Phương pháp giải:

+) Xác định hàm số \(h\left( x \right) = f\left( x \right) - g\left( x \right)\).

+) \(h\left( x \right) \ge m\,\,\forall x \in \left[ { - 3;3} \right] \Rightarrow m \le \mathop {min}\limits_{\left[ { - 3;3} \right]} h\left( x \right)\).

Lời giải chi tiết:

Dựa vào đồ thị hàm số ta thấy \(f\left( x \right) - g\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x =  - 3\,\,\left( {boi\,\,2} \right)\\x =  - 1\\x = 3\end{array} \right.\)

Khi đó \(f\left( x \right) - g\left( x \right)\) được viết dưới dạng \(h\left( x \right) = a{\left( {x + 3} \right)^2}\left( {x + 1} \right)\left( {x - 3} \right)\,\,\left( {a \ne 0} \right)\).

Ta có: \(f\left( 0 \right) =  - 1;\,\,g\left( 0 \right) =  - 2 \Rightarrow f\left( 0 \right) - g\left( 0 \right) = 1\)

\( \Rightarrow  - 27a = 1 \Leftrightarrow a =  - \frac{1}{{27}} \Leftrightarrow h\left( x \right) =  - \frac{1}{{27}}\left( {{x^4} + 4{x^3} - 6{x^2} - 36x - 27} \right)\)

\(\begin{array}{l}f(x) \ge g(x) + m\,\,\forall x \in \left[ { - 3;3} \right] \Leftrightarrow f\left( x \right) - g\left( x \right) \ge m\,\,\forall x \in \left[ { - 3;3} \right]\\ \Rightarrow m \le \mathop {\min }\limits_{\left[ { - 3;3} \right]} h\left( x \right),\,\,h\left( x \right) = f\left( x \right) - g\left( x \right) =  - \frac{1}{{27}}\left( {{x^4} + 4{x^3} - 6{x^2} - 36x - 27} \right)\end{array}\)

Ta có \(h'\left( x \right) = \frac{1}{{27}}\left( {4{x^3} + 12{x^2} - 12x - 36} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x =  - 3\\x = \sqrt 3 \\x =  - \sqrt 3 \end{array} \right.\)

\(h\left( { - 3} \right) = 0,\,\,\,h\left( 3 \right) = 0,\,\,\,h\left( {\sqrt 3 } \right) = \frac{{12 + 8\sqrt 3 }}{9};\,\,h\left( { - \sqrt 3 } \right) = \frac{{12 - 8\sqrt 3 }}{9}\).

\( \Rightarrow \mathop {\min }\limits_{\left[ { - 3;3} \right]} h\left( x \right) = \frac{{12 - 8\sqrt 3 }}{9} \Leftrightarrow m \le \frac{{12 - 8\sqrt 3 }}{9}\).

Chọn A.