Problem 33 - Calculus of a Single Variable

Question

Finding Extrema on a Closed Interval In Exercises \(17-36\) , find the absolute extrema of the function on the closed interval. $$ f(x)=\sin x,\left[\frac{5 \pi}{6}, \frac{11 \pi}{6}\right] $$

Step-by-Step Solution

Answer

The absolute maximum of \(f(x)\) on the given interval is \(1/2\) at \(x = 5\pi/6\) and \(x = 11\pi/6\), and the absolute minimum is \(-1\) at \(x = 3\pi/2\).

Step 1: Differentiate the function

Differentiate the function \(f(x) = \sin x\). The derivative of \(\sin x\) is \(\cos x\) , denoted \(f'(x) = \cos x\).

Step 2: Find the critical points

Set \(f'(x)\) equal to zero and solve for \(x\). This gives \(\cos x = 0\), and the solutions in the interval \(\left[\frac{5 \pi}{6}, \frac{11 \pi}{6}\right]\) are \(x = \frac{3\pi}{2}\).

Step 3: Create a table of function values at the critical points and interval endpoints

Evaluate \(f(x)\) at the critical point \(x = \frac{3\pi}{2}\), and at the interval endpoints \(x = \frac{5\pi}{6}\) and \(x = \frac{11\pi}{6}\). The values are: \(f(\frac{5\pi}{6}) = \frac{1}{2}\), \(f(\frac{11\pi}{6}) = \frac{1}{2}\), and \(f(\frac{3\pi}{2}) = -1\).

Step 4: Find the absolute maximum and minimum

The maximum function value in the table is \(\frac{1}{2}\), and the minimum value is \(-1\). Therefore, the absolute maximum of \(f(x)\) on the given interval is \(\frac{1}{2}\) at \(x = \frac{5\pi}{6}\) and \(x = \frac{11\pi}{6}\), and the absolute minimum is \(-1\) at \(x = \frac{3\pi}{2}\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, referring to the process of finding the derivative of a function. The derivative provides the rate at which a function is changing at any given point. In simpler terms, it tells us the slope of the tangent line to the function at that point. To find the derivative of a function like \( f(x) = \sin x \), we use established differentiation rules. For trigonometric functions, the derivative of \( \sin x \) is \( \cos x \). This is because differentiation involves analyzing how tiny changes in \( x \) affect the function, and correspondingly, how quickly the function changes. Key Points:

• Differentiation finds the rate of change of a function.
• The derivative of \( \sin x \) is \( \cos x \).
• Understanding derivatives is essential to solve optimization problems.

After differentiating, we can proceed to find other aspects like critical points, which are crucial for determining the behavior of the function.

Critical Points

Critical points play a key role in finding extrema (maximum and minimum values) of a function. A critical point exists where the derivative of a function is zero or undefined. In these instances, the function might change direction, indicate a peak, or drop to a valley.In the given problem, we differentiate \( f(x) = \sin x \) to get \( f'(x) = \cos x \). Setting \( f'(x) \) equal to zero gives us \( \cos x = 0 \), leading to critical points within the interval.To solve \( \cos x = 0 \), we need to find the values of \( x \) in the interval \( \left[\frac{5 \pi}{6}, \frac{11 \pi}{6}\right] \). We find that \( x = \frac{3\pi}{2} \) is a critical point where the function can potentially attain an extremum.Top Tips:

• Find critical points by setting the derivative to zero.
• Evaluate within the given interval to ensure accuracy.
• Critical points are potential candidates for extrema.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are essential in understanding periodic phenomena. They are used in various fields such as physics, engineering, and architecture, thanks to their cyclical nature.For this exercise, we focused on the sine function, \( f(x) = \sin x \). This function oscillates between -1 and 1, showing its periodic nature. The sine function graph is wave-like, and understanding this graph can help visualize peak and trough points which correspond to maximum and minimum values, respectively.

• Sine function oscillates between -1 and 1.
• It's periodic, with a regular cycle repeating every \(2\pi\).
• Graph understanding aids in identifying extrema visually.

For our interval \( \left[\frac{5 \pi}{6} , \frac{11 \pi}{6} \right] \), it is essential to understand where the sine function attains its peaks and valleys, aiding in the extrema analysis.

Absolute Maximum and Minimum

The concept of absolute maximum and minimum is central to optimization problems. These are the highest and lowest values a function takes on, respectively, within a given interval.To find these values within the interval \( \left[\frac{5 \pi}{6}, \frac{11 \pi}{6}\right] \), we check both the critical points and the endpoints of the interval. Evaluating \( f(x) = \sin x \) at these points: - At \( x = \frac{5\pi}{6} \) and \( x = \frac{11\pi}{6} \): \( f(x) = \frac{1}{2} \).- At the critical point \( x = \frac{3\pi}{2} \): \( f(x) = -1 \).This analysis shows that the absolute maximum within the interval is \( \frac{1}{2} \) and the absolute minimum is \(-1\). Evaluating the function at key points is crucial.Key Steps:

• Check function values at both critical points and endpoints.
• Compare these values to identify maxima and minima.
• Confirm that the interval is correctly addressed for accurate results.

Finding these extrema allows us to understand the function's behavior over the specified range.