Ta có: \(f\left(x\right)=x^2-1\)
\(\Rightarrow f\left(1-x_0\right)=\left(1-x_0\right)^2-1\)
\(=x_0^2-2x_0+1-1=x_0^2-2x_0\)
\(=x_0\left(x_0-2\right)\)
\(f\left(1-x_0\right)< 0\Leftrightarrow\)\(x_0\left(x_0-2\right)< 0\)
Mà \(x_0>x_0-2\)nên \(\hept{\begin{cases}x_0>0\\x_0-2< 0\end{cases}}\Leftrightarrow0< x_0< 2\)
Vậy \(0< x_0< 2\)thì \(f\left(1-x_0\right)\)đạt giá trị âm