Chapter 3: Problem 55
\(33-62 .\) Sketch the graph of each rational function after making a sign diagram for the derivative and finding all relative extreme points and asymptotes. $$ f(x)=\frac{2 x^{2}}{x^{2}-1} $$
Short Answer
Expert verified
The function has relative max at \( x = 0 \) and vertical asymptotes at \( x = -1, 1 \).
Step by step solution
01
Identify the Rational Function Components
The function given is \( f(x) = \frac{2x^2}{x^2 - 1} \). This is a rational function where the numerator is \( 2x^2 \) and the denominator is \( x^2 - 1 \). The first step is to identify the excluded values for \( x \) from the denominator. The function is undefined where the denominator is zero, i.e., \( x^2 - 1 = 0 \), giving the points \( x = 1 \) and \( x = -1 \). These are vertical asymptotes.
02
Find the Derivative
To analyze the behavior of the function, we differentiate it. Use the quotient rule: \( \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2} \). Here, \( u = 2x^2 \) and \( v = x^2 - 1 \). Then \( u' = 4x \) and \( v' = 2x \). Thus, \( f'(x) = \frac{(4x)(x^2 - 1) - (2x^2)(2x)}{(x^2 - 1)^2} \). Simplify to get \( f'(x) = \frac{4x^3 - 4x - 4x^3}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2} \).
03
Determine the Critical Points
Set the derivative equal to zero to find critical points: \(-4x=0\) gives \(x = 0\). It could be a relative maximum or minimum but requires further testing.
04
Create a Sign Diagram for the Derivative
Evaluate the sign of \( f'(x) = \frac{-4x}{(x^2 - 1)^2} \) in intervals divided by \( x = 0, 1, -1 \). For \( x < -1 \), \(-4x > 0 \), for \(-1 < x < 0 \), \(-4x < 0 \), for \( 0 < x < 1 \), \(-4x < 0 \), and for \( x > 1 \), \(-4x < 0 \). The derivative changes sign from positive to negative at \( x = 0 \).
05
Identify Relative Extremes and Asymptotes
The critical point \( x = 0 \) indicates a relative extremum. Since \( f'(x) \) changes from positive to negative, there is a relative maximum at \( x = 0 \). The vertical asymptotes are at \( x = 1 \) and \( x = -1 \). There aren't any horizontal asymptotes since the degree of the numerator equals the degree of the denominator.
06
Sketch the Graph
With the information derived: a relative maximum at \( x = 0 \), vertical asymptotes at \( x = 1 \) and \( x = -1 \), and behavior of the function in different intervals, we can sketch the graph. The graph increases on \( (-\infty, -1) \), decreases on \( (-1, 0) \), decreases again on \( (0, 1) \), and finally decreases on \( (1, \infty) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative Analysis
Understanding how a function behaves at different points is a crucial part of calculus. Derivative analysis helps with this by showing us where the graph is increasing or decreasing.
In this case, for the function \( f(x) = \frac{2x^2}{x^2 - 1} \), we use the derivative to analyze the function's behavior. By applying the quotient rule, we find that the derivative of the function is \( f'(x) = \frac{-4x}{(x^2 - 1)^2} \).
The sign of this derivative tells us a lot:
In this case, for the function \( f(x) = \frac{2x^2}{x^2 - 1} \), we use the derivative to analyze the function's behavior. By applying the quotient rule, we find that the derivative of the function is \( f'(x) = \frac{-4x}{(x^2 - 1)^2} \).
The sign of this derivative tells us a lot:
- When \( f'(x) > 0 \), the function is increasing.
- When \( f'(x) < 0 \), the function is decreasing.
Relative Extremes
Relative extremes, such as maxima or minima, represent points where a function changes its increasing/decreasing behavior and reaches local high or low values.
For the function \( f(x) = \frac{2x^2}{x^2 - 1} \), after finding that the derivative \( f'(x) = \frac{-4x}{(x^2 - 1)^2} \) equals zero at \( x = 0 \), we detected a potential relative extremum. By examining \( f'(x) \), the sign changes from positive to negative as \( x \) moves through zero.
This shift implies a relative maximum at \( x = 0 \). The presence of a relative maximum indicates that \( x = 0 \) is a local high point in the graph of the function.
For the function \( f(x) = \frac{2x^2}{x^2 - 1} \), after finding that the derivative \( f'(x) = \frac{-4x}{(x^2 - 1)^2} \) equals zero at \( x = 0 \), we detected a potential relative extremum. By examining \( f'(x) \), the sign changes from positive to negative as \( x \) moves through zero.
This shift implies a relative maximum at \( x = 0 \). The presence of a relative maximum indicates that \( x = 0 \) is a local high point in the graph of the function.
- No relative minima were found in this case based on our derivative analysis.
- Understanding these points helps in sketching the graph accurately.
Vertical Asymptotes
Vertical asymptotes occur where the function approaches infinity or negative infinity as the input approaches a particular value. In rational functions like \( f(x) = \frac{2x^2}{x^2 - 1} \), these occur where the denominator equates to zero.
For this function, solving \( x^2 - 1 = 0 \) gives \( x = 1 \) and \( x = -1 \) as vertical asymptotes. As \( x \) nears these values, \( f(x) \) shoots off towards infinity, resulting in a dramatic spike or drop in the graph.
For this function, solving \( x^2 - 1 = 0 \) gives \( x = 1 \) and \( x = -1 \) as vertical asymptotes. As \( x \) nears these values, \( f(x) \) shoots off towards infinity, resulting in a dramatic spike or drop in the graph.
- Vertical asymptotes are particularly important as they dictate sections of the graph where the function is not defined.
- Identifying them is crucial for accurately sketching the function's behavior, particularly how it extends and bends at extreme points.