Analyzing The Rational Function P(x) = (x+6)/(x^2-2)

by ADMIN 53 views

Introduction to Rational Functions

In the realm of mathematics, rational functions hold a significant position, serving as essential tools in various applications ranging from calculus to engineering. These functions, defined as the ratio of two polynomials, exhibit intriguing behaviors and characteristics that make them a compelling subject of study. In this article, we delve into the intricacies of a specific rational function, p(x)=x+6x2−2{ p(x) = \frac{x+6}{x^2-2} }, and explore its key features, including its domain, intercepts, asymptotes, and overall behavior. Understanding these aspects provides a comprehensive picture of the function's graphical representation and its mathematical properties.

The first step in analyzing any function, particularly a rational function, involves determining its domain. The domain of a function encompasses all possible input values (x-values) for which the function is defined. For rational functions, the domain is restricted by the denominator. Specifically, the function is undefined when the denominator equals zero, as division by zero is mathematically invalid. Therefore, to find the domain of p(x)=x+6x2−2{ p(x) = \frac{x+6}{x^2-2} }, we need to identify the values of x that make the denominator, x2−2{ x^2 - 2 }, equal to zero. Solving the equation x2−2=0{ x^2 - 2 = 0 }, we find that x=±2{ x = \pm\sqrt{2} }. These two values, 2{ \sqrt{2} } and −2{ -\sqrt{2} }, are excluded from the domain. Thus, the domain of p(x){ p(x) } consists of all real numbers except 2{ \sqrt{2} } and −2{ -\sqrt{2} }, which can be expressed in interval notation as (−∞,−2)∪(−2,2)∪(2,∞){ (-\infty, -\sqrt{2}) \cup (-\sqrt{2}, \sqrt{2}) \cup (\sqrt{2}, \infty) }. Understanding the domain is crucial as it sets the stage for further analysis, including the identification of vertical asymptotes, which occur at the excluded values.

Moving beyond the domain, intercepts provide valuable insights into the behavior of the function where it intersects the coordinate axes. The x-intercepts are the points where the function's graph crosses the x-axis, which occur when p(x)=0{ p(x) = 0 }. To find the x-intercepts of p(x)=x+6x2−2{ p(x) = \frac{x+6}{x^2-2} }, we set the numerator equal to zero, since a fraction is zero only if its numerator is zero. Thus, we solve the equation x+6=0{ x + 6 = 0 }, which yields x=−6{ x = -6 }. This means the graph of the function crosses the x-axis at the point (−6,0){ (-6, 0) }. The y-intercept, on the other hand, is the point where the graph crosses the y-axis, which occurs when x=0{ x = 0 }. To find the y-intercept, we substitute x=0{ x = 0 } into the function: p(0)=0+602−2=6−2=−3{ p(0) = \frac{0+6}{0^2-2} = \frac{6}{-2} = -3 }. Therefore, the y-intercept is at the point (0,−3){ (0, -3) }. The intercepts, along with the domain, give us a clearer picture of where the function exists and how it interacts with the coordinate axes.

Asymptotes and Behavior Analysis

Next, we investigate the asymptotes of the rational function. Asymptotes are lines that the graph of the function approaches but never touches. Rational functions can have three types of asymptotes: vertical, horizontal, and oblique (or slant). Vertical asymptotes occur at the values excluded from the domain, which we already identified as x=±2{ x = \pm\sqrt{2} }. These asymptotes indicate where the function's value approaches infinity or negative infinity as x gets closer to these values. To understand the behavior near the vertical asymptotes, we examine the limits as x approaches 2{ \sqrt{2} } and −2{ -\sqrt{2} } from both the left and the right. For example, as x approaches 2{ \sqrt{2} } from the left, the denominator x2−2{ x^2 - 2 } approaches zero from the negative side, while the numerator x+6{ x + 6 } approaches a positive value. Thus, the function approaches negative infinity. Conversely, as x approaches 2{ \sqrt{2} } from the right, the denominator approaches zero from the positive side, and the function approaches positive infinity. A similar analysis can be done for x=−2{ x = -\sqrt{2} }.

In addition to vertical asymptotes, horizontal asymptotes describe the behavior of the function as x approaches positive or negative infinity. To find horizontal asymptotes, we compare the degrees of the polynomials in the numerator and the denominator. In the case of p(x)=x+6x2−2{ p(x) = \frac{x+6}{x^2-2} }, the degree of the numerator (1) is less than the degree of the denominator (2). When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is always the line y=0{ y = 0 }. This means that as x becomes very large (positive or negative), the function's value approaches zero. There are no oblique (slant) asymptotes in this case, as these occur when the degree of the numerator is exactly one greater than the degree of the denominator.

Understanding the asymptotic behavior is crucial for sketching the graph of the function and for predicting its values for very large or very small inputs. The asymptotes act as guidelines, indicating the boundaries within which the function operates. Combined with the intercepts and the domain, they provide a solid framework for visualizing the function's overall shape and characteristics.

