Di ketahui fungsi f(x) = 2x + 1 dan ( f ₀ g ) (x + 1) = - 2x² - 4x - 1 Tentukanlah g ⁻¹ (x) dan g ⁻¹ ( -2 )

[tex]f(x)=2x+1 \\ (f\circ g)(x+1)=-2x^2-4x-1 \\ (f\circ g)(x)=-2(x-1)^2-4(x-1)-1 \\ (f\circ g)(x)=-2(x^2-2x+1)-4x+4-1 \\ (f\circ g)(x)=-2x^2+4x-2-4x+3 \\ (f\circ g)(x)=-2x^2+1 \\ f(g(x))=-2x^2+1 \\ 2g(x)+1=-2x^2+1 \\ 2g(x)=-2x^2 \\ g(x)=-x^2 \\ \\ y=-x^2 \\ x^2=-y \\ x=\pm\sqrt{-y} \\ g^{-1}(x)=\pm\sqrt{-x} \\ \\ g^{-1}(2)=\pm\sqrt{-(-2)} \\ g^{-1}(-2)=\pm\sqrt{2}[/tex]

fog(x+1)= -2 (x-1)^2 - 4(x-1) - 1
fog(x)=-2(x^-2x+1)-4x+4-1
=-2x^2 +4x-2-4x+3
= -2x^2 +1
fog(x) = f(g(x))
-2x^2+1=2(gx)+1
-2x^2+1-1=2g(x)
-2x^2 =2 g(x)
-x^2=g(x)