Comprehensive Analysis Of The Function Y = (x-1)(x-2)/√x

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This article delves into the intricacies of the function y = (x-1)(x-2)/√x, providing a comprehensive analysis suitable for students, educators, and mathematics enthusiasts alike. Our exploration will cover various aspects, including the domain, intercepts, asymptotes, critical points, intervals of increase and decrease, concavity, and the graphical representation of this fascinating function. Understanding these elements is crucial for grasping the behavior of the function and its applications in different mathematical contexts.

1. Domain of the Function

To begin our journey, it's imperative to identify the domain of the function y = (x-1)(x-2)/√x. The domain encompasses all possible input values (x-values) for which the function produces a real number output. In this particular case, we encounter a square root in the denominator, √x. This immediately imposes a restriction: the value under the square root, x, must be greater than zero to avoid imaginary numbers and division by zero. Therefore, x > 0. Consequently, the domain of the function is all positive real numbers, which can be expressed in interval notation as (0, ∞). This initial step is crucial as it sets the stage for all subsequent analyses, ensuring that we only consider values for x that are valid within the function's definition. The domain not only restricts the possible inputs but also significantly influences the function's behavior, particularly its asymptotes and overall shape. Recognizing this limitation upfront allows us to interpret the function's characteristics more accurately and avoid potential pitfalls in our analysis.

2. Intercepts: Where the Function Meets the Axes

Next, we explore the intercepts of the function y = (x-1)(x-2)/√x. Intercepts are the points where the function's graph intersects the coordinate axes, providing valuable insights into its behavior. To find the x-intercepts, we set y = 0 and solve for x. This leads us to the equation 0 = (x-1)(x-2)/√x. A fraction is equal to zero if and only if its numerator is zero. Thus, we have (x-1)(x-2) = 0, which yields the solutions x = 1 and x = 2. These are our x-intercepts. For the y-intercept, we set x = 0 and attempt to solve for y. However, substituting x = 0 into the function results in division by zero (since √0 = 0), which is undefined. Additionally, we established earlier that x = 0 is not in the domain of the function. Consequently, there is no y-intercept for this function. The absence of a y-intercept and the presence of x-intercepts at x = 1 and x = 2 provide crucial clues about the function's graph. It indicates that the graph crosses the x-axis at these two points but never intersects the y-axis. This understanding is essential for accurately sketching the graph and interpreting the function's behavior near the origin.

3. Unveiling Asymptotes: Guiding Lines of the Function

Now, let's investigate the asymptotes of the function y = (x-1)(x-2)/√x. Asymptotes are lines that the function's graph approaches but never quite touches, acting as guiding rails for its behavior. There are two primary types of asymptotes we need to consider: vertical asymptotes and horizontal asymptotes. Vertical asymptotes occur where the function approaches infinity or negative infinity, typically at points where the denominator of a rational function is zero. In our case, the denominator is √x, which is zero when x = 0. Since x = 0 is also outside the domain of the function, it is a potential location for a vertical asymptote. To confirm, we analyze the limit of the function as x approaches 0 from the right (since the domain is x > 0): lim x→0+ (x-1)(x-2)/√x. As x approaches 0 from the right, the numerator approaches (-1)(-2) = 2, while the denominator √x approaches 0. Thus, the fraction approaches positive infinity, indicating a vertical asymptote at x = 0. To find horizontal asymptotes, we examine the limit of the function as x approaches positive infinity: lim x→∞ (x-1)(x-2)/√x. To evaluate this limit, we can simplify the function by expanding the numerator: y = (x2 - 3x + 2)/√x. Then, we divide each term in the numerator by √x, resulting in y = x3/2 - 3x1/2 + 2x-1/2. As x approaches infinity, the term x3/2 dominates, causing the function to approach infinity as well. Therefore, there is no horizontal asymptote. However, we can examine for slant (oblique) asymptotes. A slant asymptote occurs if the degree of the numerator is exactly one more than the degree of the denominator. In this case, the degree of the numerator (after expanding) is 2, and the effective degree of the denominator (√x) is 1/2. The difference is 3/2, so there is no slant asymptote either. The presence of a vertical asymptote at x = 0 and the absence of horizontal and slant asymptotes provide crucial information about the function's long-term behavior. It indicates that the function grows without bound as x increases and approaches infinity as x gets closer to 0 from the right.

