\,\,\,\,\,x\ge 2  \\    -{{x}^{2}}+2x+{{x}^{2}}-1\,\,\,\,\,x \lt 2  \\ \end{matrix} \right.$ $\Rightarrow f(x)=\left\{ \begin{matrix}    2{{x}^{2}}-2x-1\,\,\,\,\,x\ge 2  \\    2x-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \lt 2  \\ \end{matrix}\Rightarrow \underset{x\to {{2}^{+}}}{\mathop{\lim }}\, \right.f(x)=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f(x)=f(2)=3$ $f\Rightarrow {f}'(x)=\left\{ \begin{matrix}    4x-2\,\,\,\,\,x \gt 2  \\    2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \lt 2  \\ \end{matrix} \right.$  در $x=2$ پیوسته است. $\Rightarrow {{{f}'}_{+}}(2)=6,{{{f}'}_{-}}(2)=2$  $\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\frac{f(2+3h)-f(2)}{\frac{1}{3}(3h)}-\frac{f(2-h)-f(2)}{-(-h)}=+3{{{f}'}_{-}}(2)+{{{f}'}_{+}}(2)=3\times 2+6=12$