Chapter 11.3, Problem 61E

To determine

Points onthe graph at which the tangent line is horizontal for the given function and its derivative. Usea graphing utility to verify your results.

Answer to Problem 61E

f(x) has horizontal tangents at points $\left(\frac{\pi }{6},\sqrt{3}+\frac{\pi }{6}\right)\text{ and }\left(\frac{5\pi }{6},\frac{5\pi }{6}-\sqrt{3}\right)$

Explanation of Solution

Given info:

  $\begin{array}{l}f(x)=2\cos x+x\\ f\text{'}(x)=-2\sin x+1\\ \text{over the interval }(0,2\pi )\end{array}$

Formula used:

The slope m of the graph of at the point (x,f(x)) is equal to the slope of itstangent line at (x,f(x)) and is given by

  $m=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

Calculation:

When the tangent line is horizontal f’(x) = 0

We have

  $\begin{array}{l}f\text{'}(x)=-2\sin x+1\\ -2\sin x+1=0\\ \sin x=\frac{1}{2}\\ x=\frac{\pi }{6}\text{ and }x=\frac{5\pi }{6}\\ f\left(\frac{\pi }{6}\right)=2\cos \left(\frac{\pi }{6}\right)+\left(\frac{\pi }{6}\right)=\sqrt{3}+\frac{\pi }{6}\\ f\left(\frac{5\pi }{6}\right)=2\cos \left(\frac{5\pi }{6}\right)+\left(\frac{5\pi }{6}\right)=\frac{5\pi }{6}-\sqrt{3}\\ \text{The points are }\left(\frac{\pi }{6},\sqrt{3}+\frac{\pi }{6}\right)\text{ and }\left(\frac{5\pi }{6},\frac{5\pi }{6}-\sqrt{3}\right)\end{array}$

f(x) has horizontal tangents $\frac{\pi }{6}\text{ and }\frac{5\pi }{6}$

The plot is given by

  

Conclusion:

Thus,f(x) has horizontal tangents at points $\left(\frac{\pi }{6},\sqrt{3}+\frac{\pi }{6}\right)\text{ and }\left(\frac{5\pi }{6},\frac{5\pi }{6}-\sqrt{3}\right)$