Question

expressing a solution on a number line
what is the solution of ( x^3 - x^2 > 9(x - 1) )
four number line diagrams (x-axis: -4 to 4, with open/closed circles at -3, 1, 3 and shaded regions) as options

Explanation:

Step1: Simplify the inequality

Start with the inequality \(x^{3}-x^{2}>9(x - 1)\).
First, expand the right - hand side: \(x^{3}-x^{2}>9x-9\).
Then, move all terms to the left - hand side: \(x^{3}-x^{2}-9x + 9>0\).
Factor the left - hand side by grouping. Group the first two terms and the last two terms:
\((x^{3}-x^{2})+(-9x + 9)>0\)
Factor out the greatest common factor from each group: \(x^{2}(x - 1)-9(x - 1)>0\)
Now, factor out \((x - 1)\) from the two terms: \((x - 1)(x^{2}-9)>0\)
We know that \(x^{2}-9=(x - 3)(x + 3)\) (using the difference of squares formula \(a^{2}-b^{2}=(a - b)(a + b)\) where \(a = x\) and \(b = 3\)). So the inequality becomes \((x - 1)(x - 3)(x+3)>0\).

Step2: Find the critical points

The critical points of the inequality \((x - 1)(x - 3)(x + 3)>0\) are the values of \(x\) that make each factor equal to zero.
Set \(x - 1=0\), we get \(x = 1\); set \(x - 3=0\), we get \(x = 3\); set \(x+3=0\), we get \(x=-3\).
These critical points divide the number line into four intervals: \((-\infty,-3)\), \((-3,1)\), \((1,3)\), and \((3,\infty)\).

Step3: Test intervals

• For the interval \((-\infty,-3)\), let's choose a test point, say \(x=-4\).

Substitute \(x = - 4\) into \((x - 1)(x - 3)(x + 3)\): \((-4-1)(-4 - 3)(-4 + 3)=(-5)\times(-7)\times(-1)=-35<0\).

• For the interval \((-3,1)\), let's choose a test point, say \(x = 0\).

Substitute \(x = 0\) into \((x - 1)(x - 3)(x + 3)\): \((0 - 1)(0 - 3)(0 + 3)=(-1)\times(-3)\times3 = 9>0\).

• For the interval \((1,3)\), let's choose a test point, say \(x = 2\).

Substitute \(x = 2\) into \((x - 1)(x - 3)(x + 3)\): \((2 - 1)(2 - 3)(2 + 3)=(1)\times(-1)\times5=-5<0\).

• For the interval \((3,\infty)\), let's choose a test point, say \(x = 4\).

Substitute \(x = 4\) into \((x - 1)(x - 3)(x + 3)\): \((4 - 1)(4 - 3)(4 + 3)=(3)\times(1)\times7 = 21>0\).

Since the inequality is \((x - 1)(x - 3)(x + 3)>0\), the solution intervals are \((-3,1)\) and \((3,\infty)\).

Now, let's analyze the number - line graphs:

• Open circles are used for strict inequalities (\(>\) or \(<\)), so we can eliminate the graphs with closed circles (the second and fourth graphs).
• In the first graph, the shaded regions are \((-3,1)\) and \((3,\infty)\) (the regions between \(-3\) and \(1\) and to the right of \(3\) with open circles at \(-3\), \(1\), and \(3\)).
• In the third graph, the shaded regions are \((-\infty,-3)\) and \((1,3)\) which does not match our solution.

Answer:

The first number - line graph (the one with open circles at \(-3\), \(1\), \(3\) and shaded regions \((-3,1)\) and \((3,\infty)\))