Answer
(a) $(f \circ g)(3)=4$
(b) $g(f(2))=-2$
(c) $g(f(5))= \text { undefined}$
(d) $(f \circ g)(-3)=3 $
(e) $(g \circ f)(-1)=2$
(f) $f(g(-1))= \text { undefined}$
Work Step by Step
According to the graph of the functions $f$ and $g$, we have
$ g(3)=-1, \quad f(-1)=4, \quad f(2) = 1, \quad g(1)= -2, \\ f(5)= -5, \quad g(-3)= -2, \quad f(-2)=3, \quad f(-1)=4, \\ g(4)=2, \quad g(-1)=-4 \, .$
So we have
(a) $(f \circ g)(3)=f(g(3))=f(-1)=4$
(b) $g(f(2))=g(1)=-2$
(c) $g(f(5))=g(-5)= \text { undefined}$ (Since $-5$ is not in the domain of $g$)
(d) $(f \circ g)(-3)= f(g(-3))=f(-2)=3 $
(e) $(g \circ f)(-1)=g(f(-1))=g(4)=2$
(f) $f(g(-1))=f(-4)= \text { undefined}$ (Since $-4$ is not in the domain of $f$).
