Calculus (3rd Edition)

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

Chapter 1 - Precalculus Review - 1.3 The Basic Classes of Functions - Exercises - Page 23: 42


$\delta f(x)=9\cdot 10^x$ $\delta f(x)=(b-1)b^x$

Work Step by Step

We are given the function: $f(x)=10^x$ Compute $\delta f(x)$: $\delta f(x)=f(x+1)-f(x)=10^(x+1)-10^x=10^x\cdot 10-10^x$ $=10^x(10-1)$ $=9\cdot 10^x$ We got: $\delta f(x)=9\cdot 10^x$ We are given the function: $f(x)=b^x$ Compute $\delta f(x)$: $\delta f(x)=f(x+1)-f(x)=b^(x+1)-b^x=b^x\cdot b-b^x$ $=b^x(b-1)$ $=(b-1)b^x$ We got: $\delta f(x)=(b-1)b^x$
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