Answer:
a) 1
b) 2
c) 2
Step-by-step explanation:
a) [tex]f(2)[/tex] means find the output for [tex]\frac{1}{2}x[/tex] when [tex]x=2[/tex].
[tex]\frac{1}{2}(2)[/tex]
[tex]\frac{2}{2}[/tex]
[tex]1[/tex]
b) [tex]f^{-1}(1)[/tex] means find the output for [tex]2x[/tex] when [tex]x=1[/tex]
[tex]2(1)[/tex]
[tex]2[/tex]
c) If [tex]f[/tex] and [tex]f^{-1}[/tex] are truly inverses then [tex]f(f^{-1}(u))=u[/tex] and [tex]f^{-1}(f(u))=u[/tex] as long as [tex]u[/tex] satisfies the domains.
So we should be able to conclude that [tex]f^{-1}(f(2))=2[/tex] with no work.
However, I will also show work.
[tex]f^{-1}(f(2))[/tex]
[tex]f^{-1}(1)[/tex] since [tex]f(2)=1[/tex] from part a.
[tex]2[/tex] since [tex]f^{-1}(1)=2[/tex] from part b.
