Turning Points On The Curve Y=(3+x^2)/(1-x) A Detailed Calculus Exploration

In the realm of calculus, identifying turning points is crucial for understanding the behavior of a curve. These points, also known as stationary points, mark where the function's slope momentarily becomes zero, transitioning from increasing to decreasing (maximum) or decreasing to increasing (minimum). In this comprehensive exploration, we will delve into the intricacies of finding the turning points on the curve defined by the equation y = (3 + x^2) / (1 - x). Our analysis will involve employing differential calculus to determine the critical points and subsequently classify them as either maxima or minima. This journey will not only provide a solution to the problem but also elucidate the underlying principles of calculus that govern the shape and characteristics of curves. Understanding the concept of turning points is pivotal in various fields, including optimization problems in engineering, economics, and computer science, where identifying maximum or minimum values is essential for decision-making and efficiency. By mastering the techniques to find and classify these points, one gains a valuable tool for analyzing and interpreting mathematical models of real-world phenomena.

Unveiling the Equation: y = (3 + x^2) / (1 - x)

The equation y = (3 + x^2) / (1 - x) represents a rational function, a type of function formed by dividing one polynomial by another. Rational functions exhibit a rich variety of behaviors, including asymptotes, discontinuities, and, of course, turning points. The numerator, 3 + x^2, is a quadratic polynomial, while the denominator, 1 - x, is a linear polynomial. The interplay between these polynomials dictates the shape and characteristics of the curve. Before diving into the calculus, it's beneficial to observe some key features of this equation. The denominator, 1 - x, equals zero when x = 1, indicating a vertical asymptote at this point. This means the function will approach infinity (or negative infinity) as x gets closer to 1, creating a break in the curve. The presence of the x^2 term in the numerator suggests a parabolic influence, but the division by (1 - x) distorts this parabolic shape, leading to a more complex curve with potential turning points. To accurately locate these turning points, we need the power of differential calculus, which allows us to analyze the rate of change of the function.

Differentiation: Finding the Slope

The cornerstone of finding turning points lies in the concept of the derivative. The derivative of a function, denoted as dy/dx, represents the instantaneous rate of change of the function at any given point. In geometric terms, the derivative gives the slope of the tangent line to the curve at that point. Turning points, by definition, occur where the slope of the tangent line is zero (or undefined). To find the derivative of y = (3 + x^2) / (1 - x), we'll employ the quotient rule, a fundamental rule in differential calculus. The quotient rule states that if y = u(x) / v(x), then dy/dx = (v(x) * du/dx - u(x) * dv/dx) / (v(x))^2. In our case, u(x) = 3 + x^2 and v(x) = 1 - x. Differentiating u(x) with respect to x yields du/dx = 2x, and differentiating v(x) with respect to x gives dv/dx = -1. Plugging these into the quotient rule, we get:

dy/dx = ((1 - x) * (2x) - (3 + x^2) * (-1)) / (1 - x)^2

Simplifying this expression, we have:

dy/dx = (2x - 2x^2 + 3 + x^2) / (1 - x)^2

dy/dx = (-x^2 + 2x + 3) / (1 - x)^2

This expression for dy/dx is crucial. It tells us the slope of the curve at any point x. To find the turning points, we need to determine where this slope is zero.

Setting the Derivative to Zero: Critical Points

Turning points occur where the derivative dy/dx is equal to zero. Therefore, we need to solve the equation:

(-x^2 + 2x + 3) / (1 - x)^2 = 0

A fraction is equal to zero if and only if its numerator is equal to zero (and the denominator is not zero). Thus, we need to solve the quadratic equation:

-x^2 + 2x + 3 = 0

Multiplying through by -1 for convenience, we get:

x^2 - 2x - 3 = 0

This quadratic equation can be factored as:

(x - 3)(x + 1) = 0

This gives us two solutions for x: x = 3 and x = -1. These values are called critical points or stationary points, as they represent potential turning points on the curve. However, we need to verify that these points are indeed turning points and not just points where the slope is momentarily zero before continuing in the same direction. Also, remember that x = 1 is not in the domain of the original function due to the vertical asymptote, so it cannot be a turning point.

Determining the y-coordinates

Now that we have the x-coordinates of the potential turning points, we need to find their corresponding y-coordinates. To do this, we substitute the x-values into the original equation, y = (3 + x^2) / (1 - x). For x = -1:

y = (3 + (-1)^2) / (1 - (-1)) = (3 + 1) / (1 + 1) = 4 / 2 = 2

So, one potential turning point is (-1, 2). For x = 3:

y = (3 + (3)^2) / (1 - 3) = (3 + 9) / (-2) = 12 / (-2) = -6

Therefore, the other potential turning point is (3, -6). We now have two candidate turning points: (-1, 2) and (3, -6). The next crucial step is to classify these points as either maximum or minimum points.

Classifying Turning Points: Maximum or Minimum

To determine whether a turning point is a maximum or a minimum, we can use the second derivative test. The second derivative, denoted as d^2y/dx2, represents the rate of change of the slope. If the second derivative is positive at a turning point, the curve is concave up, indicating a local minimum. Conversely, if the second derivative is negative, the curve is concave down, indicating a local maximum. To find the second derivative, we differentiate the first derivative, dy/dx = (-x^2 + 2x + 3) / (1 - x)^2, with respect to x. This again requires the quotient rule. Let u(x) = -x^2 + 2x + 3 and v(x) = (1 - x)^2. Then, du/dx = -2x + 2 and dv/dx = 2(1 - x)(-1) = -2(1 - x). Applying the quotient rule:

d^2y/dx2 = (v(x) * du/dx - u(x) * dv/dx) / (v(x))^2

d^2y/dx2 = ((1 - x)^2 * (-2x + 2) - (-x^2 + 2x + 3) * (-2(1 - x))) / ((1 - x)^4)

Simplifying this expression is a bit involved, but after careful algebraic manipulation, we obtain:

d^2y/dx2 = (2x + 8) / (1 - x)^3

Now we can evaluate the second derivative at our critical points.

Applying the Second Derivative Test

Let's evaluate the second derivative, d^2y/dx2 = (2x + 8) / (1 - x)^3, at our critical points x = -1 and x = 3. For x = -1:

d^2y/dx2|_(x=-1) = (2(-1) + 8) / (1 - (-1))^3 = (6) / (2)^3 = 6 / 8 = 3/4

Since 3/4 is positive, the curve is concave up at x = -1, indicating a local minimum. Thus, the point (-1, 2) is a local minimum. For x = 3:

d^2y/dx2|_(x=3) = (2(3) + 8) / (1 - 3)^3 = (14) / (-2)^3 = 14 / (-8) = -7/4

Since -7/4 is negative, the curve is concave down at x = 3, indicating a local maximum. Therefore, the point (3, -6) is a local maximum.

Conclusion: Identifying the Turning Points

Through a rigorous application of differential calculus, we have successfully identified and classified the turning points on the curve y = (3 + x^2) / (1 - x). Our analysis revealed two critical points: (-1, 2) and (3, -6). By employing the second derivative test, we definitively determined that (-1, 2) is a local minimum and (3, -6) is a local maximum. These turning points provide valuable insights into the behavior of the curve, indicating where it changes direction. The process of finding turning points is a fundamental technique in calculus with wide-ranging applications in various fields. Understanding the interplay between the function, its first derivative, and its second derivative allows us to accurately analyze and interpret the characteristics of curves and functions. This skill is not only essential in mathematics but also in fields that rely on mathematical modeling and optimization.

Therefore, the turning points on the curve y = (3 + x^2) / (1 - x) are:

• (-1, 2): minimum
• (3, -6): maximum