Question

graph the solution to the following inequality on the number line. $(x + 1)(x + 7)\leq0$

Explanation:

Step1: Find the critical points

Set \((x + 1)(x + 7)=0\).
Using the zero - product property \(a\times b = 0\Rightarrow a = 0\) or \(b = 0\).
If \(x+1=0\), then \(x=-1\); if \(x + 7=0\), then \(x=-7\).

Step2: Test intervals

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

• For the interval \((-\infty,-7)\), let \(x=-8\). Then \((-8 + 1)(-8+7)=(-7)\times(-1)=7>0\).
• For the interval \((-7,-1)\), let \(x=-4\). Then \((-4 + 1)(-4 + 7)=(-3)\times3=-9<0\).
• For the interval \((-1,\infty)\), let \(x=0\). Then \((0 + 1)(0+7)=7>0\).

Since the inequality is \((x + 1)(x + 7)\leq0\), we include the endpoints \(x=-7\) and \(x=-1\) and the interval between them.

Answer:

The solution of the inequality \((x + 1)(x + 7)\leq0\) is \(x\in[-7,-1]\). On the number - line, we draw a solid circle at \(x=-7\) and \(x=-1\) (because the inequality includes equality) and shade the line segment between them.