Chapter 4 - Exponential and Logarithmic Functions - Section 4.2 One-to-One Functions; Inverse Functions - 4.2 Assess Your Understanding - Page 292: 71

$f^{-1}(x)=\frac{2}{\sqrt {1-2x}}$

Step 1. $f(x)=\frac{x^2-4}{2x^2} \Longrightarrow y=\frac{x^2-4}{2x^2} \Longrightarrow x=\frac{y^2-4}{2y^2} \Longrightarrow y=\frac{2}{\sqrt {1-2x}} \Longrightarrow f^{-1}(x)=\frac{2}{\sqrt {1-2x}}$ Step 2. Check $f(f^{-1}(x))=\frac{(\frac{2}{\sqrt {1-2x}})^2-4}{2(\frac{2}{\sqrt {1-2x}})^2}=x$ and $f^{-1}(f(x))=\frac{2}{\sqrt {1-2(\frac{x^2-4}{2x^2})}}=x$