Math Problem Statement

Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

f(x) = 8x − 

2

x

 on [2, 4]

maximum   minimum

Solution

To find the absolute maximum and minimum values of the function f(x)=8x2xf(x) = 8x - \frac{2}{x} on the interval [2,4][2, 4], we need to follow these steps:

  1. Compute the derivative of f(x)f(x) and set it equal to zero to find critical points.
  2. Evaluate the function at the endpoints of the interval and at any critical points within the interval.
  3. Compare these values to determine the absolute maximum and minimum.

Step 1: Find f(x)f'(x)

The function is: f(x)=8x2xf(x) = 8x - \frac{2}{x} Differentiating with respect to xx: f(x)=8+2x2f'(x) = 8 + \frac{2}{x^2}

To find critical points, set f(x)=0f'(x) = 0: 8+2x2=08 + \frac{2}{x^2} = 0 However, solving 8+2x2=08 + \frac{2}{x^2} = 0

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Math Problem Analysis

Mathematical Concepts

Calculus
Optimization
Critical Points

Formulas

f(x) = 8x - 2/x
f'(x) for finding critical points
Evaluating f(x) at endpoints

Theorems

Extreme Value Theorem
Derivative for finding critical points

Suitable Grade Level

College Calculus