Math Problem Statement

Solve

Discuss the function defined as f(x) = sin(x) + 1/2√2 cos(2x) for extreme values in the interval (0, 2π).

Solution

The function f(x)=sinx+122cos2xf(x) = \sin x + \frac{1}{2\sqrt{2}} \cos 2x has the following first derivative:

f(x)=cosx22sin2xf'(x) = \cos x - \frac{\sqrt{2}}{2} \sin 2x

To find the extreme values, we solve f(x)=0f'(x) = 0 for critical points in the interval (0,2π)(0, 2\pi). The critical points are approximately:

x=0.7854,1.5708,2.3562x = 0.7854, 1.5708, 2.3562

These correspond to π4,π2,3π4\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}.

Next, you can determine if these are maxima, minima, or points of inflection by examining the second derivative or evaluating the function at these points.

Would you like details on these critical points or further analysis?

Here are some related questions to explore:

  1. What is the second derivative of the function to classify the critical points?
  2. How do we determine whether these critical points correspond to local maxima or minima?
  3. What is the behavior of the function at the boundaries of the interval (0,2π)(0, 2\pi)?
  4. How does the periodic nature of sine and cosine affect the extrema?
  5. Can the method of Lagrange multipliers be applied to constrained extrema?

Tip: Always verify if critical points lie within the specified domain when solving optimization problems.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Differentiation
Extrema

Formulas

f'(x) = cos(x) - (√2 / 2) sin(2x)
sin(2x) = 2 sin(x) cos(x)

Theorems

Critical points of a function
First derivative test

Suitable Grade Level

Grades 11-12