Consider the inequality $$ \frac{x-7}{x^{2}(x+8)}

Question

Consider the inequality $$ \frac{x-7}{x^{2}(x+8)}<0 $$ The solution of this inequality consists of one or more of the following intervals: $$ (-\infty, A),(A, B),(B, C) $$, and $$ (C, \infty) $$ where $$ A

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**Step 1. Identify the critical points by setting the numerator and denominator equal to zero.**

The inequality is  
$$
\frac{x-7}{x^2(x+8)}<0.
$$

1. Set the numerator equal to zero:
   $$
   x-7=0 \quad\Longrightarrow\quad x=7.
   $$
2. Set the denominator equal to zero:
   $$
   x^2(x+8)=0.
   $$
   This gives  
   $$
   x^2=0 \quad \text{or} \quad x+8=0.
   $$
   Hence,  
   $$
   x=0 \quad \text{or} \quad x=-8.
   $$

**Step 2. Arrange the critical points in increasing order.**

The critical points are $$x=-8$$, $$x=0$$, and $$x=7$$. In the form of the intervals  
$$
(-\infty, A),\ (A, B),\ (B, C),\ (C, \infty),
$$
we have  
$$
A=-8,\quad B=0,\quad C=7.
$$

**Step 3. Determine the sign of the expression in each interval.**

We test a sample point in each interval.

1. **Interval $$(-\infty, -8)$$**: Take $$x=-10$$.  
   - Numerator: $$x-7 = -10-7=-17$$ (negative).  
   - Denominator:  
     $$
     x^2(x+8)=(-10)^2 \cdot (-10+8)=100\cdot(-2)=-200 \quad \text{(negative)}.
     $$
   - The fraction: $$\dfrac{\text{negative}}{\text{negative}}=\text{positive}.$$  
     Since we need the expression to be less than zero, this interval does **not** satisfy the inequality.

2. **Interval $$(-8,0)$$**: Take $$x=-4$$.  
   - Numerator: $$x-7 = -4-7=-11$$ (negative).  
   - Denominator:  
     $$
     x^2(x+8)=(-4)^2\cdot(-4+8)=16\cdot4=64 \quad \text{(positive)}.
     $$
   - The fraction: $$\dfrac{\text{negative}}{\text{positive}}=\text{negative}.$$  
     This interval satisfies the inequality.

3. **Interval $$(0,7)$$**: Take $$x=1$$.  
   - Numerator: $$x-7 =1-7=-6$$ (negative).  
   - Denominator:  
     $$
     x^2(x+8)=1^2\cdot(1+8)=1\cdot9=9 \quad \text{(positive)}.
     $$
   - The fraction: $$\dfrac{\text{negative}}{\text{positive}}=\text{negative}.$$  
     This interval also satisfies the inequality.

4. **Interval $$(7,\infty)$$**: Take $$x=8$$.  
   - Numerator: $$x-7 =8-7=1$$ (positive).  
   - Denominator:  
     $$
     x^2(x+8)=8^2\cdot(8+8)=64\cdot16=1024 \quad \text{(positive)}.
     $$
   - The fraction: $$\dfrac{\text{positive}}{\text{positive}}=\text{positive}.$$  
     This interval does **not** satisfy the inequality.

**Step 4. Compile the results.**

- $$ A=-8 $$
  
- $$ B=0 $$
  
- $$ C=7 $$

Interval inclusion in the solution:

- $$(-\infty, A) = (-\infty,-8):$$ NO  
- $$(A, B) = (-8,0):$$ YES  
- $$(B, C) = (0,7):$$ YES  
- $$(C, \infty) = (7,\infty):$$ NO
$$ A = -8 $$, $$ B = 0 $$, $$ C = 7 $$. Intervals included in the solution: $$(A, B) = (-8, 0)$$: YES $$(B, C) = (0, 7)$$: YES

Consider the inequality

The solution of this inequality consists of one or more of the following intervals: , and where .
Find
Find
Find
For each interval, answer YES or NO to whether the interval is included in the solution.

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Consider the inequality The solution of this inequality consists of one or more of the following intervals: , and where . Find Find Find For each interval, answer YES or NO to whether the interval is included in the solution.