Precalculus (6th Edition) Blitzer

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

Chapter 10 - Cumulative Review Exercises - Page 1128: 31


The value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $-h-2x-2$.

Work Step by Step

We have to obtain the value of $ f\left( x+h \right)$; replace $ x $ in the provided function $ f\left( x \right)$ by $\left( x+h \right)$. Since, $ f\left( x \right)=-{{x}^{2}}-2x+1$ then $\begin{align} & f\left( x+h \right)=-{{\left( x+h \right)}^{2}}-2\left( x+h \right)+1 \\ & =-{{x}^{2}}-{{h}^{2}}-2xh-2x-2h+1 \end{align}$ Then, for $ f\left( x+h \right)-f\left( x \right)$: $\begin{align} & f\left( x+h \right)-f\left( x \right)=\left( -{{x}^{2}}-{{h}^{2}}-2xh-2x-2h+1 \right)-\left( -{{x}^{2}}-2x+1 \right) \\ & =-{{x}^{2}}-{{h}^{2}}-2xh-2x-2h+1+{{x}^{2}}+2x-1 \\ & =-h\left( h+2x+2 \right) \end{align}$ Next, divide $ f\left( x+h \right)-f\left( x \right)$ by ‘h’ to get $\frac{f\left( x+h \right)-f\left( x \right)}{h}$: $\begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{-h\left( h+2x+2 \right)}{h} \\ & =-h-2x-2 \end{align}$ Thus, the value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $-h-2x-2$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.