Problem 48 - Single Variable Calculus

Question

Find the interval(s) where \(f\) is continuous. \(f(x)=\frac{2 \cos x}{5+2 \sin x}\)

Step-by-Step Solution

Answer

The function \(f(x) = \frac{2 \cos x}{5+2 \sin x}\) is continuous on the interval \((-\infty, \infty)\) since the denominator is never zero.

Step 1: Identify the denominator of the function

We look at the function \(f(x) = \frac{2 \cos x}{5+2 \sin x}\) and identify that the denominator is \(5+2 \sin x\).

Step 2: Find where the denominator is nonzero

To determine where \(5+2 \sin x \neq 0\), we first need to analyze the equation \(5+2 \sin x = 0\). Solving for \(\sin x\): \( \begin{aligned} 5+2 \sin x &= 0 \\ 2 \sin x &= -5 \\ \sin x &= -\frac{5}{2} \end{aligned} \) Since \(\sin x\) can only range within \([-1,1]\), it's impossible for \(\sin x = -\frac{5}{2}\) in the function's domain.

Step 3: Determine intervals where the function is continuous

Since the denominator \(5+2 \sin x\) is never zero, the function \(f(x) = \frac{2 \cos x}{5+2 \sin x}\) is continuous everywhere it's defined. So, the intervals where the function is continuous are all real numbers, which can be expressed as: \[(-\infty, \infty)\]

Key Concepts

Intervals of Continuity

Understanding the intervals of continuity of a function is crucial for students embarking on calculus studies. In essence, a function is considered continuous on an interval if you can draw it without lifting your pencil from the start of the interval to its end. To determine these intervals for a given function, it's essential to examine where the function does not have any breaks, jumps, or vertical asymptotes.

For the function in question, \( f(x) = \frac{2 \cos x}{5+2 \sin x} \) evaluating its interval of continuity requires knowing where the function is defined and does not experience any undefined behaviors, such as division by zero. After the analysis in the given solution, it's established that the denominator never equates to zero, hence no division by zero issues are present. Therefore, we conclude that the function is continuous everywhere, across the entire real number line, which can be represented as \( (-\infty, \infty) \).

Sine and Cosine Functions

The sine and cosine functions, fundamental tenants of trigonometry, are periodic functions that depict wave-like patterns. Sine and cosine functions are pivotal to understanding a variety of phenomena in physics, engineering, and even finance, due to their oscillatory nature. Their mastery is vital to comprehend other concepts in mathematics such as continuity.

In the given problem, these functions appear in the form \( \cos x \) and \( \sin x \) within the continuous function \( f(x) \) which implies their values oscillate between -1 and 1. Because of their bounded nature, we can infer that for any real number \( x \) , \( \sin x \) will never exceed these bounds, which provides insight into the behavior of our function \( f(x) \) and helps determine its intervals of continuity.

Denominator Non-Zero Condition

An unwavering rule in mathematics is that division by zero is undefined. Thus, for a rational function \( f(x) = \frac{N(x)}{D(x)} \), where \( N(x) \) and \( D(x) \) are polynomial expressions, the function can only be continuous where \( D(x) eq 0 \).

Ensuring the denominator is never zero is crucial. In the exercise, we start by identifying the denominator \( 5+2 \sin x \) and proceed to investigate values of \( x \) that could potentially make it zero. The critical realization here is that \( \sin x \) will never be \( -\frac{5}{2} \) because it falls outside its maximum range of \( [-1,1] \). Therefore, the denominator non-zero condition is satisfied for all real numbers \( x \) in this case. Ensuring this condition is met confirms the continuity of the function over specified intervals or, in this exercise, across the entire domain of all real numbers.