Problem 20 - Thomas Calculus in SI Units

Question

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. $$f(x)=\left\\{\begin{array}{lr} x+1, & -1 \leq x < 0 \\ \cos x, & 0 < x \leq \frac{\pi}{2} \end{array}\right.$$

Step-by-Step Solution

Answer

The function has an absolute maximum of 1 and an absolute minimum of 0 on its domain, consistent with Theorem 1.

Step 1: Identify the Domain and Function Segments

The function \(f(x)\) is piecewise and consists of two segments. The first segment is \(f(x) = x + 1\) for \(-1 \leq x < 0\). The second segment is \(f(x) = \cos x\) for \(0 < x \leq \frac{\pi}{2}\). Identify the domain as \([-1, \frac{\pi}{2}]\).

Step 2: Analyze the First Segment

For \(-1 \leq x < 0\), the function is linear: \(f(x) = x + 1\). This segment decreases from \(f(-1) = 0\) to \(f(0^-)=1\). The value at \(x = 0\) is not included since \(x < 0\).

Step 3: Analyze the Second Segment

For \(0 < x \leq \frac{\pi}{2}\), the function is \(f(x) = \cos x\). This part of the function starts from just right of zero and reduces from \(f(0^+)\) close to \(1\) down to \(f(\frac{\pi}{2}) = 0\).

Step 4: Combine and Sketch the Graph

Draw the piecewise function consisting of two segments on the number line from \(-1to \frac{\pi}{2}\). The linear segment starts at 0 (inclusive) and ends approaching 1 (not reaching 1). The cosine segment starts close to 1 (exclusive) and decreases to 0.

Step 5: Identify Absolute Extreme Values

Consider both segments to find absolute extremum. The maximum value, 1, is approached by the first segment \(x\to 0^-\) and immediately achieved at \(x\to 0^+\) in the second segment. The minimum reported value is \(0\) at \(x = -1\) or \(x = \frac{\pi}{2}\). So the absolute maximum is 1, and the absolute minimum is 0.

Step 6: Verify with Theorem 1

Theorem 1 states that a continuous function on a closed interval has a maximum and minimum value. Each segment in \(f(x)\) is continuous, and the domain of \([-1, \frac{\pi}{2}]\) is closed with endpoints forming extrema, satisfying the theorem with absolute maximum 1 and minimum 0.

Key Concepts

Graph Sketching

Understanding how to sketch a graph of a piecewise function is essential in visualizing how functions behave over specific intervals. For the function given in the exercise, it is presented in two distinct segments over its domain of either a linear form or a trigonometric form.

• From -1 to just before 0, the function follows the line equation \(f(x) = x + 1\). This linear part decreases steadily from \(f(-1) = 0\) to approximately \(f(0^-) = 1\).
• In the interval just after 0 to \(\frac{\pi}{2}\), the behavior switches to that of a cosine function \(f(x) = \cos x\), starting close to 1 and decreasing toward \(f(\frac{\pi}{2}) = 0\).

The sketch should reflect these changes in direction and steepness. When sketching, take note of the intervals and endpoints: the linear part begins at an inclusive 0 and progresses, while the cosine section does not actually touch the initial value of 1 but instead continues from just below it. Breaking down each segment makes graph sketching more manageable and intuitive.

Absolute Extrema

Finding absolute extrema involves identifying the highest and lowest points a function reaches over its domain. For piecewise functions, we need to consider each segment individually and collectively. The linear portion \(f(x) = x + 1\) on the interval \(-1 \leq x < 0\) shows that:

• The minimum value is 0 at \(x = -1\).
• The maximum value approaches 1 as \(x\) gets closer to 0 from the left.

For the cosine segment \(f(x) = \cos x\), valid in \(0 < x \leq \frac{\pi}{2}\):

• The value moves from 1 to 0, with the maximum of 1 achieved just after 0 and the absolute minimum 0 happening at \(x = \frac{\pi}{2}\).

Thus, the absolute maximum for the entire function is 1 (attained at the division point \(x = 0^+\)), and the absolute minimum is 0 at both endpoints of the domain, ensuring each segment is thoroughly evaluated.

Theorem 1 in Calculus

Theorem 1 is often known as the Extreme Value Theorem, which states that if a function is continuous over a closed interval, then it must attain a maximum and minimum value, i.e., extrema. In the case of this exercise, each piece of the function independently satisfies the theorem.

• The segment \(f(x) = x + 1\) maintains continuity over the interval \(-1 \leq x < 0\).
• The segment \(f(x) = \cos x\) stays continuous over \(0 < x \leq \frac{\pi}{2}\).

Although there's a switch in function behavior at \(x = 0\), the key is recognizing continuity within each defined segment and assessing extrema. The consistency of this piecewise function with the theorem comes from verifying that each function portion shows continuity over its specific interval. The entire domain \([-1, \frac{\pi}{2}]\) being closed and bounded ensures that these extrema (maximum of 1 and minimum of 0) align with the theorem's prediction. Understanding Theorem 1 helps confirm that all potential max/min points are considered for a piecewise function using interval and endpoint behavior.