Use f(x) = 2 x and f-'(x) = 2x to solve the problems.f(2)=...F1 (1) =...F1 (12)) = ...

Accepted Solution

Answer:a) 1b) 2c) 2Step-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.