determine the points if any at which the graph of the function has a horizontal tangent line

Question

y=(sqrt3)x+2cos(x), 0 less than or equal to x < 2pi

1) find derivative y'

2) set y'=0 and solve for x

x1=?

x2=?

3) find the y values by substituting the values from step 2 into the original function. y1 corresponds to x1 and y2 corresponds to x2. List the points where the function has horizonatal tangent lines . submit your answers in terms of pi for pi.

x1,y1,=?

x2,y2=?

Answer 1

1)

y=√3x+2cos(x)

Find the derivative both sides with respect to x.

y' = d/dx(√3x+2cos(x))

   =√3 - 2sinx

2)

Set y' = 0

√3 - 2sinx=0

sinx = √3/2

sinx = sin π/3

General solution: x = nπ+(-1)^n(π/3)

If n = 0, x = (0)π + (- 1)⁰( π/3) =  π/3.

If n = 1, x = (1)π + (- 1)¹(π/3) = π - π/3  = 2π/3.

If n = 2, x = (2)π + (- 1)²(π/3) = 2π+ π/3  = 7π/3.

Therefore, the solutions of the given equation are x1 = π/3 and x2 = 2π/3 in the interval [0, 2π).

Answer 2

3)

Find the y values at (π/3,2π/3).

y=√3x+2cos(x)

At x =π/3

y1=√3(π/3)+2cos(π/3)

    =√3(π/3)+2(1/2)

   =√3(π/3)+1

 

At x =2π/3

y2=√3(2π/3)+2cos(2π/3)

    =√3(π/3)+2(-1/2)

   =√3(π/3)-1

Therefore, the points are (π/3,√3(π/3)+1) and (2π/3,√3(π/3)-1)