Lời giải

• \(g(x) = f(x^2 + x) \Rightarrow g'(x) = (2x + 1)f'(x^2 + x) \)
• \( g'(x) \geq 0 \)

\(\Rightarrow \begin{cases}
2x + 1 \geq 0 \\
f'(x^2 + x) \geq 0
\end{cases}
\text{ hoặc } 
\begin{cases}
2x + 1 \leq 0 \\
f'(x^2 + x) \leq 0
\end{cases}\)

\( \Leftrightarrow \) \(\begin{cases}
x \geq -\frac{1}{2} \\
-1 \leq x^2 + x \leq 1 \quad \lor \quad x^2 + x \geq 4
\end{cases} \)

hoặc \( \begin{cases}
x \leq -\frac{1}{2} \\
x^2 + x \leq -1 \quad \lor \quad 1 \leq x^2 + x \leq 4
\end{cases}\)

\( \Leftrightarrow \begin{cases} 
x \geq -\frac{1}{2} \\ 
\frac{-1 - \sqrt{5}}{2} \leq x \leq \frac{-1 + \sqrt{5}}{2} \quad \lor \quad x \leq \frac{-1 -\sqrt{17}}{2} \quad \lor \quad x \geq \frac{-1 + \sqrt{17}}{2} 
\end{cases} \)

hoặc

\( \begin{cases}
x \leq -\frac{1}{2} \\
\frac{-1 -\sqrt{17}}{2} \leq x \leq \frac{-1 - \sqrt{5}}{2} \quad \lor \quad \frac{-1 + \sqrt{5}}{2} \leq x \leq \frac{-1 + \sqrt{17}}{2}
\end{cases} \)

\( \Leftrightarrow \) \( \left[ \begin{array}{} 
-\frac{1}{2} \leq x \leq \frac{-1 + \sqrt{5}}{2} \quad \lor \quad x \geq \frac{-1 + \sqrt{17}}{2} \\
\frac{-1 - \sqrt{17}}{2} \leq x \leq \frac{-1 - \sqrt{5}}{2}
\end{array} \right. \)

Hàm \(f(x)\) có 5 cực trị.