Problem 38 - Thomas Calculus

Question

In Exercises \(37-40\) , find the function's absolute maximum and minimum values and say where they occur. $$ f(x)=x^{5 / 3}, \quad-1 \leq x \leq 8 $$

Step-by-Step Solution

Answer

The absolute maximum is 32 at \(x = 8\), and the absolute minimum is -1 at \(x = -1\).

Step 1: Understand the Problem

To find absolute maximum and minimum values of the function \( f(x) = x^{5/3} \) over the interval \(-1 \leq x \leq 8\), we must evaluate the function at critical points within the interval and the endpoints of the interval.

Step 2: Find the Derivative

First, calculate the derivative of the function \( f(x) = x^{5/3} \). The derivative is \( f'(x) = \frac{5}{3}x^{2/3} \).

Step 3: Find Critical Points

A critical point occurs when the derivative is zero or undefined. Since \( f'(x) = \frac{5}{3}x^{2/3} \), it is never zero because \( \frac{5}{3} eq 0 \) and \( x^{2/3} \) is zero only at \( x = 0 \). Thus, \( x = 0 \) is a critical point because the derivative is undefined there.

Step 4: Evaluate the Function at Critical and Endpoint Values

Substitute the critical point and interval endpoints into \( f(x) \): - At \( x = -1 \), \( f(-1) = (-1)^{5/3} = -1 \). - At \( x = 0 \), \( f(0) = 0^{5/3} = 0 \). - At \( x = 8 \), \( f(8) = 8^{5/3} = 32 \).

Step 5: Determine Absolute Maximum and Minimum

Compare the function values: - \( f(-1) = -1 \) - \( f(0) = 0 \) - \( f(8) = 32 \) The absolute maximum value is 32 at \( x = 8 \) and the absolute minimum value is -1 at \( x = -1 \).

Key Concepts

Critical Points

Critical points in calculus are the values of \(x\) where the derivative of a function is either zero or undefined. These points are essential in finding the local maximum and minimum values of a function, as well as the absolute extrema within a given interval.In our exercise, the function \(f(x) = x^{5/3}\) has its derivative \(f'(x) = \frac{5}{3}x^{2/3}\). Here, the derivative is undefined at \(x = 0\) since the expression \(x^{2/3}\) creates a division by zero issue at this point. This makes \(x = 0\) a critical point of the function.To summarize, exploring critical points can give us more insights into where the function might change direction, thus indicating potential maximum or minimum values.

Derivative

The derivative of a function helps us understand the rate at which the function's value is changing at any given point. For the function \(f(x) = x^{5/3}\), the derivative is calculated as \(f'(x) = \frac{5}{3}x^{2/3}\).By finding the derivative, we set the stage for identifying critical points. The derivative tells us how the function's slope behaves:

• If the derivative is positive, the function is increasing.
• If the derivative is negative, the function is decreasing.
• If the derivative is zero, that point could be a local max, min, or a saddle point.

In this exercise, since the derivative \(f'(x)\) is never zero, the function does not have local maxima or minima by ordinary critical point conditions. But don't forget, derivatives that don't exist (such as at \(x=0\) for radial power functions) also count as critical points!

Interval Endpoints

When finding absolute maximum and minimum values on a closed interval, it's necessary to evaluate the function at the interval endpoints. These endpoints are determined by the boundaries of the interval provided in the problem.For our exercise, the interval is \(-1 \leq x \leq 8\). Therefore, we consider \(x = -1\) and \(x = 8\) as the endpoints.This step is crucial because although critical points tell us about potential max or min values, the endpoints can still represent absolute extrema within the interval. Endpoints provide fixed positions outside the interior where the function values may be most or least.

Evaluate Function Values

Once critical points and interval endpoints are identified, the next step is to evaluate the function at these points to find the absolute maximum and minimum values.In this problem, we've computed:- \(f(-1) = (-1)^{5/3} = -1\)- \(f(0) = 0^{5/3} = 0\)- \(f(8) = 8^{5/3} = 32\)By comparing these values, we identify that the absolute maximum value is 32, occurring at \(x = 8\), and the absolute minimum value is -1, occurring at \(x = -1\). Thus, this process of evaluating function values solidifies understanding of where the largest and smallest outputs of the function appear over the entire interval.