Problem 38 - Thomas Calculus in SI Units

Question

Find the function's absolute maximum and minimum values and say where they are assumed. $$f(x)=x^{5 / 3}, \quad-1 \leq x \leq 8$$

Step-by-Step Solution

Answer

Absolute max: 32 at x=8; Absolute min: -1 at x=-1.

Step 1: Find the critical points

First, we need to find the derivative of the function \( f(x) = x^{5/3} \). The derivative of \( x^{5/3} \) using the power rule is \( f'(x) = \frac{5}{3} x^{2/3} \). Set the derivative equal to zero to find critical points: \( \frac{5}{3} x^{2/3} = 0 \). Solving for \( x \) gives \( x = 0 \). We will also check where the derivative doesn’t exist, which is also \( x = 0 \). Thus, the only critical point is at \( x = 0 \).

Step 2: Evaluate the function at the endpoints and critical points

Next, evaluate the function \( f(x) = x^{5/3} \) at the endpoints of the interval \([-1, 8]\), and at the critical point \( x = 0 \):- \( f(-1) = (-1)^{5/3} = -1 \)- \( f(0) = (0)^{5/3} = 0 \)- \( f(8) = (8)^{5/3} = 32 \)

Step 3: Determine the absolute maximum and minimum

Compare the values from Step 2. The function values we have at specific points are: \( f(-1) = -1 \), \( f(0) = 0 \), and \( f(8) = 32 \). Therefore, the lowest value is \(-1\) at \( x = -1 \), which is the absolute minimum. The highest value is \(32\) at \( x = 8 \), which is the absolute maximum.

Key Concepts

Critical Points

Critical points of a function are where the first derivative is either zero or undefined. These points are essential because they can be potential candidates for local maxima or minima. To find them, we first calculate the derivative of the function. For the function given, \( f(x) = x^{5/3} \), the derivative is \( f'(x) = \frac{5}{3} x^{2/3} \).

• Set the derivative equal to zero: \( \frac{5}{3} x^{2/3} = 0 \).
• Solving for \( x \), we find the critical point at \( x = 0 \).

Additionally, it's important to check where the derivative does not exist. For \( f'(x) = \frac{5}{3} x^{2/3} \), it doesn't exist where \( x = 0 \), reaffirming that \( x = 0 \) is a critical point. Finding critical points is essential in assessing where a function could achieve its extreme values within a specified interval.

Derivative

A derivative represents the rate of change of a function concerning its variable. It can help identify how the function behaves over a range of values, which is why it’s crucial in finding both local and absolute extrema. When given a function like \( f(x) = x^{5/3} \), we use differentiation rules, such as the power rule, to find the derivative. The power rule states that if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). Applying this, we get:

• \( f'(x) = \frac{5}{3} x^{2/3} \).

Using the derivative, we can explore the function's increase or decrease trends and highlight important areas where abrupt changes occur, such as critical points. Understanding how to differentiate is an essential skill for navigating calculus problems related to optimization and analysis of functions.

Endpoints Evaluation

Endpoints evaluation is a method used to find the absolute maximum and minimum of a function over a given interval. After identifying critical points, we must evaluate the function at these points as well as at the endpoints of the interval.

• For the function \( f(x) = x^{5/3} \), defined over \([-1, 8]\), we evaluate:
• \( f(-1) = (-1)^{5/3} = -1 \),
• \( f(0) = (0)^{5/3} = 0 \),
• \( f(8) = (8)^{5/3} = 32 \).

By comparing these values, we determine the function's behavior across the interval. The smallest value \(-1\) at \(x = -1\) is the absolute minimum, and the highest value \(32\) at \(x = 8\) is the absolute maximum. Checking endpoints is crucial as they can sometimes hold the extreme values of the function within the restricted interval.