find the coordinates of all points on the graph of f(x) = x + 2cos(x) in the interval [0, pie] at which the tangent line is horizontal.

f'(x) = 1 - 2sinx

= 0 for a horizontal slope
1-2sinx = 0
sinx = 1/2
x = π/6 or 5π/5

if x = π/6
y = π/6 + √3 = (π + 6√3)/6 ---> point (π/6 , (π+6√3)/6

if x = 5π/6
y = 5π/6 - √3 = (5π - 6√3)/6 ---> point (5π/6 , (5π-6√3)/6 )

thank you reiny

To find the coordinates of all points on the graph of f(x) = x + 2cos(x) in the interval [0, pi] at which the tangent line is horizontal, we need to determine the values of x that make the derivative of f(x) equal to zero. In other words, we need to find the critical points of f(x).

Step 1: Find the derivative of f(x)
The derivative of f(x) with respect to x can be found using the sum and chain rules of differentiation. Taking the derivative of x + 2cos(x) gives us:

f'(x) = 1 - 2sin(x)

Step 2: Set the derivative equal to zero and solve for x
To find the critical points, we set f'(x) = 0 and solve for x:

1 - 2sin(x) = 0
2sin(x) = 1
sin(x) = 1/2

Step 3: Find the values of x within the interval [0, pi] that satisfy the equation
Looking at the unit circle, we can see that sin(x) = 1/2 for x = pi/6 and x = 5pi/6.

Since we are interested in the interval [0, pi], the values of x that satisfy sin(x) = 1/2 are x = pi/6.

Step 4: Find the y-coordinates of the critical points
To find the y-coordinates of the critical points, we substitute the values of x into the original function f(x):

f(pi/6) = pi/6 + 2cos(pi/6) = pi/6 + 2(√3/2) = pi/6 + √3

Therefore, the coordinates of the points on the graph of f(x) where the tangent line is horizontal are (pi/6, pi/6 + √3).

So, there is one point with coordinates (pi/6, pi/6 + √3) on the graph of f(x) = x + 2cos(x) in the interval [0, pi] at which the tangent line is horizontal.