Thomas' Calculus 13th Edition

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

Chapter 1: Functions - Questions to Guide Your Review - Page 36: 5

Answer

See below.

Work Step by Step

Taking any two values $x_{1}$ and $x_{2}$ from the domain of $f$, where $x_{1}\lt x_{2}$, we have: if it is always the case that $f(x_{1})\lt f(x_{2})$, (for any choice of $x_{1}$ and $x_{2}$, where $x_{1}\lt x_{2}$), then the function is an increasing function. (The function value of a greater input is always greater. The function values increase as the inputs increase.) If, on the other hand, it is always the case that $f(x_{1})\gt f(x_{2})$, (for any choice of $x_{1}$ and $x_{2}$, where $x_{1}\lt x_{2}$), then the function is a decreasing function. (The function value of a greater input is always smaller. The function values decrease as the inputs increase.) Examples: $f(x)=-x \quad$ is a decreasing function. (Take any two inputs. The function value of the greater input is less than the function value of the smaller input. For example, $f(2)=-2$ is greater than $f(3)=-3)$. $ f(x)=x+1\quad$ is an increasing function. (The function value of a greater input is always greater For example, $f(2)=3$ is less than $f(3)=4)$ $f(x)=2^{-x} \quad$ is a decreasing function (For example, $f(2)=2^{-2}=\displaystyle \frac{1}{4}$ is greater than $f(3)=2^{-3}=\displaystyle \frac{1}{8})$ $f(x)=2^{x} \quad$ is an increasing function (For example, $f(2)=2^{2}=4$ is less than $f(3)=2^{3}=8)$
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