Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 31: 40

Answer

The required function is $-2\cos\left(\dfrac{3x}{2}\right)$. The period is $\dfrac{4\pi}{3}$ and amplitude is $2$ units.

Work Step by Step

Note that the maximum value of function is $2$, so, the amplitude is $2$ units. The function can be obtained using transformation of $\cos{x}$ as follows: $\underline{\textbf{Step 1}}$: To make maximum value of $2$ and minimum value of $-2$, multiply the standard function by $2$. $\underline{\textbf{Step 2}}$: From the given graph if we consider one complete wave, it starts from $x=\dfrac{-\pi}{3}$ and ends at $x=\pi$, which has length of $\pi-\dfrac{-\pi}{3}=\dfrac{4\pi}{3}$. Thus, the function repeats itself after every $\dfrac{4\pi}{3}$, so, the period is $\dfrac{4\pi}{3}$. The original wave have period of $2\pi$, so, the wave is compressed horizontally by a factor of $\dfrac{2\pi}{\frac{4\pi}{3}}=\dfrac{3}{2}$. $\underline{\textbf{Step 3}}$: Note that the given graph shows reflection of the standard function about x-axis, so, multiply the function by $-1$. Therefore, the function we obtained is $-2\cos\left(\dfrac{3x}{2}\right)$. [Remember that if a function is $f(x)$, it is compressed horizontally by a factor of $k$, then, the new function is $f(k\cdot x)$]
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