Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Approximate the values of \(x\) that give maximum and minimum values of the function on the indicated intervals. $$ f(x)=x^{2} \sin \frac{x}{2} ;[0,4 \pi] $$

Step-by-Step Solution

Answer

Use numerical or graphical methods to find critical points between [0, 4π] and compare with endpoints. Evaluate function values at these points for maximum or minimum determinations.

Step 1: Find the Derivative

To find the critical points where the maximum and minimum could occur, first, we find the derivative of the function. We have \( f(x) = x^2 \sin \frac{x}{2} \). To differentiate it, use the product rule.The product rule is \( (uv)' = u'v + uv' \). Here, \( u = x^2 \) and \( v = \sin \frac{x}{2} \).The derivative of \( u \) is \( u' = 2x \), and the derivative of \( v \) with respect to \( x \) is \( v' = \cos \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2}\cos \frac{x}{2} \).Applying the product rule:\[ f'(x) = 2x \cdot \sin \frac{x}{2} + x^2 \cdot \frac{1}{2}\cos \frac{x}{2} \]Simplifying, we get:\[ f'(x) = 2x \sin \frac{x}{2} + \frac{x^2}{2} \cos \frac{x}{2} \]

Step 2: Set the Derivative Equal to Zero

To find the critical points, set the derivative equal to zero:\[ 2x \sin \frac{x}{2} + \frac{x^2}{2} \cos \frac{x}{2} = 0 \]Factor out \( x \):\[ x \left( 2 \sin \frac{x}{2} + \frac{x}{2} \cos \frac{x}{2} \right) = 0 \]This gives us two cases to consider:1. \( x = 0 \)2. \( 2 \sin \frac{x}{2} + \frac{x}{2} \cos \frac{x}{2} = 0 \)

Step 3: Solve the Trigonometric Equation

For non-zero \( x \), solve the equation:\[ 2 \sin \frac{x}{2} + \frac{x}{2} \cos \frac{x}{2} = 0 \]Rearranging gives:\[ \sin \frac{x}{2} = -\frac{x}{4} \cos \frac{x}{2} \]This equation does not have a straightforward analytical solution, so we must use numerical methods or graphing to find approximate solutions within the interval \([0, 4\pi]\). Since we are solving within a specific interval, check for solutions that might arise, particularly around known function behavior changes, like when \(x = \pi, 2\pi, 3\pi\), etc.

Step 4: Evaluate at Critical Points and Endpoints

With potential critical points found by solving and evaluating derivative equality, evaluate the function \( f(x) \) at these points, as well as the end of the interval:1. \( x = 0 \)2. \( x \approx a \) (use numerics, for example, \( x \approx 2\pi \) might indicate key changes)3. \( x = 4\pi \)Calculate:- \( f(0) = 0^2 \sin 0 = 0 \)- \( f(4\pi) = (4\pi)^2 \sin 2\pi = 0 \)- Approximate middle critical points should be evaluated numerically (if precise values are needed).

Step 5: Compare Values

Compare the function values at each point to determine maximum and minimum values. The endpoints are likely zero, but numerical solutions or graphical analysis for trigonometric solutions may provide non-zero maximums, especially for points found between the boundaries.

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus, particularly when dealing with the differentiation of products of functions. When you have a function that is the product of two or more functions, such as \( f(x) = u(x) \times v(x) \), the product rule provides a way to find the derivative of this product. The rule is stated as follows: if you have two functions, \( u \) and \( v \), that are differentiable, then the derivative of their product \((uv)'\) is given by:

• \( (uv)' = u'v + uv' \)

To apply this rule effectively, consider each part of the product separately:

• Find \( u' \), the derivative of \( u \).
• Find \( v' \), the derivative of \( v \).
• Substitute these into the product rule formula to find \( f'(x) \).

In our problem, we have \( u = x^2 \) and \( v = \sin \frac{x}{2} \). By applying the product rule correctly, we find the derivative \( f'(x) \), essential for determining the critical points of the function. This rule simplifies the complex differentiation process by breaking it down into smaller, more manageable steps.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent are critical in both mathematics and its applications. In calculus, they often appear in problems involving periodic behavior or rotational phenomena. Understanding how these functions behave and their properties is vital:

• The sine function, \( \sin(x) \), oscillates between -1 and 1 and has a periodicity of \( 2\pi \).
• The cosine function, \( \cos(x) \), also oscillates between -1 and 1 but starts at 1 when \( x = 0 \).
• The derivatives of trigonometric functions are crucial: \( \frac{d}{dx} \sin(x) = \cos(x) \) and \( \frac{d}{dx} \cos(x) = -\sin(x) \).

In the function \( f(x) = x^2 \sin \frac{x}{2} \), the sine function is scaled by \( x/2 \), impacting its frequency and amplitude. Recognizing how these alterations affect the function's graph helps in understanding where critical points might occur, as these are places where the function's slope—or derivative—changes significantly.

Critical Points

Critical points in calculus are the \( x \)-values where the derivative of a function equals zero or is undefined. These points are important because they indicate where a function might achieve its maximum or minimum values. To locate them:

• First, differentiate the function to find \( f'(x) \).
• Set \( f'(x) = 0 \) and solve for \( x \).
• Consider the endpoints of any given interval, as these can also be locations for extrema.

In our problem, after finding \( f'(x) \), we equate it to zero: \( 2x \sin \frac{x}{2} + \frac{x^2}{2} \cos \frac{x}{2} = 0 \). Solving this gives potential critical points. Due to the trigonometric component, analytical solutions might not be possible, so numerical methods or graphing may be needed to approximate these points, particularly when solutions do not align with simple trigonometric values like \( \pi, 2\pi, \) and so on.

Numerical Methods

Numerical methods provide a practical way to find approximate solutions when analytical solving is difficult or impossible. These methods are vital when dealing with complex equations involving trigonometric functions or other non-linear behavior.

• Graphical analysis or tools like graphing calculators and software (e.g., MATLAB, GeoGebra) can give visual insights into function behavior.
• Iteration methods, such as the Newton-Raphson method, can find approximate solutions by refining estimates based on function derivatives.
• Interval bisection is useful: it repeatedly narrows down the interval where a root lies until a satisfactory approximation is reached.

In the context of our problem, using numerical methods helps solve \( 2 \sin \frac{x}{2} + \frac{x}{2} \cos \frac{x}{2} = 0 \) within the interval \([0, 4\pi]\). These approaches are essential when dealing with complicated equations that don't yield simple solutions, thus providing a way to find maximum and minimum points accurately.