Question

Imaginea functie
f de x = ( x^2-x+1):(x^2+x+1) .

Answer 1

[tex]\frac{x^2-x+1}{x^2+x+1}=y \\ \\ \frac{x^2-x+1}{x^2+x+1}\left(x^2+x+1\right)=y\left(x^2+x+1\right) \\ \\ x^2-x+1=y\left(x^2+x+1\right) \\ \\ \:x^2-x+1=y\left(x^2+x+1\right)\\x^2-x+1=y\left(x^2+x+1\right)\\x^2-x+1-y=yx^2+yx\\x^2-x+1-y-yx=yx^2\\x^2-x+1-y-yx-yx^2=yx^2-yx^2\\\left(1-y\right)x^2-\left(1-y\right)x+1-y=0\ \\left(-1-y\right)^2-4\left(1-y\right)\left(1-y\right)=\left(-y-1\right)^2-4\left(y^2-2y+1\right)=1+2y+y^2-4\left(1-2y+y^2\right)=1+2y+y^2-4+8y-4y^2=-3y^2+10y-3\\=-3y^2+10y-3 \\ \frac{1}{3}\le \:f\left(x\right)\le \:3