Cho hàm số y=f(x). Hàm số y=f'(x) có đồ thị như hì...

A

- Hướng dẫn giải

\(\begin{array}{l}\begin{array}{*{20}{l}}{f'(x) = \frac{{ - {x^3}}}{2} - \frac{{{x^2}}}{2} + 7x}\\{f(x) = \frac{{ - {x^4}}}{8} - \frac{{{x^3}}}{6} + \frac{7}{2}{x^2}}\\{g\left( x \right) = f\left( {3{x^2} - 1} \right) - \frac{9}{2}{x^4} + 3{x^2}}\\{g'(x) = 6xf'\left( {3{x^2} - 1} \right) - 18{x^3} + 6x}\\{ = 6x\left( {f'\left( {3{x^2} - 1} \right) - \left( {3{x^2} - 1} \right)} \right) = 0}\\{ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = 0}\\{f'\left( t \right) - t = 0}\end{array}} \right.}\\{f'\left( t \right) - t = 0 \Rightarrow \frac{{ - {t^3}}}{2} - \frac{{{t^2}}}{2} + 7t - t = 0}\\\begin{array}{l} \Rightarrow \left[ {\begin{array}{*{20}{l}}{t = 0}\\{t = 3}\\{x = - 4}\end{array}} \right.\\ \Rightarrow \left[ \begin{array}{l}x = 0\\3{x^2} - 1 = 0\\3{x^2} - 1 = 3\\3{x^2} - 1 = - 4\end{array} \right.\end{array}\end{array}\\ \Rightarrow x \in \left\{ { \pm \frac{{2\sqrt 3 }}{3}; \pm \frac{{\sqrt 3 }}{3};0} \right\}\end{array}\)