Problem 19 - Thomas Calculus

Question

In Exercises \(15-20,\) sketch the graph of each function and determine whether the function has any absolute extreme values on its domain. Explain how your answer is consistent with Theorem \(1 .\) $$ y=3 \sin x, \quad 0 < x < 2 \pi $$

Step-by-Step Solution

Answer

No absolute extremes exist on the open interval, but there are local extremes at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

Step 1: Understanding the Function

The function given is \( y = 3 \sin x \) over the interval \( 0 < x < 2\pi \). This function is a sine function scaled by a factor of 3.

Step 2: Graph the Function

Plot the function \( y = 3 \sin x \) on a coordinate plane. Note that the sine function oscillates between -1 and 1, so \( 3\sin x \) will oscillate between -3 and 3. The graph will complete one full wave as \( x \) increases from 0 to \( 2\pi \).

Step 3: Locate Critical Points

Find the derivative \( y' = 3\cos x \). Set the derivative equal to zero to find critical points: \( 3\cos x = 0 \). Solving this gives \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \) within the open interval \( 0 < x < 2\pi \).

Step 4: Evaluate Function at Critical Points

Evaluate the function at the critical points found: \( y\left(\frac{\pi}{2}\right) = 3 \times 1 = 3 \) and \( y\left(\frac{3\pi}{2}\right) = 3 \times (-1) = -3 \). These are the local maximum and minimum, respectively.

Step 5: Check for Absolute Extremes

As the interval \( 0 < x < 2\pi \) is open, we need to determine if there are any absolute extremes. The highest point within this domain is 3, and the lowest is -3, which are the local extremes found at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

Step 6: Refer to Theorem 1 (Justification)

Theorem 1 usually states that the Extreme Value Theorem applies to a continuous function on a closed interval. Since our interval is open, no absolute extremes exist as per the theorem because \( y = 3\sin x \) does not attain its maximum or minimum values on a closed interval \( [0, 2\pi] \). However, local extremes are found as calculated.

Key Concepts

Critical Points

In the study of calculus, finding critical points is a crucial step in identifying where a function may have local maxima or minima. Critical points occur where the derivative of a function equals zero or is undefined.
For the function given as \(y = 3\sin x\), we begin by finding its derivative: \(y' = 3\cos x\). To locate critical points, we set this derivative equal to zero:

• \(3\cos x = 0\)
• Solving \(\cos x = 0\) gives us critical points at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\) within the interval \(0 < x < 2\pi\).

At these points, the slope of the tangent to the graph is zero, indicating potential maxima or minima in the graph of the function. Evaluating \(y\) at these points shows that \(y\left(\frac{\pi}{2}\right) = 3\) and \(y\left(\frac{3\pi}{2}\right) = -3\), identifying local maximum and minimum, respectively.

Sine Function Graph

The sine function is a periodic function, and its graph, known as a sine wave, is shaped like smooth, continuous waves. In the function \(y = 3\sin x\), the sine wave is vertically stretched by a factor of 3.
This means the maximum and minimum values reached by \(y = 3\sin x\) are 3 and -3, respectively.
Graphing \(y = 3\sin x\) over the interval \(0 < x < 2\pi\), it completes one full cycle. It's essential to remember the sine wave oscillates symmetrically about the x-axis.

• The wave reaches a peak (maximum) at \(x = \frac{\pi}{2}\) and a trough (minimum) at \(x = \frac{3\pi}{2}\).
• The wave continues to oscillate periodically, extending beyond these points should the interval allow.

Understanding the shape of sine waves helps us anticipate where critical points may occur and how they relate to the function's behavior over its domain.

Extreme Value Theorem

The Extreme Value Theorem is a fundamental concept in calculus. It states that if a function is continuous over a closed interval, then it must have both a maximum and a minimum value on that interval. However, this theorem does not necessarily apply if the interval is open, like our case.
The function \(y = 3\sin x\) on the open interval \(0 < x < 2\pi\) does not meet the closed interval criterion of the Extreme Value Theorem. Therefore, no absolute maximum or minimum exists within this open interval.
Although local extremes are found at \(x = \frac{\pi}{2}\) (maximum) and \(x = \frac{3\pi}{2}\) (minimum), the theorem requires a closed interval to guarantee absolute extremes. This underscores the importance of considering interval type (closed vs. open) when applying the Extreme Value Theorem.