4. Finding Critical Points: Maxima, Minima, and Inflection Points

The next crucial step is to determine the critical points of the function y = (x-1)(x-2)/√x. Critical points are the points where the derivative of the function is either zero or undefined. These points are significant because they indicate potential local maxima, local minima, or points of inflection. To find the critical points, we first need to find the derivative of the function, y'. The function is y = (x2 - 3x + 2)/√x = x3/2 - 3x1/2 + 2x-1/2. Using the power rule for differentiation, we find: y' = (3/2)x1/2 - (3/2)x-1/2 - x-3/2. To find the critical points, we set y' = 0 and solve for x: (3/2)x1/2 - (3/2)x-1/2 - x-3/2 = 0. To simplify, we multiply through by 2x3/2 to clear the fractions and negative exponents: 3x2 - 3x - 2 = 0. This is a quadratic equation, which we can solve using the quadratic formula: x = (-b ± √(b2 - 4ac)) / (2a), where a = 3, b = -3, and c = -2. Plugging in these values, we get: x = (3 ± √(9 + 24)) / 6 = (3 ± √33) / 6. This yields two critical points: x1 = (3 + √33) / 6 and x2 = (3 - √33) / 6. Since √33 is approximately 5.74, x1 is positive (approximately 1.45), which is within our domain, and x2 is negative (approximately -0.45), which is outside our domain (x > 0). Thus, we have one critical point within the domain: x = (3 + √33) / 6. We also need to consider where the derivative is undefined. The derivative y' = (3/2)x1/2 - (3/2)x-1/2 - x-3/2 can be rewritten as y' = (3x2 - 3x - 2) / (2x3/2). The derivative is undefined when the denominator is zero, which occurs at x = 0. However, x = 0 is not in the domain of the original function, so it doesn't contribute to our critical points. Therefore, we have one critical point within the domain: x = (3 + √33) / 6. Determining the critical points is a fundamental step in understanding the function's local behavior, as it helps us identify potential maxima, minima, and inflection points. This information is crucial for sketching the graph and analyzing the function's behavior.

5. Intervals of Increase and Decrease: Mapping the Function's Slope

With the critical points in hand, we can now determine the intervals of increase and decrease for the function y = (x-1)(x-2)/√x. These intervals reveal where the function is rising (increasing) and where it is falling (decreasing), providing a crucial understanding of its overall behavior. To find these intervals, we analyze the sign of the first derivative, y', in the intervals defined by the critical points and the domain. We have one critical point within the domain, x = (3 + √33) / 6 ≈ 1.45, and the domain is (0, ∞). This divides the domain into two intervals: (0, (3 + √33) / 6) and ((3 + √33) / 6, ∞). We will test a value from each interval in the derivative y' = (3x2 - 3x - 2) / (2x3/2) to determine its sign. For the interval (0, (3 + √33) / 6), let's choose x = 1. Plugging this into y', we get: y'(1) = (3(1)2 - 3(1) - 2) / (2(1)3/2) = (3 - 3 - 2) / 2 = -1. Since y'(1) < 0, the function is decreasing on the interval (0, (3 + √33) / 6). For the interval ((3 + √33) / 6, ∞), let's choose x = 2. Plugging this into y', we get: y'(2) = (3(2)2 - 3(2) - 2) / (2(2)3/2) = (12 - 6 - 2) / (2 * 2√2) = 4 / (4√2) = 1/√2. Since y'(2) > 0, the function is increasing on the interval ((3 + √33) / 6, ∞). These results indicate that the function decreases from x = 0 to x = (3 + √33) / 6 and then increases from x = (3 + √33) / 6 to infinity. This also suggests that there is a local minimum at x = (3 + √33) / 6. Mapping the intervals of increase and decrease provides a clear picture of the function's monotonic behavior. It is an essential component in understanding the function's shape and identifying local extrema.

