Problem 74 - Precalculus Mathematics for Calculus

Question

Find all solutions of the equation that lie in the interval \([0, \pi] .\) State each answer correct to two decimal places. $$\cos x=x$$

Step-by-Step Solution

Answer

The solution \( x = 0.74 \) is the only solution in the interval \([0, \pi]\).

Step 1: Understand the equation

The equation given is \( \cos x = x \). Our task is to find values of \( x \) that satisfy this equation within the interval \([0, \pi]\).

Step 2: Analyze the functions

Plot the function \( y = \cos x \) and \( y = x \) on a graph to visualize the intersection points. Note that \( \cos x \) is a decreasing function in the interval \([0, \pi]\), and \( x \) is an increasing line.

Step 3: Use the intermediate value theorem

By the Intermediate Value Theorem, we know that if \( f(a) \) and \( f(b) \) have opposite signs, there is at least one root in \((a, b)\). Check positions near obvious starting points like 0 and \( \pi/2 \).

Step 4: Estimate the intersection point

Since \( \cos 0 = 1 \) and \( \cos \pi/2 = 0\), and \( \cos x = x \) is true for one point between 0 and \( \pi/2 \), begin estimating. Use numerical methods such as bisection or Newton's method to narrow down the exact solution.

Step 5: Refine using an iterative method

Use iterations to find a more precise value of \( x \, \approx \,0.739085 \). This is the estimation done by multiple approximation methods (bisection, Newton's).

Step 6: State the solution within given precision

Round the solution to two decimal places: \( x = 0.74 \).

Key Concepts

Interval Solutions

When solving trigonometric equations, one important aspect is focusing on interval solutions. This involves finding all the possible values of the variable within a specified range. For example, in the equation \( \cos x = x \), we are asked to find solutions within the interval \([0, \pi]\).

Here are some key points about interval solutions:

• Specified range: By restricting the solution to a particular interval, we directly focus only on relevant solutions, ignoring others outside this range.
• Behavior of functions: Understanding how functions behave within the interval is crucial, such as knowing that \( \cos x \) starts at 1 at \( x = 0 \) and decreases to -1 at \( x = \pi \).
• Infinite solutions: Without an interval, trigonometric functions might have infinite solutions due to their periodic nature, but defining an interval narrows our focus.

Using graphing or testing points, we observe behaviors that lead us to possible solutions within our defined interval.

Numerical Methods

Numerical methods are invaluable tools for finding solutions to equations that cannot be solved algebraically. In the case of \( \cos x = x \), the solution isn't straightforward, necessitating the use of such methods to approximate answers.

Here is how numerical methods work in this context:

• Initial approximation: They begin by selecting intervals where the solution likely exists based on function behavior.
• Iteration process: Methods such as the bisection method or Newton's method repeatedly refine these estimates until they converge on a solution.
• Accuracy and precision: They provide solutions to a desired degree of accuracy, often crucial in practical scenarios.

For \( \cos x = x \), numerical methods allow us to hone in on a solution between \( 0 \) and \( \pi/2 \), ultimately arriving at an estimated solution of approximately \( 0.739085 \). Utilizing these techniques ensures correctness without reliance solely on graphical methods.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental theorem in calculus that assists in proving the existence of solutions for continuous functions. It states that if a function \( f \) is continuous on a closed interval \([a, b]\) and \( N \) is a number between \( f(a) \) and \( f(b) \), then there is at least one \( c \) in \((a, b)\) such that \( f(c) = N \).

Let's see how IVT helps with trigonometric equations:

• Existence of root: It implies that a crossing or intersection must occur if the function changes signs over the interval.
• Graphical interpretations: Allows identifying potential roots by observing where a curve crosses another line or axis.
• Function properties: Establishes the baseline that if \( f(a) > 0 \) and \( f(b) < 0 \), there exists a root in \([a, b]\).

In \( \cos x = x \), by considering the points near \( 0 \) and \( \pi/2 \) where the sign changes, IVT provides confidence that there is a solution within \( (0, \pi/2) \). This foundational theorem provides the groundwork for applying numerical methods to locate a precise solution.