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

38+105= 143, 39+104=143, and you will find then adding the numbers on opposite sides, moving inwards yields the answer 138 every time. because of this, you’re going to multiply 138 times the number of pairs, which is 34. therefore, 34*138= 4692(i’m sorry if it’s the wrong numbers, but it’s the right concept)

Well if you have 20 bats you probably want 20 balls? you don’t have enough context to answer the question.