Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.3 - Trigonometric Graphs - 5.3 Exercises - Page 429: 23

Answer

amplitude=$2$ period =$\displaystyle \frac{2}{3}$ graph (black): .

Work Step by Step

See p. 424 For $y=a\sin k(x-b),\ ,\quad y=a\cos k(x-b)$, $(k>0)$ Amplitude: $|a|$, Period: $\displaystyle \frac{2\pi}{k}$, Horizontal shift: $b$ ------------------ $y=-2\cos 3\pi x$ $a=-2 , k=4\pi, b=0$ So, amplitude=$|-2|=2$ period =$\displaystyle \frac{2\pi}{3\pi}=\frac{2}{3}$ To graph, begin with $f(x)=\cos x,$ (red, dashed) horizontally compress by factor $ 3\pi$ and reflect across the x-axis (black, solid line). $g(0)=-2$ $g(\displaystyle \frac{1}{6})=-2\cos(\frac{\pi}{2})=0$ $g(\displaystyle \frac{1}{3})=-2\cos(\pi)=2$ $g(\displaystyle \frac{1}{2})=-2\cos(\frac{3\pi}{2})=0$ $g(\displaystyle \frac{2}{3})=-2\cos(2\pi)=-2.$
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