Question

Find the critical points of the function f(x)=4sin(x)cos(x) contained in the interval (0,2π).

Answer 1

The critical points of the function f(x) = 4sin(x)cos(x) in the interval (0, 2π) are found by using the double angle formula to simplify the function, differentiating, and setting the derivative to zero. The critical points are at x = π/4 and 3π/4.

Step-by-step explanation:

To find the critical points of the function f(x) = 4sin(x)cos(x) within the interval (0, 2π), we first need to take the derivative of the function and set it to zero. Using the double angle formula, sin(2x) = 2sin(x)cos(x), we can simplify the function to f(x) = 2sin(2x). Then, we differentiate f(x) to get f'(x) = 4cos(2x). To find the critical points, we solve for x when f'(x) = 0.

Critical points are the values of x where f'(x) = 4cos(2x) = 0. Thus, cos(2x) = 0 when 2x = π/2, 3π/2, and since we are considering the interval between 0 and 2π, we divide these by 2, giving x = π/4, 3π/4 as the critical points within the given interval.

Answer 2

The critical points of the function f(x)=4sin(x)cos(x) in the interval (0,2π) are found by rewriting the function using the double angle identity and then taking its derivative. The critical points are located at x=π/2 and x=3π/2.

Step-by-step explanation:

To find the critical points of the function f(x) = 4sin(x)cos(x) within the interval (0,2π), we first need to calculate the derivative of the function, which will give us the rate at which the function is changing at any point within this interval. Then, we set the derivative equal to zero and solve for x to find the critical points.

The first step is to use the double angle identity sin(2x) = 2sin(x)cos(x), which allows us to rewrite the function as f(x) = 2sin(2x). The derivative of this function with respect to x is f'(x) = 4cos(2x). Setting the derivative equal to zero gives us 4cos(2x) = 0.

We can solve for x by finding the angles for which cos(2x) is zero. These occur at π/2 + kπ where k is an integer. Within the interval (0, 2π), we get the critical points at x = π/2 and x = 3π/2 because these values of x will make the cosine term equal to zero.