If \(f:\mathbb{R}\to\mathbb{R}\) is defined by \(f(x)=\dfrac{1}{x^2+ax+9}\), what condition on \(a\) makes it a function on all of \(\mathbb{R}\)?

Correct answer: \(-6<a<6\)

The function is defined for all real \(x\) only if the denominator never becomes zero. So we need \(x^2+ax+9\ne 0\) for all \(x\). A quadratic \(x^2+ax+9\) never equals \(0\) for real \(x\) exactly when its discriminant is negative: \(a^2-4\cdot 1\cdot 9=a^2-36<0\). Thus \(a^2<36\Rightarrow -6<a<6\).