Chapter 5: Problem 30
For each of the following five functions, identify any vertical and horizontal asymptotes, and identify intervals on which the function is concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing. $$f(x)=1 /\left(1-x^{2}\right)$$
Short Answer
Expert verified
Vertical asymptotes at \( x = \pm 1 \); horizontal asymptote at \( y = 0 \); concave up and increasing on \( 0 < x < 1 \); concave up and decreasing on \( -1 < x < 0 \); concave down and decreasing on \( x > 1 \) and \( x < -1 \).
Step by step solution
01
Determine Vertical Asymptotes
The vertical asymptotes of a function can be found by determining where the function is undefined. For the function \( f(x) = \frac{1}{1-x^2} \), it is undefined where the denominator is zero. Solve \( 1-x^2 = 0 \).\[ 1-x^2 = 0 \]\[ x^2 = 1 \]\[ x = \pm 1 \]Therefore, \( x = 1 \) and \( x = -1 \) are the vertical asymptotes.
02
Determine Horizontal Asymptotes
A horizontal asymptote of a rational function \( f(x) = \frac{a_n x^n + \, ...}{b_m x^m + \, ...} \) typically occurs as \( x \to \pm \infty \). For \( f(x) = \frac{1}{1-x^2} \), there is no \( x^n \) term; therefore, the horizontal asymptote is approached when \(|x| \to \infty\), where the denominator dominates and the value approaches zero.Hence, the horizontal asymptote is \( y = 0 \).
03
Analyze Concavity and Intervals Using the Second Derivative
To determine the concavity, we need the second derivative of \( f(x) \). First, calculate the first derivative:\[ f'(x) = \frac{d}{dx}\left(\frac{1}{1-x^2}\right) = \frac{2x}{(1-x^2)^2} \]Next, find the second derivative:\[ f''(x) = \frac{d}{dx}\left(\frac{2x}{(1-x^2)^2}\right) \]Using the quotient rule and simplifying, we get:\[ f''(x) = \frac{2(1+x^2)}{(1-x^2)^3} \]
04
Determine Intervals of Concavity and Monotonicity
The sign of \( f''(x) = \frac{2(1+x^2)}{(1-x^2)^3} \) determines concavity:- **Concave Up**: When \( (1-x^2)^3 > 0 \), or \( -1 < x < 1 \).- **Concave Down**: When \( (1-x^2)^3 < 0 \), or \( x > 1 \) or \( x < -1 \).For monotonicity, use \( f'(x) = \frac{2x}{(1-x^2)^2} \):- **Increasing**: \( f'(x) > 0 \) when \( x > 0 \).- **Decreasing**: \( f'(x) < 0 \) when \( x < 0 \).Combine these to find:- **Concave up and increasing**: \( 0 < x < 1 \).- **Concave up and decreasing**: \( -1 < x < 0 \).- **Concave down and increasing**: No intervals.- **Concave down and decreasing**: \( x > 1 \) and \( x < -1 \).
05
Compile Results
**Vertical Asymptotes:** \( x = 1 \) and \( x = -1 \).**Horizontal Asymptote:** \( y = 0 \).**Concavity and Monotonicity Intervals:**- **Concave up and increasing:** \( 0 < x < 1 \).- **Concave up and decreasing:** \( -1 < x < 0 \).- **Concave down and decreasing:** \( x > 1 \) and \( x < -1 \).There are no intervals where the function is concave down and increasing.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertical Asymptotes
Vertical asymptotes mark the values of \( x \) where a function tends to infinity, and the graph shoots upwards or downwards without bound. These points occur when the function is undefined, typically when the denominator of a rational function equals zero. For the function \( f(x) = \frac{1}{1-x^2} \), the vertical asymptotes are found where the expression \( 1-x^2 = 0 \). By solving for \( x \), the values \( x = \pm 1 \) make the denominator zero, leading to the vertical asymptotes at \( x = 1 \) and \( x = -1 \). These points are where the graph approaches a vertical line, splitting the function's domain into various sections.
Horizontal Asymptotes
Horizontal asymptotes represent the values that a function approaches as \( x \) becomes extremely large or small, specifically as \( x \to \pm \infty \). For rational functions where the degree of the polynomial in the numerator is less than that of the denominator, the horizontal asymptote typically lies at \( y = 0 \). In the case of \( f(x) = \frac{1}{1-x^2} \), there is no polynomial term in the numerator that is of higher degree than the denominator. Thus, as \( x \to \pm \infty \), \( f(x) \) approaches zero, confirming that the horizontal asymptote is \( y = 0 \). This horizontal line indicates that the function will level out at \( y = 0 \) as \( x \) moves towards infinity in either direction.
Concavity
The concept of concavity helps us understand how the graph of a function curves. A function is *concave up* when it curves upwards like a cup, indicating that the slope of the tangent line is increasing. Conversely, it is *concave down* when it curves downwards like a frown, with the slope of the tangent line decreasing. To find concavity for \( f(x) = \frac{1}{1-x^2} \), we need the second derivative, \( f''(x) \), which tells us about the function's rate of change of slope.
- **Concave Up**: Occurs in the interval \(-1 < x < 1\), where \( f''(x) > 0 \).
- **Concave Down**: Occurs outside the interval, at \( x > 1 \) and \( x < -1 \), where \( f''(x) < 0 \).
Monotonicity
Monotonicity describes whether a function is consistently increasing or decreasing over an interval. For \( f(x) = \frac{1}{1-x^2} \), finding the first derivative, \( f'(x) \), helps us identify these intervals:
- **Increasing**: Where \( f'(x) > 0 \), meaning the function rises with increasing \( x \). This occurs at \( x > 0 \).
- **Decreasing**: Where \( f'(x) < 0 \), showing the function drops with increasing \( x \). This is found at \( x < 0 \).