Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine the largest interval over which the given function is continuous. $$ f(x)=\sin \frac{1}{x} ; c=0 $$

Step-by-Step Solution

Answer

The largest interval of continuity is \((-\infty, 0) \cup (0, \infty)\).

Step 1: Understand the Function

We start with the function \( f(x) = \sin \left( \frac{1}{x} \right) \). This is a composition of the sine function and the reciprocal function.

Step 2: Identify the Domain

The innermost function, \( \frac{1}{x} \), is not defined at \( x = 0 \), as division by zero is undefined. This means the function \( f(x) \) is not defined at \( x = 0 \). Therefore, the domain of \( f(x) \) is all real numbers except zero, i.e., \( x eq 0 \).

Step 3: Determine Continuity of Composite Functions

The sine function, \( \sin(t) \), is continuous for all real numbers \( t \). So, wherever \( \frac{1}{x} \) is defined, \( \sin \left( \frac{1}{x} \right) \) is continuous.

Step 4: Determine the Largest Interval of Continuity

Since the only point where \( f(x) \) is not defined and thus potentially not continuous is \( x = 0 \), the largest interval where \( f(x) \) is continuous is: \(( -fty, 0) \cup (0, fty)\). This is the entire domain of the function.

Key Concepts

Domain of a Function

When we talk about the domain of a function, we're referring to all the possible input values (x-values) that you can plug into the function without running into any problems. Functions can have restrictions, such as values that make the denominator zero or the square root of a negative number, which are not allowed in real numbers. For the function given, \( f(x) = \sin \left( \frac{1}{x} \right) \), a restriction arises because \( \frac{1}{x} \) becomes undefined when \( x = 0 \). This is because dividing by zero is not allowed. Therefore, the domain of this function includes all real numbers except zero. Hence, our function is defined on the interval \((-\infty, 0) \cup (0, \infty)\). By excluding zero, we ensure that the function has no points where it breaks or becomes undefined.

Composite Functions

Composite functions involve two functions that are composed together. You can think of this like putting one function inside another. In our exercise, \( f(x) = \sin \left( \frac{1}{x} \right) \) is a composite function. It combines the reciprocal function \( \frac{1}{x} \) and the sine function \( \sin(x) \). Here, \( \frac{1}{x} \) is the inner function, while \( \sin(t) \) is the outer function.

• The inner function, \( \frac{1}{x} \), first transforms the input \( x \).
• The outer function, \( \sin(t) \) takes this result and applies another transformation using the sine function.

Understanding how these functions work individually helps to grasp how they interact when composed. The continuity of a composite function depends on the continuity of both inner and outer functions within their domains.

Sine Function

The sine function, denoted as \( \sin(x) \), is one of the basic trigonometric functions. It's known for its wave-like pattern and is defined for all real numbers. The graph of the sine function oscillates between -1 and 1, providing periodic and smooth curves which are continuous everywhere. This continuity means there are no jumps, breaks, or holes in the graph of \( \sin(x) \).
Since the sine function is continuous for all real input values, in composite functions, the continuity of the outer function (sine, in this case) isn't usually the limiting factor. Rather, it depends on the inside, which must be defined everywhere the sine function processes it. That's why, for \( \sin \left( \frac{1}{x} \right) \), the continuity issue doesn't come from \( \sin \), but from the reciprocal function \( \frac{1}{x} \).

Limits and Continuity

The concepts of limits and continuity are often intertwined in calculus. Continuity at a point means a function doesn’t suddenly jump or break at that location. It can be assessed using limits. A function \( f(x) \) is continuous at a point \( c \) if the following three conditions are satisfied:

• The function \( f(x) \) is defined at \( x = c \).
• The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
• The limit of \( f(x) \) as \( x \) approaches \( c \) is equal to the function value \( f(c) \).

In the problem \( f(x) = \sin \left( \frac{1}{x} \right) \), the issue rises at \( x = 0 \), where \( \frac{1}{x} \) is undefined, meaning the first condition for continuity fails. This means we cannot check the other criteria of continuity at \( x = 0 \). As a result, the function is continuous everywhere in its domain except at \( x = 0 \). Thus, continuity is confirmed over \((-\infty, 0) \cup (0, \infty)\).