6. Concavity and Inflection Points: Unveiling the Curve's Bend

To further refine our understanding of the function y = (x-1)(x-2)/√x, we now investigate its concavity and inflection points. Concavity describes the curve's shape – whether it curves upwards (concave up) or downwards (concave down). Inflection points are the points where the concavity changes. To determine concavity, we need to analyze the second derivative, y''. We found the first derivative to be y' = (3/2)x1/2 - (3/2)x-1/2 - x-3/2, which can be rewritten as y' = (3x2 - 3x - 2) / (2x3/2). Now, we differentiate y' to find y''. Differentiating y' = (3/2)x1/2 - (3/2)x-1/2 - x-3/2 term by term gives: y'' = (3/4)x-1/2 + (3/4)x-3/2 + (3/2)x-5/2. To find potential inflection points, we set y'' = 0 and solve for x: (3/4)x-1/2 + (3/4)x-3/2 + (3/2)x-5/2 = 0. We can simplify this equation by multiplying through by (4/3)x5/2: x2 + x + 2 = 0. This is a quadratic equation. We calculate the discriminant, Δ = b2 - 4ac = 12 - 4(1)(2) = 1 - 8 = -7. Since the discriminant is negative, the quadratic equation has no real roots. This means there are no inflection points where y'' = 0. However, we also need to consider where y'' is undefined. y'' = (3/4)x-1/2 + (3/4)x-3/2 + (3/2)x-5/2 can be rewritten as y'' = (3x2 + 3x + 6) / (4x5/2). The second derivative is undefined when the denominator is zero, which occurs at x = 0. However, x = 0 is not in the domain of the function. Since there are no points where y'' = 0 and the only point where y'' is undefined is outside the domain, there are no inflection points. To determine the concavity, we analyze the sign of y'' in the function's domain (0, ∞). Let's choose a test value, such as x = 1. Plugging this into y'', we get: y''(1) = (3(1)2 + 3(1) + 6) / (4(1)5/2) = 12 / 4 = 3. Since y''(1) > 0, the function is concave up on the interval (0, ∞). The absence of inflection points and the consistent concavity tell us that the graph of the function maintains a consistent curve throughout its domain. This information is crucial for accurately sketching the graph and understanding the function's behavior.

7. Graphical Representation: Piecing Together the Function's Story

Finally, we can synthesize our findings and create a graphical representation of the function y = (x-1)(x-2)/√x. By piecing together the information gathered in the previous sections, we can accurately sketch the function's curve. We know the following:

  • Domain: The domain of the function is (0, ∞), meaning the graph exists only for positive x-values.
  • Intercepts: The function has x-intercepts at x = 1 and x = 2, and no y-intercept.
  • Asymptotes: There is a vertical asymptote at x = 0, and no horizontal or slant asymptotes.
  • Critical Points: There is a critical point at x = (3 + √33) / 6 ≈ 1.45.
  • Intervals of Increase and Decrease: The function is decreasing on the interval (0, (3 + √33) / 6) and increasing on the interval ((3 + √33) / 6, ∞), indicating a local minimum at x ≈ 1.45.
  • Concavity and Inflection Points: The function is concave up on the entire domain (0, ∞), and there are no inflection points.

Using this information, we can sketch the graph. The graph approaches the vertical asymptote at x = 0 from the right, descends to a local minimum near x = 1.45, passes through the x-intercepts at x = 1 and x = 2, and continues to increase concavely upward as x approaches infinity. The absence of horizontal or slant asymptotes means the function grows without bound as x increases. The graphical representation provides a visual summary of the function's behavior. It allows us to see how the various analytical components – domain, intercepts, asymptotes, critical points, and concavity – interact to shape the function's curve. This visual understanding is invaluable for both comprehending the function's properties and communicating its characteristics to others. In conclusion, by systematically analyzing the function y = (x-1)(x-2)/√x, we have gained a comprehensive understanding of its behavior. From determining the domain and intercepts to identifying asymptotes, critical points, intervals of increase and decrease, concavity, and finally, creating a graphical representation, we have explored every facet of this function. This comprehensive approach provides a solid foundation for further mathematical exploration and problem-solving in calculus and related fields.