# y = sqrt(3)x + 2cos(x), 0<=x<2pi Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.

kalau | Certified Educator

Take the derivative of the function.

y' = sqrt3 -2sin(x)

Set the derivative function equal to zero, since we want to know where x is when the slope is zero.

0 = sqrt3 -2sin(x)

2sin(x)=sqrt3

sin(x)=sqrt(3) /2

We need to find all the angles of x where the y-coordinate of the unit circle equals to \sqrt3 /2 on the domain [0,2pi).

The y-value sqrt3 / 2 will exist in the first and second quadrants of the unit circle.

The angles are:

x=pi/3,(2pi)/3

These two values will give us zero when we substitute them back into the derivative function.

To get the points, plug these two values back to the original function.

y=sqrt3 (pi/3) + 2cos(pi/3)= sqrt3 (pi/3) +1

The first point is:  (pi/3, (sqrt3 pi)/3 +1)

y=sqrt3 ((2pi)/3) + 2(cos((2pi)/3))= (2sqrt3 pi) /3 + 2(-1/2)= (2sqrt3 pi) /3-1

The second point is:  ((2pi)/3,(2sqrt3 pi) /3-1)