Problem 26 - University Calculus: Early Transcendentals

Question

Show that the functions have exactly one zero in the given interval. $$r(\theta)=2 \theta-\cos ^{2} \theta+\sqrt{2}, \quad(-\infty, \infty)$$

Step-by-Step Solution

Answer

The function has exactly one zero due to being strictly increasing across the entire real line.

Step 1: Define the Function and Derivative

Let's define the function as given: \( r(\theta) = 2\theta - \cos^2\theta + \sqrt{2} \). To determine if there is exactly one zero, we first need to find the derivative, which will help us analyze the behavior of the function. The derivative is: \( r'(\theta) = 2 + 2\cos\theta\sin\theta = 2 + \sin(2\theta) \).

Step 2: Analyze the Derivative for Monotonicity

Next, analyze \( r'(\theta) = 2 + \sin(2\theta) \). Since the sine function \(\sin(2\theta)\) oscillates between -1 and 1, \( r'(\theta) olinebreak =olinebreak 2 + \sin(2\theta) \) will oscillate between 1 and 3, meaning \( r'(\theta) > 0 \) for all \( \theta \). This indicates that the function \( r(\theta) \) is strictly increasing.

Step 3: Determine Behavior as \( \theta \to \infty \) and \( \theta \to -\infty \)

Since \( r(\theta) \) is strictly increasing and spans the interval \((-olinebreak\infty, \infty)\), we can check its behavior at the extremes. As \( \theta \to \infty \), the \( 2\theta \) term dominates, implying \( r(\theta) \to \infty \). As \( \theta \to -\infty \), the \( 2\theta \) term also dominates negatively, implying \( r(\theta) \to -\infty \).

Step 4: Apply the Intermediate Value Theorem

Because \( r(\theta) \) is continuous, strictly increasing, and spans from negative infinity to positive infinity, the Intermediate Value Theorem guarantees that the function crosses the x-axis exactly once. Thus, there must be exactly one zero.

Key Concepts

Intermediate Value Theorem

The Intermediate Value Theorem is a key concept in calculus, particularly useful for finding zeros of functions. The theorem states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), and \( N \) is any number between \( f(a) \) and \( f(b) \), there exists at least one \( c \) within \( (a, b) \) such that \( f(c) = N \).

This theorem is very handy when proving the existence of roots. If you can show that a function is continuous on an interval and that it changes sign over that interval, the Intermediate Value Theorem confirms that at least one root exists where the function crosses the x-axis.

• Ensure the function is continuous - no gaps or jumps in the graph.
• Check that the function value changes from positive to negative or vice versa.
• Conclude that a zero exists in the interval per the theorem's guarantee.

In our example, since the function is continuous over the entire real line and spans all values from negative to positive infinity, it guarantees a zero exists.

Monotonicity

Monotonicity is a useful property in understanding the behavior of functions. A function is called monotonic if it is either entirely non-increasing or non-decreasing. For a strictly increasing function, the value of the function moves consistently upward as the input increases.

In the context of our example function \( r(\theta) \), we determined monotonicity by analyzing the derivative. A positive derivative \( r'(\theta) > 0 \) indicates that the function is strictly increasing. Here's why this is helpful:

• If a function is strictly increasing, it means each input value leads to a unique output value, with never any turning back.
• This behavior assures us that there can be at most one zero.

Monotonicity helps ensure there is only one point where the function will cross the x-axis, making it a powerful tool in function analysis.

Derivative Analysis

Understanding derivatives is crucial in calculus since derivatives provide insights into the rate of change. They tell us whether a function is increasing or decreasing. Specifically, the sign of a derivative indicates:

• Positive derivative: the function is increasing.
• Negative derivative: the function is decreasing.
• Zero derivative: possible extremum (maximum or minimum).

In the provided example, the derivative is \( r'(\theta) = 2 + \sin(2\theta) \). Because the sine function oscillates between -1 and 1, this makes \( 2 + \sin(2\theta) \) oscillate between 1 and 3. Since \( r'(\theta) \) is always positive, the function \( r(\theta) \) is unequivocally increasing for all \( \theta \). Such an analysis not only confirms monotonicity but also rules out any local maxima or minima. This sharpens our understanding of how the function behaves across its domain.

Continuous Functions

A function is continuous if you can draw its graph without lifting the pencil from the paper. This implies no breaks, jumps, or holes in the graph throughout its domain. Continuity is a fundamental requirement to apply the Intermediate Value Theorem and analyze zero crossings simply.

Our function \( r(\theta) = 2\theta - \cos^2\theta + \sqrt{2} \) is continuous because it is built from basic operations (addition, multiplication, and functions like cosine) that preserve continuity across the real numbers.

• Polynomials, sine, and cosine functions are naturally continuous over all real numbers.
• The addition or multiplication of continuous functions remains continuous.

Continuous functions maintain this seamless property over intervals, ensuring that behavior analysis like zero-finding is accurate and applicable. Since \( r(\theta) \) is continuous, we can confidently apply our analysis regarding monotonicity and root finding.