Absolute Extrema on Finite Closed Intervals


In Exercises 37–40, find the function’s absolute maximum and minimum values and say where they occur.


f(x) = x⁴ᐟ³, −1 ≤ x ≤ 8

Verified step by step guidance

1. Identify the function f(x) = x^(4/3) and the closed interval [-1, 8]. We need to find the absolute maximum and minimum values of this function on the given interval.
2. Calculate the derivative of the function to find the critical points. The derivative f'(x) is found using the power rule: f'(x) = (4/3)x^(1/3).
3. Set the derivative equal to zero to find critical points: (4/3)x^(1/3) = 0. Solve for x to find the critical points within the interval.
4. Evaluate the function f(x) at the critical points found in the previous step, as well as at the endpoints of the interval, x = -1 and x = 8.
5. Compare the values of f(x) at the critical points and the endpoints to determine the absolute maximum and minimum values on the interval [-1, 8].

Key Concepts

Absolute Extrema

Absolute extrema refer to the highest and lowest values a function can achieve on a given interval. To find these values, evaluate the function at critical points and endpoints of the interval. The absolute maximum is the largest value, and the absolute minimum is the smallest value within the specified range.

Critical Points

Critical points are values of x where the derivative of the function is zero or undefined. These points are potential candidates for local maxima or minima. To find critical points, take the derivative of the function, set it equal to zero, and solve for x. Also, check where the derivative does not exist within the interval.

Evaluating Endpoints

Evaluating endpoints involves calculating the function's value at the boundaries of the interval. For the function f(x) = x⁴ᐟ³ on the interval [-1, 8], compute f(-1) and f(8). These values, along with those at critical points, help determine the absolute extrema by comparing them to find the maximum and minimum values.