Cho hàm số \(y=f(x)\) có bảng biên thiên như hình vẽHàm số \(g\left( x \right) = f\left( {2{x^2} - \frac{5}{2}x - \frac{3}{2}}

* Hướng dẫn giải

Dựa vào bảng biến thiên, suy ra \(f'\left( x \right) > 0 \Leftrightarrow \left[ \begin{array}{l}
x <  - 2\\
x > 3
\end{array} \right.\) và \(f'\left( x \right) < 0 \Leftrightarrow  - 2 < x < 3.\)

Ta có \(g'\left( x \right) = \left( {4x - \frac{5}{2}} \right)f'\left( {2{x^2} - \frac{5}{2}x - \frac{3}{2}} \right).\) Xét \(g'\left( x \right) < 0 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4x - \frac{5}{2} > 0\\
f'\left( {2{x^2} - \frac{5}{2}x - \frac{3}{2}} \right) < 0
\end{array} \right.\\
\\
\left\{ \begin{array}{l}
4x - \frac{5}{2} < 0\\
f'\left( {2{x^2} - \frac{5}{2}x - \frac{3}{2}} \right) > 0
\end{array} \right.
\end{array} \right..\)

\(\left\{ \begin{array}{l}
4x - \frac{5}{2} > 0\\
f'\left( {2{x^2} - \frac{5}{2}x - \frac{3}{2}} \right) < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > \frac{5}{8}\\
 - 2 < 2{x^2} - \frac{5}{2}x - \frac{3}{2} < 3
\end{array} \right. \Leftrightarrow 1 < x < \frac{9}{4}.\)

\(\left\{ \begin{array}{l}
4x - \frac{5}{2} < 0\\
f'\left( {2{x^2} - \frac{5}{2}x - \frac{3}{2}} \right) > 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < \frac{5}{8}\\
2{x^2} - \frac{5}{2}x - \frac{3}{2} > 3
\end{array} \right.\\
\\
\left\{ \begin{array}{l}
x < \frac{5}{8}\\
2{x^2} - \frac{5}{2}x - \frac{3}{2} <  - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x <  - 1\\
\\
\\
\frac{1}{4} < x < \frac{5}{8}
\end{array} \right..\)

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