Răspuns:
g(x)={x²-6x+11 x≥3
{x -1 x<3
cazul 1 x≥3
Scrii functia sub forma canonica
g(x)=(x-b/2a)²-Δ/4a
-Δ/4a=-[(-6)²-4*11]/4= -(36-44)/4=8/4=2
g(x)=(x-(-6)/2)²+2=
(x+3)²+2
f(x )=y
y=(x+3)²+2
y-2=(x+3)²
x+3=√(y-2)
x=√(y-2)-3
g(y)=√(y-2)-3
Treci la variabila x
g⁻¹(x)=√(x-2)-3
caz X<3
g(x )=x-1
g(x)=y
y=x-1
x=y+1
g(y)=y+1
Treci la variabila x
g⁻¹(x)=x+1
g⁻¹(x)={√(x-2)-3 x≥3
{x+1 x<3
Explicație pas cu pas:
√(x-2)-3 x≥3
{
