Problem 31 - Calculus Early Transcendentals

Question

An important question about many functions concerns the existence and location of fixed points. A fixed point of \(f\) is a value of \(x\) that satisfies the equation \(f(x)=x ;\) it corresponds to a point at which the graph off intersects the line \(y=x\). Find all the fixed points of the following functions. Use preliminary analysis and graphing to determine good initial approximations. $$f(x)=2 x \cos x \text { on }[0,2]$$

Step-by-Step Solution

Answer

Question: Find the fixed point of the function \(f(x) = 2x \cos x\) on the interval \([0, 2]\). Answer: The fixed point of the function \(f(x) = 2x \cos x\) in the interval \([0, 2]\) is \(x = \frac{\pi}{3} \approx 1.047\).

Step 1: Set up fixed point equation

We are given the function \(f(x) = 2x\cos x\). To find the fixed points of the function, we need to set the function equal to \(x\), and solve for \(x\). So, we have the equation: $$2x\cos x = x.$$

Step 2: Simplify the equation

Divide both sides of the equation by \(x\) to simplify the equation: $$2\cos x = 1.$$

Step 3: Solve for x

Now, we have a simple trigonometric equation that we can solve for \(x\). To solve for \(x\), we can find the inverse cosine of the right-hand side: $$x = \cos^{-1}\left(\frac{1}{2}\right).$$

Step 4: Calculate the fixed points

Applying the inverse cosine function to \(\frac{1}{2}\), we obtain: $$x = \frac{\pi}{3}, \quad 2 \pi - \frac{\pi}{3} = \frac{5\pi}{3}.$$

Step 5: Check if the fixed points are in the desired interval

Before determining that these are the fixed points, we must check if they lie in the interval \([0, 2]\). Since \(\frac{\pi}{3} \approx 1.047\) and \(\frac{5\pi}{3} \approx 5.236\), we can see that only \(\frac{\pi}{3}\) lies in the interval \([0, 2]\).

Step 6: Final answer

The fixed point of the function \(f(x) = 2x \cos x\) in the interval \([0, 2]\) is: $$x = \frac{\pi}{3} \approx 1.047.$$

Key Concepts

Trigonometric Equation

A trigonometric equation involves trigonometric functions, such as sine, cosine, or tangent, where the goal is often to find the variable within these functions. In this exercise, we have the trigonometric function \( \cos x \) as part of the equation \( 2x \cos x = x \), which simplifies to \( 2 \cos x = 1 \). This simplification transforms the problem into a more straightforward inverse trigonometric equation. To solve it, we search for values of \( x \) where \( \cos x = \frac{1}{2} \). The cosine function has known values for angles that can be found on the unit circle, which makes it possible to solve these equations accurately. This is where the inverse cosine function, \( \cos^{-1} \), becomes useful. It effectively helps find the angle \( x \) when the value of cosine is given. Remember, trigonometric equations can often have multiple solutions due to the periodic nature of trigonometric functions. Therefore, after finding potential solutions, they must be verified against the given constraints, like seasonal filters, to ensure they sit within specified intervals.

Graphical Analysis

Graphical analysis is a powerful visual tool to understand mathematical concepts, especially when dealing with functions. By plotting the function \( f(x) = 2x \cos x \) and the line \( y = x \), we can visually identify the points where these graphs intersect. These intersections represent the fixed points of the function. Start by plotting \( f(x) \) over the interval \([0, 2]\). Graphical analysis allows us to estimate where the function meets the line \( y = x \). These estimates arise because the line \( y = x \) acts as a benchmark, allowing us to see visually where the function equals its input value, indicating a fixed point.After you have drawn these graphs, the intersection points hint at initial approximations before performing any algebraic solution. It's crucial to remember that graphical analysis often provides a rough estimate. Thus, it should be followed up with analytical methods to confirm these approximations.

Calculus Problem-Solving

Solving calculus problems often combines graphical intuition with algebraic manipulation for confirming results. In the problem at hand, after graphically identifying where \( f(x) \approx x \), calculus steps help fine-tune the solution and ensure precision. The initial step involves setting up the fixed point condition: \( f(x) = x \). It's this equation that we start solving analytically through simplification. By dividing both sides by \( x \), provided \( x eq 0 \), we find an equation manageable by inverse trigonometric functions.Calculus problem-solving also incorporates checking intervals after solving straightforward equations. The trigonometric solution \( x = \frac{\pi}{3} \approx 1.047 \) was verified as fitting within the defined interval \([0, 2]\), proving its viability as a solution. Remember that calculus approaches often finalize with checking these constraints to validate the potential solutions derived from algebraic suggestions.