1. We are given the function $$f(x)=\sqrt{x-5}$$ and need to determine whether it is even, odd, or neither. 2. Recall the definitions: - Even function: $$f(-x)=f(x)$$ for all $$x$$ in the domain. - Odd function: $$f(-x)=-f(x)$$ for all $$x$$ in the domain. 3. The domain of $$f$$ is $$x \geq 5$$ because under the square root must be non-negative. 4. Evaluate $$f(-x)=\sqrt{-x-5}$$. To be in the domain, $$-x-5 \geq 0 \Rightarrow x \leq -5$$. 5. Since the domain of $$f(x)$$ is $$x \geq 5$$ and the domain of $$f(-x)$$ is $$x \leq -5$$, the domains do not overlap except possibly at points where both sides can be defined. 6. Check if $$f(-x)=f(x)$$ or $$f(-x)=-f(x)$$ for any $$x$$ values in the domain: - For $$x=6$$ (in the domain), $$f(6)=\sqrt{6-5}=1$$. - $$f(-6)=\sqrt{-6-5}=\sqrt{-11}$$ which is not real, thus undefined. 7. Because $$f(-x)$$ is not defined in the domain of $$f(x)$$, the function neither satisfies even nor odd conditions. **Answer:** The function $$f(x)=\sqrt{x-5}$$ is neither even nor odd due to domain restrictions preventing the symmetry conditions from being met.