- Ca un punct A(x,y) sa apartina unui Gf ⇒ f(x)=y
a. f(x)=-x+1
A(-1,2)
f(-1)=1+1=2⇒ A∈Gf
B(2,3)
f(2)=-2+1=-1⇒ B∉Gf
C(5,-1)
f(5)=-5+1=-4 ⇒ C∉Gf
D(4,-3)
f(4)=-4+3=-1⇒ D∈Gf
E(3,3)
f(3)=-3+1=-2⇒ E∉Gf
b.f(x)=-2x+6
A(3,0)
f(3)=-6+6=0⇒A∈Gf
B(1,-4)
f(1)=-2+6=4⇒ B∉Gf
C(-2,10)
f(-2)=4+6=10⇒C∈Gf
D(-1,4)
f(-1)=2+6=8⇒D∉Gf
E(2,2)
f(2)=-4+6=2⇒ E∈Gf
c. f(x)=-4x+11
[tex]A(\frac{1}{2},8)\\\\ f(\frac{1}{2})=-2+11=9[/tex]⇒ A∉Gf
[tex]B(\frac{11}{4} ,0)\\\\f(\frac{11}{4})=-11+11=0[/tex]⇒B∈Gf
[tex]C(-\frac{3}{4} ,6)\\\\f(-\frac{3}{4} )=3+11=14[/tex] ⇒C∉Gf
[tex]D(\frac{7}{3} ,\frac{5}{3} )\\\\f(\frac{7}{3} )=-\frac{28}{3} +11=\frac{5}{3}[/tex]⇒D∈Gf
[tex]E(\frac{11}{5} ,\frac{11}{5})\\\\f(\frac{11}{5})=-\frac{44}{5} +11=\frac{11}{5}[/tex]⇒E∈Gf
d.f(x)=3x-1
A(0,-1)
f(0)=0-1=-1⇒ A∈Gf
B(-1,4)
f(-1)=-3-1=-4⇒B∉Gf
[tex]C(\frac{1}{3},0) \\\\f(\frac{1}{3})=1-1=0[/tex]⇒C∈Gf
[tex]D(\frac{2}{9},\frac{1}{3})\\\\ f(\frac{2}{9} )=\frac{2}{3}-1=-\frac{1}{3}[/tex]⇒ D∉Gf
[tex]E(\frac{1}{2} ,\frac{1}{2})\\\\f(\frac{1}{2})=\frac{3}{2} -1=\frac{1}{2}[/tex]⇒E∈Gf
