Cho hàm số bậc ba y = f(x) có đồ thị như hình vẽ

Chọn A

$g\left(x\right)=f\left(f\right(x)+2)$

Có $g^{\text{'}}\left(x\right)=f^{\text{'}}\left(x\right)f^{\text{'}}\left(f\right(x)+2)$.

Ta được $g^{\text{'}}\left(x\right)=0\iff f^{\text{'}}\left(x\right)f^{\text{'}}\left(f\right(x)+2)=0\iff \left[\begin{array}{l}{f}^{\text{'}}\left(x\right)=0\left(1\right)\\ f^{\text{'}}\left(f\right(x)+2)=0\left(2\right)\end{array}\right.$

$$\begin{gathered} \left(1\right)\iff \left[\begin{array}{l}x=-1\\ x=1\end{array}\right. \\ \left(2\right)\iff \left[\begin{array}{l}f\left(x\right)+2=-1\\ f\left(x\right)+2=1\end{array}\right.\iff \left[\begin{array}{l}f\left(x\right)=-3\\ f\left(x\right)=-1\end{array}\right.\iff \left[\begin{array}{l}x=-2\\ x=1\\ x=a\in (-2;-1)\\ x=0\\ x=b\in (1;2)\end{array}\right. \end{gathered}$$

Vậy $g\text{'}\left(x\right)=0\iff \left[\begin{array}{l}x=-2\\ x=-1\\ x=1\\ x=0\\ x=a\in (-2;-1)\\ x=b\in (1;2)\end{array}\right.$

Vậy g'(x) có 6 nghiệm thực phân biệt.