How do you find the points where the graph of the function #y = (sqrt 3)x + 2 cosx# has horizontal tangents and what is the equation?

Question

Charles Anctil · Accepted Answer

#x = pi/6 , {5pi}/6 , {13pi}/6 , {17pi}/6 ...#

Tangent line Equation

#y = Y#, where #Y# is the #y#-coordinate of the point.

Points with horizontal tangents are point that have #frac{"d"y}{"d"x} = 0#. #frac{"d"y}{"d"x} = sqrt3 - 2 sinx# So solve #sqrt3 - 2 sinx = 0# #sinx = sqrt3/2# #x = pi/6 , {5pi}/6 , {13pi}/6 , {17pi}/6 ...# or equivalently #x = (2k+1/2)pi +- pi/3#, where #k in ZZ#. For each value of #x = X# that has a gradient of zero, the horizontal line #y = sqrt3 X + 2cosX# is the tangent line at the respective points.

Christopher Allers · Answer

undefined To find the points where the graph of the function y = √3x + 2cos(x) has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, let's find the derivative of the function. The derivative of y = √3x + 2cos(x) can be found using the sum rule and the chain rule:

dy/dx = d(√3x)/dx + d(2cos(x))/dx
       = √3 + (-2sin(x))

Next, we set the derivative equal to zero and solve for x:

√3 - 2sin(x) = 0

To find the values of x, we isolate sin(x):

2sin(x) = √3
sin(x) = √3/2

The values of x where sin(x) = √3/2 are π/3 and 2π/3.

Therefore, the points where the graph of the function has horizontal tangents are (π/3, f(π/3)) and (2π/3, f(2π/3)), where f(x) represents the function y = √3x + 2cos(x).

The equation of the horizontal tangent lines at these points can be found by substituting the x-values into the original function and solving for y.