Đáp án:
\(x \in \left( {2 - 2\sqrt {10} ;2 + 2\sqrt {10} } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
g'\left( x \right) = - \dfrac{1}{2}f'\left( {1 - \dfrac{x}{2}} \right) + 4\\
= - \dfrac{1}{2}.\left[ {{{\left( {1 - \dfrac{x}{2}} \right)}^2} - 2\left( {1 - \dfrac{x}{2}} \right)} \right] + 4\\
= - \dfrac{1}{2}\left[ {1 - 2x + \dfrac{{{x^2}}}{4} - 2 + x} \right] + 4\\
= - \dfrac{1}{2}\left( {\dfrac{{{x^2}}}{4} - x - 1} \right) + 4\\
= - \dfrac{{{x^2}}}{8} + \dfrac{x}{2} + \dfrac{1}{2} + 4\\
= - \dfrac{{{x^2}}}{8} + \dfrac{x}{2} + \dfrac{9}{2}
\end{array}\)
Để g(x) đồng biến
\(\begin{array}{l}
\to g'\left( x \right) > 0\\
\to - \dfrac{{{x^2}}}{8} + \dfrac{x}{2} + \dfrac{9}{2} > 0\\
\to x \in \left( {2 - 2\sqrt {10} ;2 + 2\sqrt {10} } \right)
\end{array}\)
