Solving R(x) = (x-3)/(x+5) > 0 A Step By Step Guide
In this article, we will delve into the process of finding the solutions to the inequality r(x) > 0, where the function r(x) is defined as (x-3)/(x+5). This involves understanding how rational functions behave and how their signs change across different intervals. To effectively tackle this, we'll break down the problem into manageable steps, focusing on identifying critical points and analyzing the sign of the function in the intervals they create. This exploration will not only help solve the specific problem at hand but also enhance your skills in dealing with rational inequalities in general. Let's embark on this mathematical journey to unravel the solutions.
Before diving into the specifics of r(x) = (x-3)/(x+5) > 0, it’s crucial to grasp the fundamental concepts of rational functions and inequalities. A rational function is essentially a function that can be expressed as the quotient of two polynomials. The function r(x) perfectly fits this description, with (x-3) and (x+5) being simple linear polynomials. When dealing with inequalities involving rational functions, we're essentially trying to find the values of x for which the function is either greater than, less than, greater than or equal to, or less than or equal to a certain value (in our case, 0). The key to solving these inequalities lies in understanding how the sign of the function changes across different intervals, which are typically determined by the function's critical points.
Critical points are the values of x where the function is either zero or undefined. For a rational function like r(x), these points occur where the numerator is zero (which makes the function zero) or where the denominator is zero (which makes the function undefined). In our case, the numerator (x-3) becomes zero when x = 3, and the denominator (x+5) becomes zero when x = -5. These two values, 3 and -5, are the critical points that will guide our analysis. They divide the number line into three intervals: x < -5, -5 < x < 3, and x > 3. By examining the sign of r(x) in each of these intervals, we can determine the solutions to the inequality r(x) > 0. This methodical approach ensures we capture all possible solutions while avoiding common pitfalls in inequality solving.
To effectively solve the inequality r(x) > 0, the initial and crucial step is to identify the critical points of the function r(x) = (x-3)/(x+5). These critical points serve as the boundaries that delineate the intervals where the function's behavior—specifically, its sign—remains consistent. As mentioned earlier, the critical points of a rational function are the values of x that make either the numerator or the denominator equal to zero. These points are pivotal because they are the only places where the function can change its sign. The numerator, (x-3), becomes zero when x = 3. This is a critical point because it's where the function r(x) equals zero. The denominator, (x+5), becomes zero when x = -5. This is another critical point, but of a different nature: it's where the function r(x) is undefined, leading to a vertical asymptote on the graph of the function.
These two critical points, x = 3 and x = -5, are the cornerstones of our analysis. They divide the number line into three distinct intervals: x < -5, -5 < x < 3, and x > 3. Within each of these intervals, the function r(x) will maintain a consistent sign—either positive or negative. The reason for this consistency is that between critical points, neither the numerator nor the denominator can change signs without passing through zero, which would mean encountering another critical point. Therefore, by analyzing the sign of r(x) in each interval, we can determine where r(x) is greater than zero and, thus, solve our inequality. This step of identifying critical points is not just a mechanical procedure; it’s a fundamental aspect of understanding the behavior of rational functions and a key to accurately solving inequalities involving them.
With the critical points x = -5 and x = 3 identified, the next pivotal step in solving the inequality r(x) > 0 is to analyze the sign of the function r(x) = (x-3)/(x+5) within the intervals these points define. These intervals are x < -5, -5 < x < 3, and x > 3. To determine the sign of r(x) in each interval, we can select a test value within that interval and evaluate r(x) at that point. The sign of r(x) at the test value will be the sign of r(x) throughout the entire interval.
Let's start with the interval x < -5. We can choose a test value, say x = -6, which is less than -5. Evaluating r(-6) gives us (-6-3)/(-6+5) = (-9)/(-1) = 9, which is positive. Therefore, r(x) > 0 for all x in the interval x < -5. Next, consider the interval -5 < x < 3. A convenient test value here is x = 0. Evaluating r(0) yields (0-3)/(0+5) = -3/5, which is negative. Thus, r(x) < 0 for all x in the interval -5 < x < 3. Finally, for the interval x > 3, we can pick x = 4 as a test value. Calculating r(4) gives us (4-3)/(4+5) = 1/9, which is positive. Consequently, r(x) > 0 for all x in the interval x > 3. This methodical analysis of the sign of r(x) in each interval is crucial because it directly leads us to the solution set of the inequality. By understanding where r(x) is positive, negative, or zero, we can accurately determine the values of x that satisfy r(x) > 0.
Having analyzed the sign of r(x) in each interval, we are now in a position to determine the solution set for the inequality r(x) > 0. Our analysis revealed that r(x) is positive in the intervals x < -5 and x > 3. This means that the values of x in these intervals are the solutions to our inequality. It's important to consider the critical points themselves when forming the solution set. The critical point x = -5 is where the denominator of r(x) is zero, making the function undefined. Therefore, x = -5 cannot be included in the solution set. The critical point x = 3 is where the numerator of r(x) is zero, making r(x) = 0. Since we are looking for values of x where r(x) is strictly greater than zero (i.e., r(x) > 0), x = 3 is also not included in the solution set.
Therefore, the solution set consists of all x less than -5 and all x greater than 3. In interval notation, this is expressed as (-∞, -5) ∪ (3, ∞). This solution set accurately represents all the values of x for which the function r(x) = (x-3)/(x+5) is greater than zero. This final step is where all the previous analyses converge to provide a clear and concise answer to the problem. Understanding how to correctly interpret the sign analysis and translate it into a solution set is a fundamental skill in solving inequalities involving rational functions.
In conclusion, we have successfully navigated the process of solving the inequality r(x) > 0, where r(x) = (x-3)/(x+5). We began by understanding the nature of rational functions and the importance of critical points. We then identified these critical points as x = -5 and x = 3, which divide the number line into intervals. By carefully analyzing the sign of r(x) in each interval, we determined that r(x) is positive when x < -5 and x > 3. This led us to the solution set (-∞, -5) ∪ (3, ∞), which represents all values of x that satisfy the inequality. The key takeaways from this exercise are the methodical approach to solving rational inequalities: identifying critical points, analyzing signs in intervals, and forming the solution set with careful consideration of the original inequality’s conditions.
This process is not only applicable to this specific problem but serves as a blueprint for tackling a wide range of rational inequalities. Understanding the behavior of functions, especially rational functions, and applying logical steps to solve inequalities are essential skills in mathematics. The ability to break down a problem into manageable parts, such as identifying critical points and analyzing intervals, is a valuable strategy in mathematical problem-solving. By mastering these skills, you can confidently approach more complex problems and deepen your understanding of mathematical concepts. Remember, practice and a clear understanding of the underlying principles are the keys to success in mathematics.
The final answer is (D) and
