Question

graph the solution to the following inequality on the number line.
$(x - 1)(x + 7)\geq0$

Explanation:

Step1: Find the critical points

Set \((x - 1)(x + 7)=0\).
Using the zero - product property \(x-1 = 0\) gives \(x = 1\), and \(x+7=0\) gives \(x=-7\).

Step2: Test the intervals

We have three intervals: \((-\infty,-7)\), \((-7,1)\), and \((1,\infty)\).

• For \(x=-8\) (in the interval \((-\infty,-7)\)): \((-8 - 1)(-8 + 7)=(-9)\times(-1)=9\geq0\).
• For \(x = 0\) (in the interval \((-7,1)\)): \((0 - 1)(0 + 7)=(-1)\times7=-7<0\).
• For \(x = 2\) (in the interval \((1,\infty)\)): \((2 - 1)(2 + 7)=1\times9 = 9\geq0\).

Since the inequality is \(\geq0\), we include the critical points \(x=-7\) and \(x = 1\).

Answer:

The solution of the inequality \((x - 1)(x + 7)\geq0\) is \(x\leq-7\) or \(x\geq1\). On the number - line, we put a closed circle at \(x=-7\) and \(x = 1\), and shade the regions to the left of \(x=-7\) (including \(x=-7\)) and to the right of \(x = 1\) (including \(x = 1\)).