Problem 133 For the following exercises, det... [FREE SOLUTION]

Chapter 12: Problem 133

For the following exercises, determine why the function $f$ is discontinuous at a given point $a$ on the graph. State which condition fails. $$ f(x)=\frac{x}{|x|}, \quad a=0 $$

Short Answer

Expert verified

The function is discontinuous at 0 because $f(0)$ is undefined, and the limit does not exist.

Step by step solution

Understand the Function

The function given is $f(x) = \frac{x}{|x|}$. This is known as the sign function, and it has different expressions depending on the sign of $x$. When $x > 0$, $f(x) = 1$, when $x < 0$, $f(x) = -1$, and when $x = 0$, the function is undefined because we cannot divide by zero.

Identify the Problem Point

The point in question is $a = 0$. We need to determine the behavior of the function $f(x)$ around this point and gather information about it being discontinuous at $x = 0$.

Check the Definition of Continuity

A function is continuous at a point $a$ if the following three conditions are met: 1. $f(a)$ is defined, 2. the limit of $f(x)$ as $x$ approaches $a$ exists, and 3. the limit of $f(x)$ as $x$ approaches $a$ is equal to $f(a)$.

Verify Condition 1 (Function Defined)

Check if $f(0)$ is defined. Since $f(x) = \frac{x}{|x|}$ results in division by zero when $x = 0$, $f(0)$ is not defined. Thus, the first condition for continuity fails.

Check Condition 2 (Limit Exists)

Examine the one-sided limits as $x$ approaches 0. For $x \to 0^+$, $f(x) = 1$, so $\lim_{{x \to 0^+}} f(x) = 1$. For $x \to 0^-$, $f(x) = -1$, so $\lim_{{x \to 0^-}} f(x) = -1$. Since the right-hand and left-hand limits are not equal, the overall limit as $x$ approaches 0 does not exist. Condition 2 also fails.

Conclude on Discontinuity

Since condition 1 fails, meaning $f(0)$ is not defined, and condition 2 also fails because the limits from the left and right are not equal, the function $f(x) = \frac{x}{|x|}$ is discontinuous at $x = 0$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sign Function

The sign function, often denoted as $ \text{sgn}(x) $, plays a vital role in understanding the behavior of the function $ f(x) = \frac{x}{|x|} $. This function effectively indicates the sign of a number $ x $, reflecting whether $ x $ is positive, negative, or zero. Here's a quick overview:

For $ x > 0 $, $ \text{sgn}(x) = 1 $, suggesting a positive value.
For $ x < 0 $, $ \text{sgn}(x) = -1 $, indicating a negative value.
For $ x = 0 $, $ \text{sgn}(x) $ is often treated as undefined, particularly from a mathematical standpoint of ratios like $ \frac{x}{|x|} $.

The sign function adopts a piecewise nature, assigning different values based on the sign of $ x $. This makes it useful, yet it introduces a discontinuity when transitioning at $ x = 0 $. Such a function is crucial for identifying changes in a direction or a magnitude of values.

Limits and Continuity

Understanding limits and continuity is central to the discussion of any function's behavior at specific points, such as $ x = 0 $ in our $ f(x) = \frac{x}{|x|} $ scenario. A function is continuous at a point $ a $ if it meets three criteria:

The function's value at $ a $ is defined $ (f(a)) $.
The limit of the function as $ x $ approaches $ a $ exists.
The limit equals the function value at $ a $: $ \lim_{{x \to a}} f(x) = f(a) $.

For our example function, $ f(x) = \frac{x}{|x|} $, continuity fails because:

$ f(0) $ is undefined due to division by zero.
The limit as $ x \to 0 $ varies because the one-sided limits, $ \lim_{{x \to 0^+}} f(x) = 1 $ and $ \lim_{{x \to 0^-}} f(x) = -1 $, differ.

These discrepancies illustrate why $ f(x) $ is discontinuous at $ x = 0 $, underscoring the importance of examining limits for understanding function behavior.

Piecewise Functions

Piecewise functions, like the sign function $ f(x) = \frac{x}{|x|} $, are constructed by different expressions over various intervals.They are incredibly useful when a single algebraic expression cannot describe the function across its domain. In this particular case,

For $ x > 0 $, the function follows one expression: $ f(x) = 1 $.
For $ x < 0 $, a different expression applies: $ f(x) = -1 $.
At $ x = 0 $, it is undefined, causing discontinuity.

Piecewise functions are perfect for modeling real-world situations where conditions change at specific points. However, they can complicate analyses of continuity, as seen with the discontinuity at $ x = 0 $ for $ f(x) = \frac{x}{|x|} $.Understanding how these functions behave at different segments helps grasp the idea of discontinuities and their implications in diverse contexts.

91影视

For the following exercises, determine why the function \(f\) is discontinuous at a given point \(a\) on the graph. State which condition fails. $$ f(x)=\frac{x}{|x|}, \quad a=0 $$

Short Answer

Step by step solution

Understand the Function

Identify the Problem Point

Check the Definition of Continuity

Verify Condition 1 (Function Defined)

Check Condition 2 (Limit Exists)

Conclude on Discontinuity

Key Concepts

Sign Function

Limits and Continuity

Piecewise Functions

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