Problem 23 - Essential Calculus

Question

Use a graphing utility to graph the function on the closed interval [a,b]. Determine whether Rolle's Theorem can be applied to \(f\) on the interval and, if so, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=0\). $$ f(x)=\frac{x}{2}-\sin \frac{\pi x}{6}, \quad[-1,0] $$

Step-by-Step Solution

Answer

Rolle's Theorem can be applied to the function if the function is continuous and differentiable on the interval and f(-1) equals f(0). Now find the values of \(c\) by equating the derivative of the function to zero and solving for \(x\) in the interval (-1,0). Note: The derivative of the function \(f'(x)=\frac{1}{2} - \frac{\pi}{6}\cos(\frac{\pi x}{6})\)

Step 1: Graph the function

Use the equation \(f(x)=\frac{x}{2}-\sin \frac{\pi x}{6}\) to graph the function on the interval [-1,0]. Analyze the graph visually.

Step 2: Check conditions of the Rolle's Theorem

The function f(x) is both continuous and differentiable on [-1,0]. Then, check if the function value at the two end points is equal or not, meaning solve for f(-1) and f(0). If they are equal then Rolle's Theorem can be applied.

Step 3: Apply Rolle's Theorem

If Rolle's Theorem conditions are satisfied, then take the derivative of the function \(f'(x)=\frac{1}{2} - \frac{\pi}{6}\cos(\frac{\pi x}{6})\), and solve for the equation f'(x) = 0, to find the values of c in the interval (-1,0).

Key Concepts

Graphing Utility

A graphing utility is a powerful tool that helps visualize mathematical functions. In this exercise, we graph the function \(f(x) = \frac{x}{2} - \sin \frac{\pi x}{6}\) on the closed interval \([-1, 0]\). Using a graphing utility allows us to see the behavior of the function clearly.

• Input the function equation into the graphing tool.
• Set the interval from -1 to 0.
• Observe the curve and its characteristics, such as intersections and slopes.

This visual representation helps determine if the conditions for Rolle's Theorem are met.

Continuous Function

A continuous function is one where small changes in the input result in small changes in the output, without any jumps or breaks. For Rolle's Theorem, the function \(f(x) = \frac{x}{2} - \sin \frac{\pi x}{6}\) needs to be continuous on the entire interval \([-1, 0]\).

• Check the function for any discontinuities in the interval.
• Both polynomial terms and sine functions are continuous on their own. Their combination should be continuous as well.

Ensuring continuity is crucial for applying calculus theorems effectively.

Derivative

A derivative represents the rate of change of a function. In Rolle's Theorem, we use the derivative to find points where the slope is zero. For our function, the derivative is \(f'(x) = \frac{1}{2} - \frac{\pi}{6}\cos\left(\frac{\pi x}{6}\right)\).

• Derivatives provide information about increasing or decreasing trends.
• Find points where \(f'(x) = 0\) to identify critical points within the interval.

Solving for these points helps us understand where the function’s slope becomes horizontal.

Closed Interval

A closed interval \([a, b]\) includes all points \(a\) to \(b\) and the endpoints themselves. For Rolle’s Theorem, our interval is \([-1, 0]\). The conditions are:

• The function must be continuous on \([-1, 0]\).
• The function must be differentiable on \((-1, 0)\).
• The function values at the endpoints \(f(-1)\) and \(f(0)\) must be equal.

Confirming these ensures we can apply Rolle's Theorem to find special points where the derivative equals zero.