Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.

Question

y = (√3) x + 2 cos(x), 0 ≤ x < 2𝜋

Find a derivative y'

Set y' = 0 and solve for x

x₁ = (smaller x-value)

x₂ = (larger x-value)

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 horizontal tangent lines. Submit your answers in terms of pi for 𝜋.

(x₁, y₁) =

(x₂, y₂) =

Answer 1

y = (√3) x + 2 cos(x), 0 ≤ x < 2𝜋

x values where y = (√3) x + 2 cos(x) has a slope of zero

y' = √3 – 2 sin(x) = 0 means that sin(x) = √3/2

x₁ = 𝜋/3 (smaller x-value)

x₂ = 2𝜋/3 (larger x-value)

y₁ =  (√3)(𝜋/3) + 2 cos(𝜋/3) =  √3𝜋/3 + 2(1/2) = √3𝜋/3 + 1 = (√3𝜋 + 3)/3

y₂ =  (√3)(2𝜋/3) + 2 cos(2𝜋/3) = 2√3𝜋/3 + 2(–1/2) = 2√3𝜋/3 – 1 = (2√3𝜋 – 3)/3

y = (√3) x + 2 cos(x) has horizontal tangent lines at

(x₁, y₁) = (𝜋/3, (√3𝜋 + 3)/3)

(x₂, y₂) = (2𝜋/3, (2√3𝜋 – 3)/3)