If \(f(x)=\sqrt{x^2+6x+8}\), what is its domain?

Correct answer: \(( -\infty,-4]\cup[-2,\infty)\)

For the square root to be real, the radicand must be non-negative. Since \(x^2+6x+8=(x+4)(x+2)\), we solve \((x+4)(x+2)\ge 0\). This gives \(x\le -4\) or \(x\ge -2\). Hence, the domain is \(( -\infty,-4]\cup[-2,\infty)\). Option B is incorrect because the product is negative between \(-4\) and \(-2\). Exam tip: For a radical function, require the expression inside the root to be at least zero and use a sign chart.