Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.3 Exercises - Page 345: 80

$ \frac{x-3}{2}$

To find the inverses, replace $f(x)$ by $x$ and $ x$ by $ f^{-1}(x)$ and solve for $ f^{-1}(x)$. Thus $ x = f^{-1}(x)+4 ⇒ f^{-1}(x) = x-4$ And $x= 2g^{-1}(x)-5⇒ g^{-1}(x) = \frac{x+5}{2} $ So, $(f^{-1} o g^{-1})(x) = f^{-1}(g^{-1}(x)) = f^{-1}(\frac{x+5}{2}) = \frac{x+5}{2}-4 = \frac{x-3}{2}$