Calculus (3rd Edition)

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

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 18: 34

Answer

The required proof is given below.

Work Step by Step

(a) For this part we are given that the function $f(x)$ is linear with slope, $m$. Thus, we can write $f(x)=m\cdot x +c$, where $c$ is the y-intercept of the function. Now, for any $h$, $f(x+h)=m\cdot (x+h)+c=m\cdot x +m\cdot h +c$ [using defined function] $\Rightarrow f(x+h)=(m\cdot x +c)+m\cdot h$ [ by shifting terms] $\Rightarrow f(x+h)=f(x)+m\cdot h$ [As, $f(x)=m\cdot x+c$] $\Rightarrow f(x+h)-f(x)=m\cdot h$. This is valid for all $x$ and $h$. Hence, proved. (b) Now, for all $x$ and $h$, we have $f(x+h)-f(x)=m\cdot h$. Thus, $\dfrac{f(x+h)-f(x)}{h}=m$ $\Rightarrow \dfrac{f(x+h)-f(x)}{(x+h)-x}=m$ $\Rightarrow \dfrac{\Delta f}{\Delta x}=m$, which is constant. Since, the change in function with respect to change in $x$ is constant, which is only possible for a linear function. Therefore, $f(x)$ is a linear function.
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