Question

attach file ,please!

Answer 1

i could have sworn i just did this problem

Answer 2

but no matter we do it again

Answer 3

\[g\circ f\] first
\[g\circ f(1)=g(f(1))=g(1)=0\] so we have (1,0)
\[g\circ f(0)=g(f(0))=g(-3)=1\] so two ordered pairs for
\[g\circ f\] are
\[\{(1,0),(0,1)\}\]

Answer 4

now for
\[f\circ g\]
\[f\circ g(1)=f(g(1))=f(0)=-3\] giving (1,-3)
\[f\circ g(-3)=f(g(-3))=f(1)=1\] giving (-3,1) and finally
\[f\circ g(2)=f(g(2))=f(1)=1\]giving (2,1)
ordered pairs for
\[f\circ g\] are \[\{(1,-3),(-3,1),(2,1)\}\]