Answer
Solution set: $(-\infty,-3) \cup \left(\frac{1}{2},3\right)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$2x^3-18x \lt x^2-9$,
$2x^3-18x-x^2+9 \lt 0$,
$2x^3-x^2-18x+9 \lt 0$,
$x^2(2x-1)-9(2x-1) \lt 0$,
$(x^2-9)(2x-1) \lt 0$,
$(x-3)(x+3)(2x-1) \lt 0$
$f(x)=(x-3)(x+3)(2x-1)$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(x-3)(x+3)(2x-1)=0$
$x=3$ or $x=-3$ or $x=\frac{1}{2}$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \lt 0 ? \\
& &(a-3)(a+3)(2a-1)& \\
(-\infty,-3) & -6 & (-)(-)(-) & T\\
(-3,\frac{1}{2}) & 0 & (-)(+)(-) & F\\
(\frac{1}{2},3) & 2 & (-)(+)(+) & T\\
(3,\infty) & 4 & (+)(+)(+) & F
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-\infty,-3) \cup \left(\frac{1}{2},3\right)$
