Problem 66 - Precalculus with Limits

Question

In Exercises \(63-70\) , use the function and its derivative to determine any points on the graph of \(f\) at which the tangent line is horizontal. Use a graphing utility to verify your results. $$ \begin{array}{l}{f(x)=x-2 \sin x, \quad f^{\prime}(x)=1-2 \cos x} \\ {\text { over the interval }(0,2 \pi)}\end{array} $$

Step-by-Step Solution

Answer

The points on the graph of \(f\) at which the tangent line is horizontal are \((\frac{\pi}{3}, \frac{\pi}{3}-\sqrt{3})\) and \((\frac{5\pi}{3}, \frac{5\pi}{3}-\sqrt{3})\)

Step 1: Solve the Derivative Equation for x

The derivative \(f'(x) = 1-2\ \cos\ x\) should be set to zero to find the x values where the tangent line will be horizontal.\nSet up the equation: \(1 - 2\ \cos\ x = 0\).\nSolving the equation will give: \(\cos\ x =0.5\).\nThe solutions to the equation on the interval from 0 to \(2\pi\) are \(x=\frac{\pi}{3}, \frac{5\pi}{3}\).

Step 2: Find the Corresponding y Values

Plug these x values back into the original equation to find the corresponding y values. So y when \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\) is: \(y=f(x)= x-2 \sin\ x\).\nWhen \(x=\frac{\pi}{3}\), \(y= \frac{\pi}{3}-2 \sin(\frac{\pi}{3}) = \frac{\pi}{3}-\sqrt{3}\).\nSimilarly when \(x=\frac{5\pi}{3}\), \(y= \frac{5\pi}{3}-2 \sin(\frac{5\pi}{3}) = \frac{5\pi}{3}-\sqrt{3}\).

Step 3: Combine the x and y Values to Form Coordinates for the Points where the Tangent Line is Horizontal

So the points where the slope of the tangent line to the graph of the function is horizontal, are \((\frac{\pi}{3}, \frac{\pi}{3}-\sqrt{3})\) and \((\frac{5\pi}{3}, \frac{5\pi}{3}-\sqrt{3})\) on the interval from \(0\) to \(2\pi\).

Key Concepts

Derivative of a Function

Understanding the derivative of a function is crucial in calculus. It measures how a function's output value changes as its input value changes. Think of it as the function's instantaneous rate of change or