Question

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

Short answer

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

Step-by-step solution

Identify points of discontinuity

The function given is \( f(x) = \frac{\cos x}{x} \). This is a rational function, which means it is continuous on its domain wherever it's defined. The primary point of concern is where \( x = 0 \), since division by zero is undefined. Thus, \( x = 0 \) is a point of discontinuity.

Determine the domain of continuity

Since \( x = 0 \) causes a discontinuity, we need to exclude it from the domain. Therefore, the function is continuous for all \( x eq 0 \). This gives us two intervals: \( (-\infty, 0) \) and \( (0, \infty) \) where the function remains continuous.

Find the largest interval

Combine the two intervals of continuity. Since \( f(x) \) is continuous on both \( (-\infty, 0) \) and \( (0, \infty) \), these two intervals can be combined (excluding \( x = 0 \)) into the largest possible interval. In interval notation, this is \( (-\infty, 0) \cup (0, \infty) \).

Key concepts

Discontinuity

In mathematical terms, a discontinuity is a point at which a function is not continuous. When we talk about continuity, we refer to how smoothly a function behaves across its entire domain. This means that if you can trace the graph of the function without lifting your pencil, then the function is continuous at every point in that interval. However, there are points, often where a function is undefined, that disrupt this smoothness. These are known as discontinuities.

There are several types of discontinuities:

• Removable Discontinuity: This occurs when a point on the graph is undefined but can be "repaired" by redefining the function at that point.
• Jump Discontinuity: This happens when there is a sudden "jump" in the function's values as you move from left to right across the point.
• Infinite Discontinuity: This occurs when the function approaches infinity as it nears the point of discontinuity, often seen in rational functions where there is division by zero.

In the problem at hand, the function \( f(x) = \frac{\cos x}{x} \) faces a discontinuity at \( x = 0 \). This is because division by zero is undefined, leading to an infinite discontinuity, which must be excluded from the domain for the function to remain continuous elsewhere.

Domain

The domain of a function refers to the complete set of possible input values (\( x \)-values) for which the function is defined. For a function to be continuous, every \( x \)-value within the domain should map to exactly one \( y \)-value.

For rational functions like \( f(x) = \frac{\cos x}{x} \), the domain is all real numbers except for those values that make the denominator zero. An equation such as \( \frac{1}{x} \) shows a clear example where the function is undefined at \( x = 0 \). Therefore, the domain must exclude these singularities or points of discontinuity.

Understanding the domain is crucial for determining where a function is continuous. In this exercise, the domain of \( f(x) \) includes all real numbers except zero, expressed in interval notation as \( (-\infty, 0) \cup (0, \infty) \). This means the function is continuous on both sides of \( x = 0 \), except at \( x = 0 \) itself.

Rational Functions

Rational functions are a class of functions that appear as the ratio of two polynomials. These functions can often look simple, but their behavior can be quite complex, especially around points where the denominator is zero.

For the rational function \( f(x) = \frac{\cos x}{x} \), its general form is \( \frac{p(x)}{q(x)} \), where \( p(x) = \cos x \) and \( q(x) = x \). It is important to check where the denominator \( q(x) \) equals zero since these are potential points of discontinuity. In this example, \( q(x) = x \), sets \( x = 0 \) as an undefined point, creating a discontinuity.

Despite these discontinuities, rational functions behave predictably across most of their domains. They are generally continuous everywhere except where the denominator is zero. This highlights why rational functions are continuous over their domain, represented by combining intervals where they remain defined and smooth, without any undefined points or holes.