Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.2 - Combining Functions; Shifting and Scaling Graphs - Exercises 1.2 - Page 20: 66

Answer

$y=1-\dfrac{x^3}{8}$

Work Step by Step

Given the graph of $y=f(x)$ and a real number $c>1$, we obtain the graph of $y=f(\dfrac{x}{c})$ by horizontally stretching the graph of $y=f(x)$ by a factor of $c$. Hence to horizontally stretch the graph of $y=1-x^3$ by a factor of 2, we replace each occurrence of $x$ in this equation by $\dfrac{x}{2}$, and, thus, obtaining the equation $$y=1-\dfrac{x^3}{(2)^3},$$ or equivalently, $$y=1-\dfrac{x^3}8.$$
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