Ta có : f( x - 1 ) = 2 - ( x - 1 )2
=> \(f_{\left(x-1\right)}=2-x^2+2x-1\)
=> \(f_{\left(x-1\right)}=-x^2+2x+1\)
Ta có : \(f_{\left(1-x\right)}=2-\left(1-x\right)^2\)
=> \(f_{\left(1-x\right)}=2-1+2x-x_2\)
=> \(f_{\left(1-x\right)}=1+2x-x^2\)
Vậy \(f_{\left(x-1\right)}=f_{\left(1-x\right)}=\left(-x^2+2x+1\right)\) ( đpcm )
Ta có : \(f\left(x-1\right)=2-\left(x-1\right)^2\)
\(f\left(1-x\right)=2-\left(1-x\right)^2=2-\left(x-1\right)^2\)
\(\Rightarrow f\left(x-1\right)=f\left(1-x\right)\)
(ĐPCM)