Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.5 - Zeros of Polynomial Functions - Exercise Set - Page 380: 98

Answer

a) The value of $\left( f\circ g \right)\left( x \right)$ is $-{{x}^{2}}-10x-21$. b) The value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $-2x-h,\text{ }h\ne 0$.

Work Step by Step

(a) Consider the functions, $f\left( x \right)=4-{{x}^{2}}$ and $g\left( x \right)=x+5$ The value of the function $\left( f\circ g \right)\left( x \right)$ is, $\left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right)$ Replace $x$ by $g\left( x \right)$ in the function $f\left( x \right)=4-{{x}^{2}}$. $f\left( g\left( x \right) \right)=4-{{\left[ g\left( x \right) \right]}^{2}}$ Substitute $g\left( x \right)=x+5$ in the function $f\left( g\left( x \right) \right)=4-{{\left[ g\left( x \right) \right]}^{2}}$. $f\left( g\left( x \right) \right)=4-{{\left( x+5 \right)}^{2}}$ Use the property ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. $\begin{align} & f\left( g\left( x \right) \right)=4-\left( {{x}^{2}}+10x+25 \right) \\ & =4-{{x}^{2}}-10x-25 \\ & =-{{x}^{2}}-10x-21 \end{align}$ Therefore, the value of $\left( f\circ g \right)\left( x \right)$ is $-{{x}^{2}}-10x-21$. (b) Consider the function, $f\left( x \right)=4-{{x}^{2}}$ Substitute $x+h$ for $x$ in the function $f\left( x \right)=4-{{x}^{2}}$. $f\left( x+h \right)=4-{{\left( x+h \right)}^{2}}$ Consider the expression, $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ Substitute $f\left( x+h \right)=4-{{\left( x+h \right)}^{2}}$ and $f\left( x \right)=4-{{x}^{2}}$ in the expression $\frac{f\left( x+h \right)-f\left( x \right)}{h}$. $\frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{\left[ 4-{{\left( x+h \right)}^{2}} \right]-\left( 4-{{x}^{2}} \right)}{h}$ Use the property ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. $\begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{\left[ 4-\left( {{x}^{2}}+2xh+{{h}^{2}} \right) \right]-4+{{x}^{2}}}{h} \\ & =\frac{4-{{x}^{2}}-2xh-{{h}^{2}}-4+{{x}^{2}}}{h} \\ & =\frac{-2xh-{{h}^{2}}}{h} \\ & =-2x-h \end{align}$ Therefore, the value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $-2x-h,\text{ }h\ne 0$.
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