The number of integers satisfying the inequality is `x/(x+6)<=1/x`

Options

A. 7
B. 8
C. 9
D. 3

Answer

To solve the inequality \( \frac{x}{x+6} \leq \frac{1}{x} \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \frac{x}{x+6} \leq \frac{1}{x} \]
We can rearrange this to bring all terms to one side:
\[ \frac{x}{x+6} - \frac{1}{x} \leq 0 \]

### Step 2: Find a Common Denominator
The common denominator for the fractions is \( x(x+6) \). We rewrite the left-hand side:
\[ \frac{x^2 - (x+6)}{x(x+6)} \leq 0 \]
This simplifies to:
\[ \frac{x^2 - x - 6}{x(x+6)} \leq 0 \]

### Step 3: Factor the Numerator
Next, we factor the quadratic expression in the numerator:
\[ x^2 - x - 6 = (x - 3)(x + 2) \]
Thus, we rewrite the inequality as:
\[ \frac{(x - 3)(x + 2)}{x(x + 6)} \leq 0 \]

### Step 4: Identify Critical Points
The critical points occur where the numerator or denominator equals zero:
- From \( x - 3 = 0 \) we get \( x = 3 \)
- From \( x + 2 = 0 \) we get \( x = -2 \)
- From \( x = 0 \) (denominator)
- From \( x + 6 = 0 \) we get \( x = -6 \)

Thus, the critical points are \( -6, -2, 0, 3 \).

### Step 5: Test Intervals
We will test the intervals determined by these critical points:
- \( (-\infty, -6) \)
- \( (-6, -2) \)
- \( (-2, 0) \)
- \( (0, 3) \)
- \( (3, \infty) \)

**Choose test points:**
1. For \( x = -7 \) (in \( (-\infty, -6) \)):
\[ \frac{(-7 - 3)(-7 + 2)}{-7(-7 + 6)} = \frac{(-10)(-5)}{(-7)(-1)} > 0 \]
2. For \( x = -4 \) (in \( (-6, -2) \)):
\[ \frac{(-4 - 3)(-4 + 2)}{-4(-4 + 6)} = \frac{(-7)(-2)}{(-4)(2)} > 0 \]
3. For \( x = -1 \) (in \( (-2, 0) \)):
\[ \frac{(-1 - 3)(-1 + 2)}{-1(-1 + 6)} = \frac{(-4)(1)}{(-1)(5)} < 0 \]
4. For \( x = 1 \) (in \( (0, 3) \)):
\[ \frac{(1 - 3)(1 + 2)}{(1)(1 + 6)} = \frac{(-2)(3)}{(1)(7)} < 0 \]
5. For \( x = 4 \) (in \( (3, \infty) \)):
\[ \frac{(4 - 3)(4 + 2)}{(4)(4 + 6)} = \frac{(1)(6)}{(4)(10)} > 0 \]

### Step 6: Determine the Solution Set
From the tests, we find:
- The expression is negative in the intervals \( (-2, 0) \) and \( (0, 3) \).
- The critical points \( -6 \) and \( 0 \) are excluded because they make the denominator zero.
- The critical point \( -2 \) is included because the inequality is less than or equal to zero.

Thus, the solution set is:
\[ (-6, -2] \cup (0, 3] \]

### Step 7: Find Integer Solutions
Now we find the integers in these intervals:
- From \( (-6, -2] \): The integers are \( -5, -4, -3, -2 \) (4 integers).
- From \( (0, 3] \): The integers are \( 1, 2, 3 \) (3 integers).

### Total Integer Solutions
Adding these, we have:
\[ 4 + 3 = 7 \]

### Final Answer
The number of integers satisfying the inequality is **7**.

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