Problem 44 - Single Variable Calculus

Question

Let \(f(x)=x^{2} \sin x\). a. Show that \(f\) satisfies the hypotheses of Rolle's Theorem on the interval \([0, \pi]\). b. Use a calculator or a computer to estimate all value(s) of \(c\) accurate to five decimal places that satisfy the conclusion of Rolle's Theorem. c. Plot the graph of \(f\) and the (horizontal) tangent lines to the graph of \(f\) at the point(s) \((c, f(c))\) for the value(s) of \(c\) found in part (b).

Step-by-Step Solution

Answer

In summary, the function \(f(x) = x^2 \sin(x)\) satisfies the hypotheses of Rolle's Theorem on the interval \([0, \pi]\) since it is continuous and differentiable on this interval, and \(f(0) = f(\pi) = 0\). The derivative of the function is \(f'(x) = x^2\cos(x) + 2x\sin(x)\). To find the value(s) of c that satisfy the conclusion of Rolle's Theorem, we need to solve the equation \(f'(c) = 0\), which requires numerical methods. Finally, we can plot the function and its tangent lines at the point(s) \((c, f(c))\), showing where the function satisfies Rolle's Theorem in the interval \([0, \pi]\).

Step 1: Show that f(x) is continuous on the closed interval [0, π]

To show that f(x) = x^2 sin(x) is continuous on the interval [0, π], we need to show that the function is continuous at every point in the interval. Both x^2 and sin(x) are continuous functions on this interval, and since the product of continuous functions is also continuous, it follows that f(x) is continuous on the closed interval [0, π].

Step 2: Show that f(x) is differentiable on the open interval (0, π)

Since x^2 and sin(x) are also differentiable functions, their product is differentiable as well. Consequently, the function f(x) = x^2 sin(x) is differentiable on the open interval (0, π).

Step 3: Prove that f(0) = f(π)

We will now evaluate the function at the end-points of the interval: \(f(0) = 0^2\sin(0) = 0\) \(f(\pi) = \pi^2\sin(\pi) = 0\) So, f(0) = f(π) = 0.

Step 4: Calculate the derivative of f(x)

Now, we will compute the derivative of f(x) using the product rule for differentiation: \(\frac{d}{dx} [x^2\sin(x)] = x^2 (\frac{d}{dx}\sin(x)) + \sin(x) (\frac{d}{dx} x^2)\) Applying the derivatives of the individual functions, we have: \(f'(x) = x^2\cos(x) + 2x\sin(x)\)

Step 5: Find the value(s) of c such that f'(c) = 0

To find the value(s) of c that satisfy the conclusion of Rolle's Theorem, we need to solve the equation f'(c) = 0: \(c^2\cos(c) + 2c\sin(c) = 0\) This equation cannot be solved by analytical means, so we will use a calculator or computer to find the numerical value(s) of c accurate to five decimal places.

Step 6: Plot the function f(x) and the tangent lines at the point(s) (c, f(c))

Using a graphing calculator or computer software, plot the function f(x) = x^2 sin(x) and the horizontal tangent lines at the point(s) (c, f(c)). The Tangent lines will have a slope of 0 since f'(c) = 0. The graph will visualize the function with its horizontal tangent lines, showing where it satisfies Rolle's Theorem in the interval [0, π].

Key Concepts

Continuity

Continuity is a crucial requirement in applying Rolle's Theorem. A function is continuous over an interval if there are no breaks, jumps, or holes in its graph within that interval. For the function \( f(x) = x^2 \sin(x) \), to check continuity over the interval \([0, \pi]\):

• Both \( x^2 \) and \( \sin(x) \) are known to be continuous functions.
• When two continuous functions are multiplied, the result is also continuous.

Thus, \( f(x) \) is continuous on \([0, \pi]\). Ensuring continuity is the first step in verifying that the conditions of Rolle’s Theorem are met. Continuous functions behave predictably, providing the groundwork for further analysis.

Differentiability

Differentiability means that a function can have its derivative calculated at every point within an interval. The requirement for a function to be differentiable is stricter than just being continuous. For \(f(x) = x^2 \sin(x)\) on the interval \((0, \pi)\):

• Both \( x^2 \) and \( \sin(x) \) are differentiable, which means their derivatives exist.
• The product of two differentiable functions is also differentiable.

Thus, \( f(x) \) is differentiable over the open interval \((0, \pi)\). Differentiability is essential for applying Rolle's Theorem because it ensures that the derivative, \( f'(x) \), which we solve for, exists at each point within our specified interval.

Derivative

A derivative represents the function's rate of change. For Roll's Theorem, finding the derivative helps locate points where the slope of the function is zero. For the function \( f(x) = x^2 \sin(x) \), we derive it using the product rule:
Given two functions, \( u(x) = x^2 \) and \( v(x) = \sin(x) \):

• \( u'(x) = 2x \)
• \( v'(x) = \cos(x) \)

By applying the product rule, \((uv)' = u'v + uv'\), the derivative \( f'(x) \) becomes:
\[ f'(x) = x^2 \cos(x) + 2x \sin(x) \]
This formula gives us a way to find where the slope of \( f(x) \) is zero, which is crucial for determining where Rolle’s Theorem holds.

Numerical Methods

Numerical methods are techniques used to approximate solutions to equations that are difficult to solve analytically. After obtaining the derivative \( f'(x) = x^2 \cos(x) + 2x \sin(x) \), finding solutions to \( f'(c) = 0 \) requires these methods because closed-form solutions might not be possible.

• Use a calculator or computer algorithm, such as Newton's method or bisection method, to approximate the root(s).
• Accurate approximations, like to five decimal places, allow us to find the values of \( c \) that satisfy the theorem.

These techniques help bridge the gap between abstract mathematical theory and practical, real-world applications. They are especially useful in calculus and applied settings, allowing us to work with complex functions and equations efficiently.