Solve the nonlinear inequality. Express the solution using interval notation.
1/x + 3 + 1/x + 4 ≤ 0

We are tasked with solving the nonlinear inequality:

\[
\frac{1}{x+3} + \frac{1}{x+4} \leq 0
\]

### Step 1: Combine the terms on the left-hand side.
To combine the fractions, we need a common denominator. The common denominator is \((x+3)(x+4)\), so:

\[
\frac{1}{x+3} + \frac{1}{x+4} = \frac{(x+4) + (x+3)}{(x+3)(x+4)} = \frac{2x + 7}{(x+3)(x+4)}
\]

Thus, the inequality becomes:

\[
\frac{2x + 7}{(x+3)(x+4)} \leq 0
\]

### Step 2: Solve the inequality.
To solve \(\frac{2x + 7}{(x+3)(x+4)} \leq 0\), we need to determine where the expression equals zero and where it is positive or negative.

#### Critical Points:
- The numerator \(2x + 7 = 0\) gives the critical point \(x = -\frac{7}{2}\).
- The denominator gives the critical points \(x = -3\) and \(x = -4\), where the expression is undefined.

Thus, the critical points are \(x = -\frac{7}{2}, x = -3, x = -4\).

### Step 3: Determine the sign of the expression in each interval.
We now check the sign of the expression in each of the following intervals determined by the critical points:
- \((-\infty, -4)\)
- \((-4, -3)\)
- \((-3, -\frac{7}{2})\)
- \((- \frac{7}{2}, \infty)\)

We can test each interval by picking a number from that interval and substituting it into the expression \(\frac{2x + 7}{(x+3)(x+4)}\).

1. **Interval \((-\infty, -4)\)**: Pick \(x = -5\).
   \[
   \frac{2(-5) + 7}{(-5 + 3)(-5 + 4)} = \frac{-10 + 7}{(-2)(-1)} = \frac{-3}{2} < 0
   \]

2. **Interval \((-4, -3)\)**: Pick \(x = -3.5\).
   \[
   \frac{2(-3.5) + 7}{(-3.5 + 3)(-3.5 + 4)} = \frac{-7 + 7}{(-0.5)(0.5)} = \frac{0}{-0.25} = 0
   \]

3. **Interval \((-3, -\frac{7}{2})\)**: Pick \(x = -3.25\).
   \[
   \frac{2(-3.25) + 7}{(-3.25 + 3)(-3.25 + 4)} = \frac{-6.5 + 7}{(-0.25)(0.75)} = \frac{0.5}{-0.1875} < 0
   \]

4. **Interval \((- \frac{7}{2}, \infty)\)**: Pick \(x = 0\).
   \[
   \frac{2(0) + 7}{(0 + 3)(0 + 4)} = \frac{7}{12} > 0
   \]

### Step 4: Construct the solution set.
We want the expression to be less than or equal to zero. Based on the sign analysis:

- The expression is negative in the intervals \((-\infty, -4)\) and \((-3, -\frac{7}{2})\).
- The expression is zero at \(x = -\frac{7}{2}\).
- The points \(x = -3\) and \(x = -4\) are excluded because the expression is undefined at those points.

Thus, the solution is:

\[
(-\infty, -4) \cup (-3, -\frac{7}{2}]
\]

This is the solution in interval notation.

Would you like further clarification or more details on any step? Here are five related questions for further exploration:

1. How do you solve inequalities involving rational functions in general?
2. What is the process for finding the sign of a rational function on different intervals?
3. How does one handle undefined points when solving inequalities?
4. What is the role of critical points in solving inequalities?
5. Why is the solution excluded at \(x = -3\) and \(x = -4\)?

**Tip:** When solving rational inequalities, always check where the function is undefined and carefully analyze the sign of the expression in different intervals.

Solve the nonlinear inequality. Express the solution using interval notation. 1/(x+3) + 1/(x+4) ≤ 0

Learn how to solve the nonlinear inequality 1/(x+3) + 1/(x+4) ≤ 0 with a detailed step-by-step solution. This problem covers key mathematical concepts, including inequalities and rational functions, and involves combining fractions and sign analysis. Suitable for students in Grades 10-12.

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