Problem 80 - Precalculus with Limits

Question

In Exercises \(79-84,\) (a) use a graphing utility to graph the function and approximate the maximum and minimum points on the graph in the interval \([0,2 \pi),\) and (b) solve the trigonometric equation and demonstrate that its solutions are the \(x\) -coordinates of the maximum and minimum points of \(f\) . (Calculus is required to find the trigonometric equation.) $$\begin{array}{ll} \qquad {\text { Function }} & {\text { Trigonometric Equation }} \\ {f(x)=\cos ^{2} x-\sin x} & {-2 \sin x \cos x-\cos x=0}\end{array}$$

Step-by-Step Solution

Answer

The approximate maximum and minimum points from graphing the function are the solutions of the trigonometric equation.

Step 1: Graphing the function

Use a graphing utility to plot the function \( f(x) = \cos^2 x \sin x \) within the interval [0,2). Roughly approximate the maximum and minimum points on the graph based solely on the visual representation.

Step 2: Solving the Trigonometric Equation

By looking closely at the equation \( -2 \sin x \cos x - \cos x = 0 \), it could be simplified by factoring out \(- \cos x\). As a result, it simplifies to \(- \cos x \) * \(\ 2 \sin x + 1 \) = 0. Solve this equation by setting each factor equal to zero, and find the solutions \( x \).

Step 3: Comparing the solutions

Compare the solutions from the trigonometric equation to the x-coordinates of the maximum and minimum points of the graph function. If they match, this confirms the solutions from the trigonometric equation are indeed the x-coordinates of the maximum and minimum points of \( f \).