Find the critical numbers of the function f(x)=2sinx+x that lie in the interval (0,2pi)?

I have the answers, but not sure how to display the work.

2pi

To find the critical numbers of a function, we need to determine where its derivative is either zero or undefined. In this case, the function is f(x) = 2sin(x) + x.

Step 1: Find the derivative of f(x):
The derivative of f(x) with respect to x is given by f'(x) = 2cos(x) + 1.

Step 2: Find where f'(x) = 0 or is undefined:
To find where f'(x) = 0, we set the derivative equal to zero and solve for x:
2cos(x) + 1 = 0
2cos(x) = -1
cos(x) = -1/2

The solutions for cos(x) = -1/2 lie in the second and third quadrants, and can be found using the unit circle or trigonometric identities. The two solutions in the interval (0, 2π) are x = 2π/3 and x = 4π/3.

Step 3: Determine if there are any points where f'(x) is undefined:
In this case, f'(x) is defined for all values of x since cos(x) is always defined.

Step 4: Identify critical numbers in the interval (0, 2π):
The critical numbers in this interval are the values of x where f'(x) = 0, which are x = 2π/3 and x = 4π/3.

Therefore, the critical numbers of the function f(x) = 2sin(x) + x that lie in the interval (0, 2π) are x = 2π/3 and x = 4π/3.

well, surely you must in fact have done some work. rather than mumble some excuses, why not show what you did? Then we can analyze it for accuracy and presentability.

The critical numbers are where
f'(x) = 0
or
f'(x) is undefined.

So, what does that give you?