To further refine our understanding of the function's behavior, we can analyze its first derivative to determine intervals of increasing and decreasing behavior and to identify any local maxima or minima. The first derivative, denoted as p′(x){ p'(x) }, gives us the slope of the tangent line at any point on the function's graph. If p′(x)>0{ p'(x) > 0 }, the function is increasing; if p′(x)<0{ p'(x) < 0 }, the function is decreasing; and if p′(x)=0{ p'(x) = 0 }, we have a critical point, which could be a local maximum, a local minimum, or a saddle point. To find the first derivative of p(x)=x+6x2−2{ p(x) = \frac{x+6}{x^2-2} }, we use the quotient rule:

p′(x)=(x2−2)(1)−(x+6)(2x)(x2−2)2=x2−2−2x2−12x(x2−2)2=−x2−12x−2(x2−2)2{ p'(x) = \frac{(x^2-2)(1) - (x+6)(2x)}{(x^2-2)^2} = \frac{x^2 - 2 - 2x^2 - 12x}{(x^2-2)^2} = \frac{-x^2 - 12x - 2}{(x^2-2)^2} }

Setting p′(x)=0{ p'(x) = 0 } to find critical points, we need to solve −x2−12x−2=0{ -x^2 - 12x - 2 = 0 }. This is a quadratic equation, which can be solved using the quadratic formula:

x=−b±b2−4ac2a=12±(−12)2−4(−1)(−2)2(−1)=12±144−8−2=12±136−2=−6±34{ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{(-12)^2 - 4(-1)(-2)}}{2(-1)} = \frac{12 \pm \sqrt{144 - 8}}{-2} = \frac{12 \pm \sqrt{136}}{-2} = -6 \pm \sqrt{34} }

Thus, the critical points are x=−6−34≈−11.83{ x = -6 - \sqrt{34} \approx -11.83 } and x=−6+34≈−0.17{ x = -6 + \sqrt{34} \approx -0.17 }. We can create a sign chart for p′(x){ p'(x) } to determine the intervals of increasing and decreasing behavior. By testing values in the intervals (−∞,−6−34){ (-\infty, -6-\sqrt{34}) }, (−6−34,−6+34){ (-6-\sqrt{34}, -6+\sqrt{34}) }, and (−6+34,∞){ (-6+\sqrt{34}, \infty) }, we find that the function is decreasing on (−∞,−6−34){ (-\infty, -6-\sqrt{34}) }, increasing on (−6−34,−6+34){ (-6-\sqrt{34}, -6+\sqrt{34}) }, and decreasing on (−6+34,∞){ (-6+\sqrt{34}, \infty) }. This indicates a local minimum at x=−6−34{ x = -6 - \sqrt{34} } and a local maximum at x=−6+34{ x = -6 + \sqrt{34} }. The y-values of these local extrema can be found by plugging these x-values back into the original function, p(x){ p(x) }.

Concavity and Inflection Points

Finally, we examine the second derivative to analyze the concavity of the function and to identify any inflection points. The second derivative, denoted as p′′(x){ p''(x) }, gives us the rate of change of the slope. If p′′(x)>0{ p''(x) > 0 }, the function is concave up; if p′′(x)<0{ p''(x) < 0 }, the function is concave down; and if p′′(x)=0{ p''(x) = 0 }, we have a potential inflection point, where the concavity changes. To find the second derivative, we differentiate p′(x){ p'(x) }:

p′(x)=−x2−12x−2(x2−2)2{ p'(x) = \frac{-x^2 - 12x - 2}{(x^2-2)^2} }

Using the quotient rule again:

p′′(x)=(x2−2)2(−2x−12)−(−x2−12x−2)(2)(x2−2)(2x)(x2−2)4{ p''(x) = \frac{(x^2-2)^2(-2x-12) - (-x^2-12x-2)(2)(x^2-2)(2x)}{(x^2-2)^4} }

Simplifying the expression:

p′′(x)=(x2−2)[(x2−2)(−2x−12)−4x(−x2−12x−2)](x2−2)4{ p''(x) = \frac{(x^2-2)[(x^2-2)(-2x-12) - 4x(-x^2-12x-2)]}{(x^2-2)^4} }

p′′(x)=(−2x3−12x2+4x+24)−(−4x3−48x2−8x)(x2−2)3{ p''(x) = \frac{(-2x^3-12x^2+4x+24) - (-4x^3-48x^2-8x)}{(x^2-2)^3} }

p′′(x)=2x3+36x2+12x+24(x2−2)3{ p''(x) = \frac{2x^3+36x^2+12x+24}{(x^2-2)^3} }

Setting p′′(x)=0{ p''(x) = 0 }, we need to solve the cubic equation 2x3+36x2+12x+24=0{ 2x^3 + 36x^2 + 12x + 24 = 0 }. This equation is more complex and may require numerical methods or software to find the roots. The roots of this equation will give us the potential inflection points. We would then create a sign chart for p′′(x){ p''(x) } to determine the intervals of concavity. By testing values in the intervals determined by the roots and the excluded values from the domain, we can identify where the function is concave up or concave down.

Conclusion

In conclusion, analyzing the rational function p(x)=x+6x2−2{ p(x) = \frac{x+6}{x^2-2} } involves a multi-faceted approach, beginning with understanding its domain and intercepts, proceeding through the identification of asymptotes and the analysis of increasing and decreasing intervals using the first derivative, and culminating in the examination of concavity and inflection points using the second derivative. This comprehensive analysis provides a deep understanding of the function's behavior and allows us to accurately sketch its graph. Rational functions, such as the one we have explored, are fundamental in various mathematical and scientific applications, making their thorough understanding essential. By systematically analyzing these functions, we can gain valuable insights into their properties and behaviors.