Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Cumulative Review Exercises - Page 1181: 10

Answer

See below:
1570764208

Work Step by Step

Consider the provided function, $ y=\frac{1}{2}\sec 2\pi x $ By use the reciprocal function, $ y=\frac{1}{2}\cos 2\pi x $ Now substitute different values of $ x $ in the provided function to find the values of $ y $. As the interval is given, put values of $ x $ in the range of the interval, $\text{0}\le x\le 2$ To find the key points of the function, put $ x=0$ in the provided function, $\begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot 0 \right) \\ & =\frac{1}{2}\cdot 1 \\ & =\frac{1}{2} \end{align}$ And, put $ x=\frac{1}{4}$ $\begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{1}{4} \right) \\ & =\frac{1}{2}\cdot 0 \\ & =0 \end{align}$ And, put $ x=\frac{1}{2}$ $\begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{1}{2} \right) \\ & =\frac{1}{2}\cdot \left( -1 \right) \\ & =-\frac{1}{2} \end{align}$ And, put $ x=\frac{3}{4}$ $\begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{3}{4} \right) \\ & =\frac{1}{2}\cdot 0 \\ & =0 \end{align}$ And, put $ x=1$ $\begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot 1 \right) \\ & =\frac{1}{2}\cdot 1 \\ & =\frac{1}{2} \end{align}$ And, put $ x=\frac{5}{4}$ $\begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{5}{4} \right) \\ & =\frac{1}{2}\cdot \left( 0 \right) \\ & =0 \end{align}$ And, put $ x=\frac{3}{2}$ $\begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{3}{2} \right) \\ & =\frac{1}{2}\cdot \left( -1 \right) \\ & =-\frac{1}{2} \end{align}$ And, put $ x=\frac{7}{4}$ $\begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot \frac{7}{4} \right) \\ & =\frac{1}{2}\cdot \left( 0 \right) \\ & =0 \end{align}$ And, put $ x=2$ $\begin{align} & y=\frac{1}{2}\cos 2\pi x \\ & =\frac{1}{2}\cos \left( 2\pi \cdot 2 \right) \\ & =\frac{1}{2}\cdot \left( 1 \right) \\ & =\frac{1}{2} \end{align}$
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