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Chapter 4: Problem 4
Welche der Funktionen \(f: x \mapsto f(\boldsymbol{x})\) sind an der Stelle\(x_{0}\) bestimmt, welche unbestimmt divergent? Geben Sie gegcbenenfalls an, ob\(f\) dort eine Unendlichkeitsstelle mit oder ohne Zcichenwechsel besitzt. a) \(f(x)=\frac{1}{x+2}, x_{0}=-2\) b) \(f(x)=\frac{x}{(x-3)^{2}}, x_{0}=3\) c) \(f(x)=|x| \cdot\left(x-\frac{1}{x}\right), x_{0}=0\) d) \(f(x)=\tan \frac{1}{2} x_{3}, x_{0}=x\) e) \(f(x)=\left\\{\begin{array}{ll}\frac{1}{x-1} & \text { für } x>1 \\ 2 &\text { für } x \leq 1^{+}\end{array} x_{0}=1\right.\)
Short Answer
Expert verified
Function (a) is undefined at \( x_{0}=-2 \) with an infinity point without sign change. Function (b) is undefined at \( x_{0}=3 \) with an infinity point without sign change. Function (c) is undefined at \( x_{0}=0 \) with an infinity point with sign change. Function (d) is defined at \( x_{0}=x \). Function (e) is defined at \( x_{0}=1 \).
Step by step solution
01
Identifying the value of the function at the given point
Plug the given value of \( x_{0} \) into the function and try to evaluate. If it's possible to find a finite, real-valued number as the outcome, the function is defined at that point. If the outcome is a division by zero, or leads to an indeterminate form ±∞, the function is undefined at the point or divergent.
02
Examining Function (a) \( f(x)=\frac{1}{x+2} \), \( x_{0}=-2 \)
Substitute \( x_{0} \) into the function to get \( f(-2)=\frac{1}{-2+2} \). This calculation results in division by zero, which implies that the function is undefined at \( x_{0}=-2 \) and it is an infinity point without a sign change.
03
Examining Function (b) \( f(x)=\frac{x}{(x-3)^{2}} \), \( x_{0}=3 \)
Substitute \( x_{0} \) into the function to get \( f(3)=\frac{3}{(3-3)^{2}} \). This calculation results in division by zero, which implies that the function is undefined at \( x_{0}=3 \) and it is an infinity point without a sign change.
04
Examining Function (c) \( f(x)=|x| \cdot\left(x-\frac{1}{x}\right) \), \( x_{0}=0 \)
Substitute \( x_{0} \) into the function to get \( f(0)=|0| \cdot\left(0-\frac{1}{0}\right) \). This calculation results in a division by zero form, which implies that the function is undefined at \( x_{0}=0 \) and it is an infinity point with sign change.
05
Examining Function (d) \( f(x)=\tan \frac{1}{2} x_{3} \), \( x_{0}=x \)
This function is a constant function with the variable \( x_{3} \). Set \( x_{0} = x_{3} \) and put into the function to get \( f(x_{0}) = \tan \frac{1}{2} x_{0} \). The tangent function is defined for all real numbers. Hence, the function is defined at \( x_{0} = x \).
06
Examining Function (e) \( f(x)=\left\{\begin{array}{ll}\frac{1}{x-1} & \text { for } x>1 \\ 2 & \text { for } x \leq 1^{+}\\\end{array}\right.\), \( x_{0}=1 \)
For \( x_{0} = 1 \), use the second part of the piecewise function to get \( f(1) = 2 \). So, the function is defined at \( x_{0} = 1 \).
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms
When exploring limits and continuity in calculus, you may come across expressions that don't instantly reveal a clear outcome. These are often referred to as indeterminate forms. Examples include 0/0 or \( \infty/\infty \). While it may look like evaluating such forms will lead nowhere, with the correct methods, such as L'Hospital's Rule, you can often find a well-defined limit. In the context of the provided exercise, for instance, when we come across a scenario like \( f(x)=|x| \cdot\left(x-\frac{1}{x}\right) \), at \( x_0=0 \), substituting \( x_0 \) into the function leads to an undefined form, \( 0\cdot(0-\frac{1}{0}) \). This signals that further investigation is required to determine the behavior around \( x_0 \) using more sophisticated limit techniques.
It is essential to handle indeterminate forms carefully in calculus, as they can lead to different types of unbounded behavior or identify points of discontinuity within a function.
Unbounded Behavior
When a function's value grows larger and larger without bound as it approaches a specific point, or decreases to negative infinity, we refer to this as unbounded behavior. In simpler terms, the function heads off to infinity or negative infinity near that point. During the examination of functions like \( f(x)=\frac{1}{x+2} \) at \( x_0=-2 \), or \( f(x)=\frac{x}{(x-3)^2} \) at \( x_0=3 \) from the exercise, substituting \( x_0 \) leads to division by zero. This is a classic signal of unbounded behavior - in these cases, the functions don't just become large; they become infinitely large, indicating a vertical asymptote at those points.
Understanding unbounded behavior is key to analyzing and interpreting the behavior of functions within their domains, as well as predicting their graphs. Identifying points where functions go to infinity helps in pinpointing discontinuities and asymptotes which are critical aspects of the function's character.
Piecewise Functions
Among the many types of functions you will encounter are piecewise functions, which are defined by different expressions over different intervals. They can be tricky since the function's behavior can change abruptly at specified points. These points, often where the formula switches, are essential to check for continuity and limits. In the given exercise, function (e), \( f(x)=\begin{cases}\frac{1}{x-1} & \text{for } x>1 \ 2 & \text{for } x \leq 1^{+}\end{cases} \), is a perfect example of a piecewise function. When \( x_0=1 \) is substituted, we pick the second expression to find its value, which yields \( f(1) = 2 \).
The analysis of piecewise functions often requires a separate limit and continuity check on each piece. Look for common complications like jumps or holes in the graph, particularly around the points where the function's definition changes. This is not only key to solving calculus problems but also to understanding real-world scenarios where phenomena have different behaviors depending on the conditions or ranges.
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