Problem 14 - Calculus - AP Edition

Question

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. $$f(x)=\sin x-\cos x ;[0, \pi]$$

Step-by-Step Solution

Answer

Maximum: \(\sqrt{2}\) at \(x=\frac{3\pi}{4}\); Minimum: \(-1\) at \(x=0\).

Step 1: Identify Critical Points

To find critical points, we first differentiate the function \(f(x) = \sin x - \cos x\). The derivative is \(f'(x) = \cos x + \sin x\). Set the derivative equal to zero to find critical points: \(\cos x + \sin x = 0\). Solve for \(x\):\[\tan x = -1 \\Rightarrow x = \frac{3\pi}{4}\] since it must be within \([0, \pi]\).

Step 2: Evaluate Function at Critical Points

Evaluate \(f(x)\) at the critical point and at the endpoints of the interval. Compute \(f(0), f(\pi),\) and \(f\left(\frac{3\pi}{4}\right)\):\[f(0) = \sin(0) - \cos(0) = 0 - 1 = -1 \f(\pi) = \sin(\pi) - \cos(\pi) = 0 + 1 = 1 \f\left(\frac{3\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right) - \cos\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} - \left(-\frac{\sqrt{2}}{2}\right) = \sqrt{2} \approx 1.414\]

Step 3: Determine the Absolute Maximum and Minimum

Compare the function values from Step 2. The values are:- \(f(0) = -1\)- \(f(\pi) = 1\)- \(f\left(\frac{3\pi}{4}\right) = \sqrt{2}\)The absolute maximum value is \(f\left(\frac{3\pi}{4}\right) = \sqrt{2}\) (approximately 1.414), and it occurs at \(x = \frac{3\pi}{4}\). The absolute minimum value is \(f(0) = -1\), occurring at \(x = 0\).

Key Concepts

Critical Points

In calculus, critical points are crucial in finding the extremum points of a differentiable function. To locate these points for a function like \(f(x) = \sin x - \cos x\), you'll need to find its derivative and set it to zero. The derivative here is \(f'(x) = \cos x + \sin x\). By solving the equation \(\cos x + \sin x = 0\), you identify where the slope of the tangent to the curve is zero, indicating a potential extremum.
This process resulted in the critical point \(x = \frac{3\pi}{4}\). Remember, finding critical points requires checking the range you're investigating, as not all solutions might be within your interval.

Absolute Maximum and Minimum

Absolute maximum and minimum values of a function on a closed interval are the highest and lowest values the function achieves over that specific range. For \(f(x) = \sin x - \cos x\) on \([0, \pi]\), you begin by evaluating \(f\) at critical points and endpoints of the interval.

• At \(x = 0\), \(f(0) = -1\).
• At \(x = \frac{3\pi}{4}\), \(f\left(\frac{3\pi}{4}\right) \approx 1.414\).
• At \(x = \pi\), \(f(\pi) = 1\).

Comparing these results, \(f\left(\frac{3\pi}{4}\right)\) is the absolute maximum because it is the highest value, and \(f(0)\) is the absolute minimum as it is the lowest.

Derivative

The derivative is a fundamental concept that describes the rate at which a function changes. For this problem, the derivative of \(f(x) = \sin x - \cos x\) is \(f'(x) = \cos x + \sin x\).

• The derivative was used to find critical points by setting \(f'(x) = 0\).
• This step ensures that the slope is zero, indicating a possible extremum like a maximum or minimum.

Understanding how derivatives help determine the shape and behavior of a function is essential for solving optimization problems like this.

Trigonometric Functions

Trigonometric functions play a pivotal role in calculus, especially when dealing with periodic phenomena. This exercise involved the sine and cosine functions in the function \(f(x) = \sin x - \cos x\).

• These functions are differentiable over their entire domains, allowing the application of calculus tools like derivatives.
• The critical point found amidst trigonometric functions \(x = \frac{3\pi}{4}\) aligns with specific angles where ratios, such as \(\tan x = -1\), denote special relationships between sine and cosine.

Trigonometric identities and their derivatives are invaluable for analyzing and comprehending periodic functions effectively.