Question

what is the solution of $x^{3}-x^{2}>9(x - 1)$

Explanation:

Step1: Rearrange the inequality

\[x^{3}-x^{2}-9(x - 1)>0\]
Factor out \(x^{2}\) from the first two - terms: \(x^{2}(x - 1)-9(x - 1)>0\). Then factor out \((x - 1)\): \((x - 1)(x^{2}-9)>0\). Using the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), we can rewrite \(x^{2}-9\) as \((x + 3)(x - 3)\). So the inequality becomes \((x - 1)(x + 3)(x - 3)>0\).

Step2: Find the critical points

Set each factor equal to zero: \(x-1 = 0\) gives \(x = 1\), \(x + 3=0\) gives \(x=-3\), and \(x - 3=0\) gives \(x = 3\). These are the critical points that divide the number line into the intervals \((-\infty,-3)\), \((-3,1)\), \((1,3)\), and \((3,\infty)\).

Step3: Test the intervals

For \(x<-3\), let \(x=-4\). Then \((-4 - 1)(-4 + 3)(-4 - 3)=(-5)\times(-1)\times(-7)=-35<0\).
For \(-3For \(1For \(x>3\), let \(x = 4\). Then \((4 - 1)(4 + 3)(4 - 3)=(3)\times7\times(1)=21>0\).

Answer:

The solution of the inequality \(x^{3}-x^{2}>9(x - 1)\) is \(-33\), which is represented on the number line as an open - circle at \(x=-3\), an open - circle at \(x = 1\), an open - circle at \(x = 3\), and shading the intervals \((-3,1)\) and \((3,\infty)\). So the correct number - line graph is the one with open circles at \(x=-3\), \(x = 1\), and \(x = 3\), and shading between \(-3\) and \(1\) and to the right of \(3\).