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Find the critical numbers, if any, of function (f) on the interval ([-π/2,π/2]) given (f(x) = 2x - 2 cos(x)).

a) (0, π/2)
b) (-π/2, π/2)
c) (0)
d) DNE

User Litek
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Final answer:

After taking the derivative of the function f(x) = 2x - 2cos(x), we set it equal to zero and found the critical number within the interval [-π/2, π/2] to be -π/2. Including the endpoints, the critical numbers are -π/2 and π/2, so the correct answer is b) (-π/2, π/2).

Step-by-step explanation:

To find the critical numbers of the function f(x) = 2x - 2cos(x) on the interval [-π/2, π/2], we need to find the values of x where the derivative of the function is zero or undefined. First, we take the derivative of the function:

f'(x) = d/dx (2x - 2cos(x)) = 2 + 2sin(x)

Setting f'(x) to zero gives us the equation:

2 + 2sin(x) = 0

Solving for x:

sin(x) = -1

We find that x = -π/2 within the given interval. Since the derivative exists and is continuous on the entire interval, there are no points where it is undefined. Therefore, the critical number is just -π/2.

However, we must consider the endpoints of the interval as well. f'(-π/2) = 0 and f'(π/2) = 4 because sin(π/2) = 1. So, the critical numbers, including the endpoints, are -π/2 and π/2.

Therefore, the correct answer is option b) (-π/2, π/2).

User Xiomara
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