Problem 14 - Calculus Early Transcendentals

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

The absolute maximum is \(\sqrt{2}\) at \(x = \frac{3\pi}{4}\), and the minimum is \(-1\) at \(x = 0\).

Step 1: Find the derivative of f(x)

The function given is \( f(x) = \sin x - \cos x \). We need to find the derivative \( f'(x) \), which will allow us to find the critical points. The derivative is calculated as follows:\[ f'(x) = \frac{d}{dx}(\sin x) - \frac{d}{dx}(\cos x) = \cos x + \sin x \]

Step 2: Find critical points

To find the critical points, set the derivative \( f'(x) = \cos x + \sin x \) equal to zero:\[ \cos x + \sin x = 0 \]Solving the equation \( \sin x = -\cos x \) gives \( \tan x = -1 \). In the interval \([0, \pi]\), \( x = \frac{3\pi}{4} \) is the solution to \( \tan x = -1 \).

Step 3: Evaluate f(x) at the critical points and endpoints

Now, calculate the values of \( f(x) \) at the critical points and the endpoints of the interval \([0, \pi]\):- \( x = 0 \): \( f(0) = \sin 0 - \cos 0 = -1 \)- \( x = \frac{3\pi}{4} \): \( f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{4}) - \cos(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \)- \( x = \pi \): \( f(\pi) = \sin \pi - \cos \pi = 1 \)

Step 4: Determine the absolute maximum and minimum

Among the values \(-1\), \(\sqrt{2}\), and \(1\), the maximum is \(\sqrt{2}\), which occurs at \( x = \frac{3\pi}{4} \), and the minimum is \(-1\), which occurs at \(x = 0\).

Key Concepts

Critical Points

Understanding critical points is crucial in calculus, especially for finding extreme values such as maximums or minimums of a function. A critical point occurs where the derivative of a function is zero or undefined.
This means the slope of the tangent line to the function at that point is horizontal (it doesn't slant up or down).To find critical points, you follow these steps:

• First, compute the derivative of the function.
• Next, set this derivative equal to zero and solve for the variable.
• Lastly, check for any points where the derivative is undefined.

In our example, we found the derivative to be \( f'(x) = \cos x + \sin x \) and set it to zero.
Solving \( \cos x + \sin x = 0 \) gives us the critical point \( x = \frac{3\pi}{4} \).
This shows us where we have potential maxima or minima.
Always remember to consider the domain of the function and check endpoint values as well, as was done with the interval \([0, \pi]\).

Derivative

The derivative of a function is a fundamental concept in calculus.
It measures the rate at which a function is changing at any given point, effectively giving us the slope of the tangent line to the function at that point.
Derivatives are particularly useful when you need to identify critical points or understand the behavior of a function.To find the derivative of a function, you need to know some basic derivative rules like:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \(-\sin x \).

In the provided exercise, we calculated the derivative of \( f(x) = \sin x - \cos x \) by using these rules:
So, \( f'(x) = \cos x + \sin x \).
This derivative was then used to find points where the function’s slope is zero, by setting it equal to zero and solving the equation.
Derivatives can also show us where a function is increasing or decreasing based on the sign of \( f'(x) \).

Trigonometric Functions

Trigonometric functions like sine and cosine are essential in calculus for modeling periodic phenomena, among other applications.
These functions have well-known properties and derivatives:

• The sine function, \( \sin x \), varies between -1 and 1, repeating every \( 2\pi \) radians.
• Similarly, the cosine function, \( \cos x \), ranges from -1 to 1, also repeating every \( 2\pi \) radians.

In this exercise, the function \( f(x) = \sin x - \cos x \) combines both.
Its behavior depends on both the sine and cosine through their ranges and periodicity.
To interpret these functions, recall the key points within their period, such as \( x = 0, \frac{\pi}{2}, \pi \), to help compute values effectively.Understanding how \( \sin x \) and \( \cos x \) interact in expressions is vital when solving problems involving trigonometric functions like finding intersections or solving equations such as \( \tan x = -1 \), which relates sine and cosine directly.