Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.3 - New Functions from Old Functions - 1.3 Exercises - Page 43: 8

Answer

(a) The graph of $y=2\sin x$(red) is obtained by stretching the graph of $y=\sin x$(blue) vertically by the factor of $2$. (Upper portion of the figure) (b) The graph of $y=1+\sqrt{x}$(red) is obtained from the graph of $y=\sqrt{x}$(blue) by translating it vertically upwards for $1$ unit. (Lower portion of the figure)
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Work Step by Step

(a) Observe carefully the Figure 6. The graph of $2\sin x$ will have the same zeros as $\sin x$ i.e. it will intercept $x$ axis at $\ldots-3\pi,-2\pi,-\pi,0,\pi,2\pi,3\pi\ldots$ but every value of the function would double, i.e. every point would be twice as far from the $x$ axis compared to the graph of $\sin x$. Thus the maxima and the minima between two zeros will have the value of $2$ compared to $1$ for $\sin x$. So to get the graph of $2\sin x$ from the graph of $\sin x$ we have to stretch its' graph vertically by the factor of $2$. The graphs of $y=\sin x$ (blue) and $y=2\sin x$ (red) are shown in the upper portion of the figure below. (b) Here every value on the $y$ axis is greater for $1$ unit. Thus, we obtain the graph of $y=1+\sqrt{x}$ by moving upwards for one unit the graph of $y=\sqrt{x}$ i.e. we perform the vertical upwards translation for $1$ unit. The graphs of $y=\sqrt{y}$ (blue) and $y=1+\sqrt{y}$(red) are shown on the lower portion of the graph below.
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