How do you find the points on the curve #y=x+2cosx# that have a horizontal tangent line?

Question

Scarlett Allison · Accepted Answer

See below.

We know that horizontal tangents occur where the derivative equals #0#. So we first need to differentiate the function.

#dy/dx (x+2cos(x)= 1-2sin(x)#

We need to find values of #x# that give #1-2sin(x)=0#

#:.#

#sin(x)= 1/2=> x= -(7pi)/6 , - (11pi)/6 , pi/6 , (5pi)/6#

For:

#-2pi<= x <= 2pi#

Graph of #y = x+cos(x)#

Emma Aldridge · Answer

undefined To find the points on the curve y=x+2cosx that have a horizontal tangent line, we need to find the values of x where the derivative of y with respect to x equals zero.

First, we find the derivative of y with respect to x, which is dy/dx = 1 - 2sinx.

Next, we set the derivative equal to zero and solve for x: 1 - 2sinx = 0.

Simplifying the equation, we have 2sinx = 1.

Dividing both sides by 2, we get sinx = 1/2.

The solutions for sinx = 1/2 are x = π/6 + 2πn and x = 5π/6 + 2πn, where n is an integer.

Substituting these values of x back into the original equation y = x + 2cosx, we can find the corresponding y-values.

Therefore, the points on the curve y=x+2cosx that have a horizontal tangent line are (π/6 + 2πn, π/6 + 2cos(π/6 + 2πn)) and (5π/6 + 2πn, 5π/6 + 2cos(5π/6 + 2πn)), where n is an integer.