Math Problem Statement

The function f(x)=8x+2x^-1 has one local minimum and one local maximum. This function has a local maximum at x= with value

and a local minimum at x= with value

Solution

To find the local maximum and minimum of the function f(x)=8x+2x1f(x) = 8x + 2x^{-1}, let's go through the steps:

  1. Find the derivative f(x)f'(x) to locate critical points: f(x)=8x+2x1f(x) = 8x + 2x^{-1} Differentiating with respect to xx: f(x)=82x2=82x2f'(x) = 8 - 2x^{-2} = 8 - \frac{2}{x^2}

  2. Set f(x)=0f'(x) = 0 to find critical points: 82x2=08 - \frac{2}{x^2} = 0 Solving for xx: 2x2=8\frac{2}{x^2} = 8 x2=28=14x^2 = \frac{2}{8} = \frac{1}{4} x=±12x = \pm \frac{1}{2}

    So, the critical points are x=12x = \frac{1}{2} and x=12x = -\frac{1}{2}.

  3. Determine if these points are maxima or minima using the second derivative f(x)f''(x): f(x)=4x3f''(x) = \frac{4}{x^3}

    • At x=12x = \frac{1}{2}: f(12)=4(12)3=418=32>0f''\left(\frac{1}{2}\right) = \frac{4}{\left(\frac{1}{2}\right)^3} = \frac{4}{\frac{1}{8}} = 32 > 0 Since f(x)>0f''(x) > 0, x=12x = \frac{1}{2} is a local minimum.

    • At x=12x = -\frac{1}{2}: f(12)=4(12)3=418=32<0f''\left(-\frac{1}{2}\right) = \frac{4}{\left(-\frac{1}{2}\right)^3} = \frac{4}{-\frac{1}{8}} = -32 < 0 Since f(x)<0f''(x) < 0, x=12x = -\frac{1}{2} is a local maximum.

  4. Calculate the function values at these points:

    • For the local minimum at x=12x = \frac{1}{2}: f(12)=812+2112=4+4=8f\left(\frac{1}{2}\right) = 8 \cdot \frac{1}{2} + 2 \cdot \frac{1}{\frac{1}{2}} = 4 + 4 = 8

    • For the local maximum at x=12x = -\frac{1}{2}: f(12)=8(12)+2112=44=8f\left(-\frac{1}{2}\right) = 8 \cdot \left(-\frac{1}{2}\right) + 2 \cdot \frac{1}{-\frac{1}{2}} = -4 - 4 = -8

Summary of Results:

  • The local maximum occurs at x=12x = -\frac{1}{2} with value 8-8.
  • The local minimum occurs at x=12x = \frac{1}{2} with value 88.

Would you like a more detailed explanation on any of these steps?


  1. What is the significance of the second derivative test for identifying maxima and minima?
  2. How would the result change if we modified the function to f(x)=8x+3xf(x) = 8x + \frac{3}{x}?
  3. What happens to f(x)f(x) as xx approaches 0?
  4. How do we interpret the local maxima and minima on a graph?
  5. Can this method be applied to find inflection points?

Tip: When differentiating functions with terms like x1x^{-1}, rewriting them as fractions or powers simplifies the derivative process.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Critical Points
Second Derivative Test

Formulas

f(x) = 8x + 2x^-1
f'(x) = 8 - 2x^-2
f''(x) = 4/x^3

Theorems

Second Derivative Test

Suitable Grade Level

Grades 11-12