Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.1 Review of Functions - 1.1 Exercises: 82

Answer

For $f(x)=x^{n}$, where n is a positive odd integer, the range will be all real numbers, because one end of the equation will go to -$\infty$, while the other will go to +$\infty$. Therefore, since the function is continuous it will take on every value on integer (-$\infty$,+$\infty$). For $f(x)=x^{n}$, where n is a positive even integer, the range will be all nonnegative real numbers because the equation will approach $\infty$ as x goes to -$\infty$ as well as +$\infty$ and will be zero when x=0, since $0^{anything}=zero$

Work Step by Step

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