Chapter 2, Functions - Section 2.8 - One-to-One Functions and their Inverses - 2.8 Exercises - Page 263: 67

$f^{-1}(x) = \frac{x^{2} - 5}{8}, x\geq0$

$f(x) = \sqrt {5 + 8x}$ $y = \sqrt {5 + 8x}$ $y^{2} = ({\sqrt {5 + 8x}})^{2}$ $y^{2} = 5 + 8x$ $y^{2} - 5 = 8x$ $\frac{y^{2} - 5}{8} = x$ $\frac{x^{2} - 5}{8} = f^{-1}(x)$ $f^{-1}(x) = \frac{x^{2} - 5}{8}, x\geq0$