\(f^2\left(1+2x\right)=x-f^3\left(1-2x\right)\)
Thay \(x=0\) vào ta được:
\(f^2\left(1\right)=-f^3\left(1\right)\Leftrightarrow f^2\left(1\right)\left[1+f\left(1\right)\right]=0\Rightarrow\left[{}\begin{matrix}f\left(1\right)=0\\f\left(1\right)=-1\end{matrix}\right.\)
Đạo hàm 2 vế:
\(\Rightarrow4f\left(1+2x\right).f'\left(1+2x\right)=1+6f^2\left(1-2x\right).f'\left(1-2x\right)\)
Thay \(x=0\) vào ta được:
\(4f\left(1\right).f'\left(1\right)=1+6f^2\left(1\right).f'\left(1\right)\)
- Nếu \(f\left(1\right)=0\Rightarrow0=1\) (loại)
- Nếu \(f\left(1\right)=-1\Rightarrow-4f'\left(1\right)=1+6f'\left(1\right)\Rightarrow f'\left(1\right)=-\frac{1}{10}\)
