Question

Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line. $$ y=\sqrt{3} x+2 \cos x, \quad 0 \leq x<2 \pi $$

Short answer

The function \(y=\sqrt{3} x+2 \cos x\) has horizontal tangents at the points \((\pi/3, \pi/\sqrt{3} + 1)\) and \((2\pi/3, 2\pi/\sqrt{3} - 1)\).

Step-by-step solution

Compute the derivative of the function

We want to find the derivative of the function \(y=\sqrt{3} x+2 \cos x\). Here we use the fact that the derivative of a sum of functions is the sum of their derivatives, and the derivatives of \(x\) and \(\cos x\) are \(1\) and \(-\sin x\), respectively:\[y'(x) = d(\sqrt{3} x)/dx + d(2\cos x)/dx = \sqrt{3} -2\sin x\].

Solve the equation y'(x) = 0

A tangent line to the graph of \(y\) is horizontal precisely when the derivative of \(y\) at that point is zero, i.e., \(y'(x)=0\). We get the equation:\[\sqrt{3} - 2\sin x = 0.\]This can be solved for \(\sin x\) as follows: \[\sin x = \sqrt{3}/2.\] The solutions of this equation within the interval \([0,2\pi)\) are \(x=\pi/3\) and \(x= 2\pi/3\).

Determine the corresponding y-values

To determine the points on the graph where the tangent line is horizontal, we need to compute the y-coordinates that correspond to the found x-coordinates. These are given by\[y(\pi/3) = \sqrt{3} \cdot (\pi/3) + 2\cos(\pi/3) = \pi/\sqrt{3} + 1,\]and\[y(2\pi/3) = \sqrt{3} \cdot (2\pi/3) + 2\cos(2\pi/3) = 2\pi/\sqrt{3} - 1.\]