Question

f:R->R, f(x)=-x+2
Sa se rezolve ecuatia: f²(x)-f(f(x))= -2

Answer 1

$f^{2} (x)=(- x)^{2} -4x+ 2^{2}$

$f^{2}(x)= x^{2} -4x+4$

f(f(x))=f(-x+2) <=> x-2+2=x

$f^{2}(x)-f(f(x))= x^{2} -5x+4$=-2 <=>$x^{2} -5x+6=0$

Δ=25-24=1  =>x1=3,x2=2

Answer 2

f(x)=-x+2 => f²(x)=(-x+2)²=(2-x)²=4-4x+x² si f(f(x))=f(-x+2)=-(-x+2)+2=x-2+2=x .Atunci: f²(x)-f(f(x))=-2 <=> 4-4x+x²-x=-2 <=> x²-5x+4=-2 <=> x²-5x+4+2=0 <=> x²-5x+6=0 <=> a=1,b=-5,c=6.delta=25-24=1 => x1=(5-1)/2=4/2=2 si x2=(5+1)/2=6/2=3. deci x€{2;3}