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Magazine Chapter 5: Problem 81
Sketch the curves. Identify clearly any interesting features, including local maximum and minimum points, inflection points, asymptotes, and intercepts. $$ y=(x-1) /\left(x^{2}\right) $$
Short Answer
Expert verified
The curve has a local maximum at \( x = \frac{1}{2} \), a vertical asymptote at \( x = 0 \), and a horizontal asymptote at \( y = 0 \). The x-intercept is \( (1,0) \).
Step by step solution
01
Find the Domain
The function is defined as \[ y = \frac{x-1}{x^2} \].The domain of this function excludes values of \( x \) where the denominator is zero. So, \( x^2 eq 0 \) implies \( x eq 0 \). Hence, the domain is \( x \in (-\infty, 0) \cup (0, \infty) \).
02
Find Intercepts
To find the y-intercept, set \( x = 0 \). However, the function is not defined at \( x = 0 \), so there is no y-intercept. For the x-intercept, set \( y = 0 \). The numerator \( x-1 = 0 \) gives \( x = 1 \). Thus, the x-intercept is at \( (1, 0) \).
03
Find Asymptotes
Vertical asymptote occurs where the denominator is zero, \( x = 0 \). Horizontal asymptote can be found by evaluating \( \lim_{x \to \pm\infty} y \). Since the degree of the denominator is greater than the numerator, the horizontal asymptote is \( y = 0 \).
04
Calculate Derivative and Critical Points
Finding the derivative to locate local maxima and minima:\[ y' = \frac{d}{dx} \left( \frac{x-1}{x^2} \right) = \frac{-1(2x)(x-1) - 1}{x^4} = \frac{2x - 1}{x^3} \].Set \( y' = 0 \), we get \( 2x - 1 = 0 \) which gives \( x = \frac{1}{2} \). Evaluate behavior around this point for maxima/minima.
05
Second Derivative and Inflection Points
Take the second derivative to analyze concavity:\[ y'' = \frac{d}{dx} \left( \frac{2x - 1}{x^3} \right) = \frac{(6x^2) - (2)}{x^5} \].Set \( y'' = 0 \), which gives no real solutions, suggesting no inflection points due to critical denominator behavior.
06
Analyze Critical Points and Sketching
Analyze the sign changes of \( y' \) around \( x = \frac{1}{2} \). Before \( x = \frac{1}{2} \), \( y' > 0 \) and after \( x = \frac{1}{2} \), \( y' < 0 \), indicating a local maximum. Sketch noting:- x-intercept at \( (1,0) \).- Vertical asymptote at \( x = 0 \).- Horizontal asymptote at \( y = 0 \).- Local maximum at \( x = \frac{1}{2} \).
07
Final Sketch and Summary of Features
Draw the curve noting: - A vertical asymptote at \( x = 0 \) leads to a sharp decline from positive to negative infinity.- Left of \( x = 0 \), the curve approaches \( y = 0 \) from negative direction.- The curve intercepts the x-axis at \( (1,0) \).- Local maximum at \( x = \frac{1}{2} \) above x-axis, then decreases towards \( y = 0 \) as \( x \to \infty \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of a Function
Determining the domain of a function involves identifying all possible input values (or x-values) for which the function is defined. In the exercise, we have the function \[ y = \frac{x-1}{x^2} \]
Here, the function's denominator can never be zero, as division by zero is undefined. Thus, any x-value that makes the denominator zero鈥攕uch as in this case, \( x = 0 \)鈥攊s excluded from the domain.
Here, the function's denominator can never be zero, as division by zero is undefined. Thus, any x-value that makes the denominator zero鈥攕uch as in this case, \( x = 0 \)鈥攊s excluded from the domain.
- The domain for this function is \(x \in (-\infty, 0) \cup (0, \infty)\).
Asymptotes
An asymptote is a line that the function approaches but never touches. For the function \[y = \frac{x-1}{x^2}\], we can have:
- Vertical Asymptotes: These occur where the denominator equals zero. Since the denominator is \( x^2 \), a vertical asymptote exists at \( x = 0 \).
- Horizontal Asymptotes: Determined by evaluating the behavior of the function as \( x \) approaches positive or negative infinity. Here, the degree of the denominator (2) is greater than that of the numerator (1), which implies a horizontal asymptote at \( y = 0 \).
Critical Points
Critical points occur where the derivative of a function equals zero or is undefined. Such points help us identify potential local maxima or minima.For the function \( y = \frac{x-1}{x^2} \), the first derivative is calculated as:\[ y' = \frac{2x - 1}{x^3} \]
Setting \( y' = 0 \) helps us find critical points. Solving \( 2x - 1 = 0 \) gives us \( x = \frac{1}{2} \). This point is among the x-values where the function changes direction, indicating either a local maximum or minimum.
Setting \( y' = 0 \) helps us find critical points. Solving \( 2x - 1 = 0 \) gives us \( x = \frac{1}{2} \). This point is among the x-values where the function changes direction, indicating either a local maximum or minimum.
Local Maxima and Minima
Local maxima and minima refer to the highest or lowest points in a particular region of a graph.To identify these, investigate the sign changes in the first derivative around critical points.
- For \( x = \frac{1}{2} \), simplify the interpretation by checking the derivative signs. Here, \( y' > 0 \) before \( x = \frac{1}{2} \), and \( y' < 0 \) afterwards, indicating that this point is a local maximum.
Inflection Points
Inflection points occur where the curve changes concavity, which involves examining the second derivative.For our function,\[ y'' = \frac{(6x^2) - (2)}{x^5} \]Set \( y'' = 0 \) and solve to locate potential inflection points. However, the given second derivative does not yield real solutions due to the absence of valid zeros.
- No real inflection points exist in this function, largely because the solutions would require the denominator not to vanish.
