Thread: For what values of x does the graph of f have a horizontal tangent?

I cannot understand...



F(x)=x-2sinx

I did..f'(x)=1-2cosx=0
cosx=1/2

And I stop there.
?

Originally Posted by TeresaFurst28

I cannot understand...



F(x)=x-2sinx

I did..f'(x)=1-2cosx=0
cosx=1/2

And I stop there.
?

f(x) = x-2sin(x) => f'(x) = 1-2cos(x)

Horizontal tangent when f'(x) = 0, hence we have

1-2cos(x) = 0 <=> cos(x) = 1/2

Now we want to find for which values of x does cos(x) = 1/2. Sketch a triangle were the hypotenuse have the length 2, the catheters (correct English?) have the length 1 and sqrt(3). The angles are then going to be pi/2 = 90°, pi/3 = 60° or pi/6 = 30°. By using the relationship you find that the value for x is pi/3. Therefore x = pi/3 (or x = 60°) is one solution, but this is not all of the solutions.

Since cos(x) is periodic by 2*pi (360°) we have a solution for every x = pi/3+2pi*n were n is any integer. You should also know that cos(x) = cos(-x) therefore x = -pi/3+2pi*n is also a valid solution.

Originally Posted by TeresaFurst28

I cannot understand...



F(x)=x-2sinx

I did..f'(x)=1-2cosx=0
cosx=1/2

And I stop there.
?

please post calculus questions in the calculus forum.

also, you need to learn the unit circle ...