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Chapter 2: Problem 31

Give an example to show that the product of two functions \(f\) and \(g\) may be continuous at a number \(a\) where \(f\) is continuous at \(a\) but \(g\) is discontinuous at \(a\).

Short Answer

Expert verified
An example is \( f(x)=x \) and \( g(x)=\begin{cases} 1 & x eq 0, 0 & x = 0 \end{cases} \). The product \( f(x) \cdot g(x) \) is continuous at 0 while \( g \) is not.

Step by step solution

01

- Define the functions

Choose two functions, one of which is continuous at a point and the other is discontinuous at the same point. For example, let: - \( f(x) = x \), which is continuous at all points.- \( g(x) = \begin{cases} 1 & \text{if } x eq 0, 0 & \text{if } x = 0 \end{cases} \), which is discontinuous at \( x = 0 \).
02

- Define the point of interest

Identify the point where we need to check the continuity of the product. In this case, we choose the point \( a = 0 \).
03

- Compute the product of the functions

The product of the functions \( f(x) \) and \( g(x) \) is:\( (f \cdot g)(x) = f(x) \cdot g(x) \). Using the definitions from Step 1, we have:\( (f \cdot g)(x) = x \cdot \begin{cases} 1 & \text{if } x eq 0 0 & \text{if } x = 0 \end{cases} \).
04

- Simplify the product

Simplify \( (f \cdot g)(x) \) to:\( (f \cdot g)(x) = \begin{cases} x & \text{if } x eq 0, 0 & \text{if } x = 0 \end{cases} \).
05

- Check continuity at \( x = 0 \)

To check if \( (f \cdot g)(x) \) is continuous at \( x = 0 \), we need to find the limit as \( x \) approaches 0 and compare it with the value \( (f \cdot g)(0) \): Limit: \( \lim_{x \to 0} (f \cdot g)(x) = \lim_{x \to 0} x = 0 \).Value at 0: \( (f \cdot g)(0) = 0 \). Since the limit equals the value at the point, \( (f \cdot g) \) is continuous at \( x = 0 \).

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Key Concepts

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

continuous functions
Continuous functions are a fundamental concept in calculus and analysis. A function is said to be continuous at a point if its value does not abruptly change at that point. To be more precise, if a function \( f \) is continuous at a point \( a \), the following conditions must hold:
- The function \( f \) is defined at \( a \).
- The limit of \( f(x) \) as \( x \) approaches \( a \) exists.
- The limit of \( f(x) \) as \( x \) approaches \( a \) is equal to \( f(a) \).
For instance, the function \( f(x) = x \) is continuous at every point in its domain. As shown in the exercise, when we evaluated this function at \( x = 0 \), both the limit and the function value at \( x = 0 \) were equal, which confirms its continuity. This concept is crucial because it ensures that small changes in input result in small changes in output, maintaining a smooth graph without jumps or breaks.
discontinuous functions
Discontinuous functions exhibit at least one point where the function value makes a sudden jump, creates a gap, or is undefined. This leads to a lack of 'smoothness' in the graph of the function. A function \( g \) is discontinuous at a point \( b \) if any of the following are true:
- \( g(b) \) is not defined.
- The limit of \( g(x) \) as \( x \) approaches \( b \) does not exist.
- The limit exists but does not equal \( g(b) \).
In the exercise, the function \( g(x) \) is given as:
\

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Most popular questions from this chapter

Prove that the function is discontinuous at the number \(a\). Then determine if the discontinuity is removable or essential. If the discontinuity is removable, define \(f(a)\) so that the discontinuity is removed.\(f(x)=\frac{\sqrt[3]{x+1}-1}{x} ; a=0\)

Draw a sketch of the graph of the function; then by observing where there are breaks in the graph, determine the values of the independent variable at which the function is discontinuous and show why Definition 2.5.1 is not satisfied at each discontinuity.\(h(x)=\frac{x+3}{x^{2}+x-6}\)

Find the value of the limit and when applicable indicate the limit theorems being used.\(\lim _{t \rightarrow 0} \frac{2-\sqrt{4-t}}{t}\)

Establish the limit by using Definition 2.1.1; that is, for any \(\epsilon>0\), find a \(\delta>0\), such that \(|f(x)-L|<\epsilon\) whenever \(0<|x-a|<\delta$$\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4}=\frac{1}{4}\)

Find the value of the limit and when applicable indicate the limit theorems being used.\(\lim _{x \rightarrow 0} \frac{\sqrt{x+2}-\sqrt{2}}{x}\)
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