Ta có \[g\left[ x \right] = f\left[ {\left| {{e^{2x}} - 2x - 2} \right|} \right] = f\left[ {\sqrt {{{\left[ {{e^{2x}} - 2x - 2} \right]}^2}} } \right]\]
\[\begin{array}{l} \Rightarrow g'\left[ x \right] = \sqrt {{{\left[ {{e^{2x}} - 2x - 2} \right]}2}} '.f'\left[ {\sqrt {{{\left[ {{e{2x}} - 2x - 2} \right]}2}} } \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2\left[ {{e{2x}} - 2x - 2} \right]\left[ {2{e^{2x}} - 2} \right]}}{{2\sqrt {{{\left[ {{e^{2x}} - 2x - 2} \right]}2}} }}f'\left[ {\sqrt {{{\left[ {{e{2x}} - 2x - 2} \right]}2}} } \right]\\g'\left[ x \right] = 0 \Leftrightarrow \left[ \begin{array}{l}{e{2x}} - 2x - 2 = 0\\2{e^{2x}} - 2 = 0\\f'\left[ {\sqrt {{{\left[ {{e^{2x}} - 2x - 2} \right]}^2}} } \right] = 0\end{array} \right.\end{array}\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}{e^{2x}} - 2x - 2 = 0\\x = 0\\\sqrt {{{\left[ {{e^{2x}} - 2x - 2} \right]}2}} = a < - 1\,\,\,\left[ {Loai} \right]\\\sqrt {{{\left[ {{e{2x}} - 2x - 2} \right]}2}} = b \in \left[ { - 1;0} \right]\,\,\left[ {Loai} \right]\\\sqrt {{{\left[ {{e{2x}} - 2x - 2} \right]}2}} = c \in \left[ {0;1} \right]\\\sqrt {{{\left[ {{e{2x}} - 2x - 2} \right]}^2}} = d > 1\end{array} \right.\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}{e^{2x}} - 2x - 2 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ 1 \right]\\x = 0\\{e^{2x}} - 2x - 2 = c \in \left[ {0;1} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ 2 \right]\\{e^{2x}} - 2x - 2 = - c \in \left[ { - 1;0} \right]\,\,\,\,\,\,\,\,\,\,\,\left[ 3 \right]\\{e^{2x}} - 2x - 2 = d,\,\,d > 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ 4 \right]\\{e^{2x}} - 2x - 2 = - d,\,\, - d < - 1\,\,\,\,\,\,\,\,\,\left[ 5 \right]\end{array} \right.\]
Xét hàm số \[h\left[ x \right] = {e^{2x}} - 2x - 2\] ta có \[h'\left[ x \right] = 2{e^{2x}} - 2 = 0 \Leftrightarrow x = 0\].
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