Chapter 4 - Inverse, Exponential, and Logarithmic Functions - Summary Exercises on Functions: Domains and Defining Equations - Exercises - Page 486: 11

$(-\infty,-1]\cup[8,\infty)$.

The domain requirement for $f(x)=\sqrt {x^2-7x-8}=\sqrt {(x+1)(x-8)}$ is $(x+1)(x-8)\geq0$ which gives $x\geq8$ (both factors are positive or zero) or $x\leq-1$ (both factors are negative or zero) , in interval notation $(-\infty,-1]\cup[8,\infty)$.