What is the x-coordinate for the minimum point in the function f(x) = 4 cos(2x − π) from x = 0 to x = π?

When you are looking at a graph, a minimum point would be where the curve is decreasing, then begins to increase. Right at the point where it switches, the slope is a horizontal line, or 0. We can take the derivative is f(x), then look for all the x values where the slope (which is equal to the first derivative) is equal to zero.

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

The 2 comes from the derivative of the inside, 2x-pi.

So now set the derivative equal to 0.

-8sin(2x-pi) = 0

We can drop the -8 by dividing both sides by -8.

sin(2x-pi) = 0

This can be rewritten as arcsin(0) = 2x-pi

So when theta equals 0, what is the value of sin(theta)? At an angle of 0, there is just a horizontal line pointing to the right on the unit circle with length of 1. Sine is y/h, but there is no y value so it is just 0. If arcsin(0) = 0, we can now set 2x-pi = 0

2x = pi

x = pi/2

This is a critical number. To find the minimum value between 0 and pi, we need to find the y values for the endpoints and the critical number.

f(0) = -4

f(pi/2) = 4

f(pi) = -4

So the minimum points are at x=0 and x=pi