Find the local maxima and local minima of the functions:

• (i) f(x) = sin(2x), when 0 < x < π
• (ii) f(x) = (sin(2x) − x), when −π/2 ≤ x ≤ π/2

To find the local maxima and minima of the given functions, we need to analyze their derivatives and critical points.

Function 1: f(x) = sin(2x)

First, we differentiate the function:

f'(x) = 2cos(2x)

Next, we set the derivative equal to zero to find critical points:

• 2cos(2x) = 0
• cos(2x) = 0

This occurs when:

• 2x = (2n + 1)π/2, where n is an integer
• x = (2n + 1)π/4

Within the interval [0, π], the critical points are:

• x = π/4
• x = 3π/4

To determine whether these points are maxima or minima, we can use the second derivative test:

f''(x) = -4sin(2x)

Evaluating at the critical points:

• At x = π/4: f''(π/4) = -4sin(π/2) = -4 (local maximum)
• At x = 3π/4: f''(3π/4) = -4sin(3π/2) = 4 (local minimum)

Function 2: f(x) = sin(2x) - x

Now, we differentiate this function:

f'(x) = 2cos(2x) - 1

Setting the derivative to zero gives:

• 2cos(2x) - 1 = 0
• cos(2x) = 1/2

This occurs when:

• 2x = π/3 + 2nπ or 2x = 5π/3 + 2nπ
• x = π/6 + nπ or x = 5π/6 + nπ

Considering the interval [-π/2, π/2], the relevant critical points are:

• x = π/6
• x = -π/6

Using the second derivative test:

f''(x) = -4sin(2x)

Evaluating at the critical points:

• At x = π/6: f''(π/6) = -4sin(π/3) = -4(√3/2) < 0 (local maximum)
• At x = -π/6: f''(-π/6) = -4sin(-π/3) = 4(√3/2) > 0 (local minimum)

Summary of Results

For the first function:

• Local maximum at x = π/4
• Local minimum at x = 3π/4

For the second function:

• Local maximum at x = π/6
• Local minimum at x = -π/6