Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.3 - Limits and Continuity - Exercise Set - Page 1163: 68


The value of $\frac{s\left( a+h \right)-s\left( a \right)}{h}$ is $-32a-16h+48$.

Work Step by Step

Consider the function $ s\left( t \right)=-16{{t}^{2}}+48t+160$, Find the value of $\frac{s\left( a+h \right)-s\left( a \right)}{h}$ using the definition of the function. $\begin{align} & \frac{s\left( a+h \right)-s\left( a \right)}{h}=\frac{-16{{\left( a+h \right)}^{2}}+48\left( a+h \right)+160-\left( -16{{a}^{2}}+48a+160 \right)}{h} \\ & =\frac{-16\left( {{a}^{2}}+{{h}^{2}}+2ah \right)+48\left( a+h \right)+160+16{{a}^{2}}-48a-160}{h} \\ & =\frac{-16{{a}^{2}}-16{{h}^{2}}-32ah+48a+48h+160+16{{a}^{2}}-48a-160}{h} \\ & =\frac{-16{{h}^{2}}-32ah+48h}{h} \end{align}$ Factoring out h from the numerator, $\begin{align} & \frac{s\left( a+h \right)-s\left( a \right)}{h}=\frac{h\left( -16h-32a+48 \right)}{h} \\ & =-32a-16h+48 \end{align}$ Thus, the value of $\frac{s\left( a+h \right)-s\left( a \right)}{h}$ is $-32a-16h+48$.
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