The domain of \( f(x)=\sqrt{x^{2}-3 x-18} \) is (Type your answer in interval notation. Simplify your answer. Use

To find the domain of the function \( f(x)=\sqrt{x^{2}-3x-18} \), we need to determine the values of \( x \) for which the expression inside the square root is non-negative, as the square root of a negative number is not a real number. The expression inside the square root is \( x^{2}-3x-18 \). We need to find the values of \( x \) for which this expression is greater than or equal to 0. Let's solve the inequality \( x^{2}-3x-18 \geq 0 \) to find the domain of the function. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x^{2}-3x-18\geq 0\) - step1: Rewrite the expression: \(x^{2}-3x-18=0\) - step2: Factor the expression: \(\left(x-6\right)\left(x+3\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x-6=0\\&x+3=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=6\\&x=-3\end{align}\) - step5: Determine the test intervals: \(\begin{align}&x<-3\\&-36\end{align}\) - step6: Choose a value: \(\begin{align}&x_{1}=-4\\&x_{2}=2\\&x_{3}=7\end{align}\) - step7: Test the chosen value: \(\begin{align}&x<-3\textrm{ }\textrm{is the solution}\\&-36\textrm{ }\textrm{is the solution}\end{align}\) - step8: Include the critical value: \(\begin{align}&x\leq -3\textrm{ }\textrm{is the solution}\\&x\geq 6\textrm{ }\textrm{is the solution}\end{align}\) - step9: The final solution is \(x \in \left(-\infty,-3\right]\cup \left[6,+\infty\right):\) \(x \in \left(-\infty,-3\right]\cup \left[6,+\infty\right)\) The solution to the inequality \( x^{2}-3x-18 \geq 0 \) is \( x \in (-\infty,-3] \cup [6,+\infty) \). Therefore, the domain of the function \( f(x)=\sqrt{x^{2}-3x-18} \) is \( x \in (-\infty,-3] \cup [6,+\infty) \) in interval notation.