Problem 26 - 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. $$ f(x)=x^{2 / 3} ;[-8,8] $$

Step-by-Step Solution

Answer

The minimum value is 0 at \( x = 0 \), and the maximum value is 4 at \( x = -8 \) and \( x = 8 \).

Step 1: Understand the Problem

We are tasked with finding the extreme values of the function \( f(x) = x^{2/3} \) on the interval \([-8, 8]\). Extreme values include local maxima, local minima, and can also be endpoints of the interval.

Step 2: Calculate the Derivative of the Function

The first step to find critical points is to compute the derivative of the function. The derivative of \( f(x) = x^{2/3} \) using the power rule is \( f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3} \frac{1}{x^{1/3}} \).

Step 3: Identify Critical Points

Critical points occur where the derivative is zero or undefined. Since \( f'(x) = \frac{2}{3x^{1/3}} \), this function is undefined for \( x = 0 \), giving us a critical point at \( x = 0 \).

Step 4: Evaluate the Function at Critical Points and Endpoints

Evaluate \( f(x) \) at critical points and the endpoints of the interval: \(-8\), \(0\), and \(8\). Calculate: \( f(-8) = (-8)^{2/3} = 4 \), \( f(0) = 0^{2/3} = 0 \), \( f(8) = 8^{2/3} = 4 \).

Step 5: Determine Extreme Values

Compare values from Step 4. The smallest value is \(0\) at \( x = 0 \), making it the global minimum within the interval. The largest value is \(4\) at both \( x = -8 \) and \( x = 8 \), these are global maxima.

Key Concepts

Critical Points

In calculus, understanding critical points is key to determining extreme values. Critical points of a function occur at values of x where the derivative is zero or undefined. This is because it is at these points where the function's slope changes, potentially indicating peaks or valleys. For the function \( f(x) = x^{2/3} \), the derivative \( f'(x) = \frac{2}{3}x^{-1/3} \) is undefined at \( x = 0 \). This is because the expression \( 1/x^{1/3} \) is undefined when \( x = 0 \). Therefore, \( x = 0 \) is a critical point. In practical terms, critical points are the starting places for identifying local maxima or minima in functions.

Derivative Calculus

Derivative calculus is essential for understanding how functions change. It's the mathematical tool we use to find critical points and further analyze functions. To find the derivative of a function like \( f(x) = x^{2/3} \), we apply the power rule. This rule states that if \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \). Applying this to our function, we get \( f'(x) = \frac{2}{3}x^{-1/3} \). This shows how the function's rate of change depends on x. Recognizing that the derivative tells us the slope of the function at any given point is crucial. A zero or undefined slope often signals a critical point, thus aiding in locating potential extreme values.

Local Maxima and Minima

To identify local maxima and minima, you need to recognize where the function reaches its highest or lowest points in a given interval. These points can occur at critical points or endpoints. For our function \( f(x) = x^{2/3} \) on the interval \([-8, 8]\), we checked the values at the endpoints \( x = -8 \) and \( x = 8 \), as well as at the critical point \( x = 0 \). By evaluating these, we found that the function yields \( f(-8) = 4 \), \( f(0) = 0 \), and \( f(8) = 4 \). Therefore, the local minimum occurs at \( x = 0 \) with a value of 0, and the local maxima occur at \( x = -8 \) and \( x = 8 \) with a value of 4. This thorough examination of each relevant point ensures the correct assessment of both local maxima and minima, helping us understand the function's behavior on the interval.