Find the domain of the function f(x) = (2x + 9) / (x^3 + 3x^2 - x - 3). Give your answer in interval notation.

Step-by-step solution

1. 1

   Reduce the problem to finding the zeros of the cubic. A rational function is defined everywhere its denominator is nonzero, so the domain is $\mathbb{R}$ minus the roots of $x^{3}+3x^{2}-x-3$. The numerator $2x+9$ never restricts anything. With four terms and no common factor, grouping is the natural tool rather than the rational root test.

2. 2

   Factor by grouping. Split into the first two and last two terms and pull out the obvious factors:

   $x^{3}+3x^{2}-x-3=x^{2}(x+3)-1(x+3)=\left(x^{2}-1\right)(x+3)$

   The grouping works precisely because both pairs leave the same factor $(x+3)$.

3. 3

   Finish with the difference of squares. Since $x^{2}-1=(x-1)(x+1)$, the denominator factors completely into linear pieces:

   $x^{3}+3x^{2}-x-3=(x-1)(x+1)(x+3)$

4. 4

   Set each factor to zero to list the excluded values.

   $x-1=0\ \Rightarrow\ x=1,\qquad x+1=0\ \Rightarrow\ x=-1,\qquad x+3=0\ \Rightarrow\ x=-3$

   None of these makes the numerator zero as well ($2x+9$ equals $11$, $7$ and $3$ there), so all three are true vertical asymptotes, not holes.

5. 5

   Assemble the domain in interval notation. Removing three points from the real line leaves four open intervals, written in increasing order:

   $(-\infty,-3)\cup(-3,-1)\cup(-1,1)\cup(1,\infty)$

6. 6

   Verify the factorisation numerically. Evaluating $(x-1)(x+1)(x+3)-\left(x^{3}+3x^{2}-x-3\right)$ at $x=0,1,2,3$ gives $0$ every time, so the three excluded values are the complete list.