Problem 28 - Calculus with Concepts in Calculus

Question

Find all extreme values (if any) of the given function on the given interval. Determine at which numbers in the interval these values occur. $$ f(x)=\cos \pi x ;\left(\frac{1}{3}, 1\right] $$

Step-by-Step Solution

Answer

The minimum value is -1 at \( x = 1 \). No maximum value within the open interval \( \left(\frac{1}{3}, 1\right] \).

Step 1: Identify the Critical Points

To find the critical points of the function \( f(x) = \cos(\pi x) \), we first need to find its derivative. The derivative is \( f'(x) = -\pi \sin(\pi x) \). Critical points occur where \( f'(x) = 0 \) or where \( f'(x) \) is undefined. Set \(-\pi \sin(\pi x) = 0 \) and solve for \( x \).

Step 2: Solve for Critical Points

Solving \(-\pi \sin(\pi x) = 0 \) gives \( \sin(\pi x) = 0 \). This occurs when \( \pi x = n\pi \) where \( n \) is an integer, leading to \( x = n \). In the interval \( \left(\frac{1}{3}, 1\right] \), the only integer is \( n = 1 \), corresponding to \( x = 1 \).

Step 3: Evaluate the Function at Critical Points and Endpoints

Evaluate \( f(x) = \cos(\pi x) \) at the critical point and endpoints. For \( x = 1 \), \( f(1) = \cos(\pi) = -1 \). At \( x = \frac{1}{3} \), \( f\left(\frac{1}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \). Only evaluate at the right endpoint, since \( x = \frac{1}{3} \) is not included in the interval by a strict inequality.

Step 4: Determine Extreme Values

In the interval \( \left(\frac{1}{3}, 1\right] \), \( f(x) \) takes the value \( -1 \) at \( x = 1 \), and the highest value it takes as it approaches \( x = \frac{1}{3} \) is \( \frac{1}{2} \). Therefore, \( -1 \) is the minimum value on the interval at \( x = 1 \).

Key Concepts

Critical Points

Critical points are essential in determining where a function might have its extreme values, such as maxima or minima, within a given interval. To find these points, we look at where the derivative of the function is zero or undefined. When the derivative equals zero, it indicates a potential turning point, where the function's slope changes from positive to negative or vice versa.

In the exercise, the function given is \( f(x) = \cos(\pi x) \). The derivative of this function is \( f'(x) = -\pi \sin(\pi x) \). By setting this derivative equal to zero, \( -\pi \sin(\pi x) = 0 \), we can solve for \( x \). Simplifying gives \( \sin(\pi x) = 0 \), leading to \( \pi x = n\pi \) where \( n \) is an integer. Therefore, \( x = n \) are potential critical points.

In the restricted interval \( \left(\frac{1}{3}, 1\right] \), the only integer value satisfying this condition is \( n = 1 \), giving \( x = 1 \) as a critical point.

Derivative of Trigonometric Functions

Understanding the derivative of trigonometric functions is crucial in calculus when working with periodic functions like sine and cosine. For the function \( f(x) = \cos(\pi x) \), using the chain rule is necessary due to the inner function \( \pi x \).

The derivative of \( \cos(u) \) is \( -\sin(u) \), and thus applying the chain rule, where \( u = \pi x \) leads to \( \frac{du}{dx} = \pi \), gives us:

• \( f'(x) = -\pi \sin(\pi x) \)

Here, \(-\pi\) acts as a scaling factor changing the amplitude of \( \sin(\pi x) \). Understanding how these derivatives are formed is key to solving problems where trigonometric functions are involved, especially when looking for critical points or analyzing the behavior of functions over intervals.

Interval Analysis

Interval analysis involves evaluating the behavior of a function over a specific range of input values to find its extreme values, such as maximum or minimum values. The interval provided in this exercise is \( \left(\frac{1}{3}, 1\right] \). This notation tells us that \( x \) can take values starting just beyond \( \frac{1}{3} \) yet including 1.

The process begins by identifying critical points within the interval using the derivative. In our example with \( f(x) = \cos(\pi x) \), \( x = 1 \) is the only critical point within this interval. We then evaluate the function at these critical points and the endpoints of the interval. However, since \( \frac{1}{3} \) is not included in the interval according to the strict inequality \((\frac{1}{3}
Evaluations show:

• At \( x = 1 \), \( f(1) = \cos(\pi) = -1 \), which is the minimum value of the function on the interval.
• Approaching \( x = \frac{1}{3} \), \( f(x) \) approaches \( \cos(\frac{\pi}{3}) = \frac{1}{2} \).

Thus, these computations confirm the extreme values on the given interval.