We have [tex]f\left(f^{-1}(x)\right) = x[/tex] for inverse functions [tex]f(x)[/tex] and [tex]f^{-1}(x)[/tex]. Then if [tex]f(x) = 2x+5[/tex], we have
[tex]f\left(f^{-1}(x)\right) = 2f^{-1}(x) + 5 = x \implies f^{-1}(x) = \dfrac{x-5}2[/tex]
Then
[tex]f^{-1}(8) = \dfrac{8-5}2 = \boxed{\dfrac32}[/tex]
