Given the function f (x) = 3cos x for the interval 0 < x < 2pi, determine the concavity of the function, between the critical values

f'(x) = -3sinx

f'=0 at x = 0, π and 2π

f"(x) = -3cosx
f"(0) < 0 so f is concave down in (0,π/2)
f"(π) > 0 so f is concave up in (π/2,3π/2)
f"(2π) < 0 so f is concave down in (3π/2,2π)

To determine the concavity of the function f(x) = 3cos(x) for the interval 0 < x < 2π, we need to find the second derivative of the function.

First, let's find the first derivative of f(x) = 3cos(x):

f'(x) = -3sin(x)

Next, let's find the second derivative of f(x) = 3cos(x) by differentiating the first derivative:

f''(x) = -3cos(x)

Now, we have the second derivative f''(x) = -3cos(x) for the function f(x) = 3cos(x).

To determine the concavity of the function, we need to analyze the sign of the second derivative.

Since the second derivative f''(x) = -3cos(x) is negative for all values of x in the given interval (0 < x < 2π), the function is concave down between the critical values.

To determine the concavity of a function, we need to find the second derivative of the function and then analyze its sign.

The first step is to find the first derivative of the function f(x) = 3cos(x). The derivative of cos(x) is -sin(x), so the first derivative of f(x) is f'(x) = 3(-sin(x)) = -3sin(x).

Next, we need to find the second derivative by differentiating the first derivative. The derivative of -sin(x) is -cos(x), so the second derivative of f(x) is f''(x) = -cos(x).

To determine the concavity of the function, we need to analyze the sign of the second derivative. If the second derivative is positive, the function is concave up (increasing concavity), and if the second derivative is negative, the function is concave down (decreasing concavity).

For the function f''(x) = -cos(x), the sign of cos(x) changes between the intervals of -pi/2 and pi/2. In the interval 0 < x < 2pi, the function crosses the critical points of -pi/2 and pi/2.

For x < -pi/2 or x > pi/2, cos(x) is negative, so f''(x) is negative, indicating that the function is concave down.

For -pi/2 < x < pi/2, cos(x) is positive, so f''(x) is positive, indicating that the function is concave up.

Therefore, between the critical values of -pi/2 and pi/2, the function f(x) = 3cos(x) is concave up.