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Solve each rational inequality and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x+3}{x+4}<0 $$

Short Answer

Expert verified
The solution set for the given inequality \(\frac{x+3}{x+4} < 0\) is \((-4, -3)\).

Step by step solution

01

Identifying and Evaluating the Critical Points

Evaluate where the fraction equals to 0 and where it is undefined. A fraction equals 0 when the numerator is 0. Therefore, set \(x + 3 = 0\), which gives \(x = -3\). The fraction is undefined when the denominator is 0. Hence, set \(x + 4 = 0\) which gives \(x = -4\).
02

Creating the Sign Line

The critical points, \(x = -3\) and \(x = -4\), split the number line into three intervals: \((-∞, -4)\), \((-4, -3)\), and \((-3, +∞)\). Choose a number from each interval to be a test point: choose -5 for \((-∞, -4)\), -3.5 for \((-4, -3)\), and 0 for \((-3, +∞)\). Substitute these values into the original inequality, and determine if the inequality is true or false for each interval.
03

Determining the Solution Set

After performing the tests, for \((-∞, -4)\), the inequality is false. For \((-4, -3)\), the inequality holds true. And for \((-3, +∞)\), the inequality is again false. So, the solution set is \((-4, -3)\). Keep in mind that the inequality is \(<0\) and not \(\leq 0\), so the end points -4 and -3 are not included and we indicate that with parentheses in the interval notation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points in Inequalities
Understanding critical points is essential when solving rational inequalities. These are the values which make the numerator equal zero or the denominator undefined, causing a change in the sign of the function. For the inequality \(\frac{x+3}{x+4}<0\), we determine the critical points by finding when the expression either becomes zero or is undefined.

For the numerator \(x+3=0\), solving gives us \(x=-3\). This is a critical point because our inequality may change from being true to false, or vice versa, once we cross this value on the number line. For the denominator \(x+4=0\), solving gives us \(x=-4\). When \(x=-4\), the expression is undefined since division by zero is not possible. Again, this is a critical point for our inequality. These points partition the number line into intervals. By taking these into account, we can then make more informed decisions about where the inequality holds true.
Sign Line Method
Once we've identified the critical points, we can use the sign line method to determine the solution set of the inequality. It's like a game of true or false played out on the number line. We draw a number line and plot the critical points on it. These points split the number line into distinct intervals. The inequality \(\frac{x+3}{x+4}<0\) divides the line into three intervals: \( (-\infty, -4) \), \( (-4, -3) \), and \( (-3, +\infty) \).

We choose 'test points' from each interval, ideally the simplest numbers to work with, to substitute into the inequality. For example, for the interval \( (-\infty, -4) \), we might choose \(x=-5\); for \( (-4, -3) \), \(x=-3.5\) could work; and for \( (-3, +\infty) \), let's use \(x=0\). Following this, we determine the sign of the expression for each interval: positive means the inequality is false, and negative implies it is true for that interval. We can visualise this process by drawing '+' or '-' signs above the respective intervals on our 'sign line'.

This graphical representation quickly shows us where the expression is less than zero, leading us to the solution of the inequality.
Interval Notation
In mathematics, efficiently communicating the solution of inequalities is done using interval notation. It's a concise way to describe sets of numbers along a number line. For the inequality we solved, \(\frac{x+3}{x+4}<0\), the solution set is \( (-4, -3) \). This notation tells us the set includes all numbers between -4 and -3, without including -4 and -3 themselves, because we're dealing with a strict inequality \( <0 \), not \( \leq0 \).

The use of parentheses \( ( ) \) indicates that the endpoints are not included in the set, also referred to as open intervals. Had we used brackets \[ [ ] \], those endpoints would be included, denoting closed intervals. By expressing solutions in interval notation, we ensure that they are interpreted unambiguously, which is very important for clear communication in mathematics.

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Most popular questions from this chapter

The functions $$f(x)=0.0875 x^{2}-0.4 x+66.6$$ and $$g(x)=0.0875 x^{2}+1.9 x+11.6$$ model a car's stopping distance, \(f(x)\) or \(g(x),\) in feet, traveling at \(x\) miles per hour. Function \(f\) models stopping distance on dry pavement and function g models stopping distance on wet pavement. The graphs of these functions are shown for \(\\{x | x \geq 30\\} .\) Notice that the figure does not specify which graph is the model for dry roads and which is the model for wet roads. Use this information to solve. (GRAPH CANNOT COPY). a. Use the given functions to find the stopping distance on dry pavement and the stopping distance on wet pavement for a car traveling at 55 miles per hour. Round to the nearest foot. b. Based on your answers to part (a), which rectangular coordinate graph shows stopping distances on dry pavement and which shows stopping distances on wet pavement? c. How well do your answers to part (a) model the actual stopping distances shown in Figure 3.43 on page \(411 ?\) d. Determine speeds on wet pavement requiring stopping distances that exceed the length of one and one-half football fields, or 540 feet. Round to the nearest mile per hour. How is this shown on the appropriate graph of the models?

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