## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 1 - Review Exercises - Page 303: 100

#### Answer

See the explanation below.

#### Work Step by Step

(a) Consider the function, \begin{align} & \left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right) \\ & =f\left( 4x-1 \right) \\ & ={{\left( 4x-1 \right)}^{2}}+3 \end{align} On simplification, we get, \begin{align} & \left( f\circ g \right)\left( x \right)={{\left( 4x-1 \right)}^{2}}+3 \\ & =16{{x}^{2}}-8x+1+3 \\ & =16{{x}^{2}}-8x+4 \end{align} Therefore, the value of $\left( f\circ g \right)\left( x \right)$ for the functions $f\left( x \right)={{x}^{2}}\text{+3 and }g\left( x \right)=4x-1$ is $16{{x}^{2}}-8x+4$. (b) Consider the function \begin{align} & \left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right) \\ & =g\left( {{x}^{2}}+3 \right) \\ & =4\left( {{x}^{2}}+3 \right)-1 \\ & =4{{x}^{2}}+11 \end{align} Therefore, the value of $\left( g\circ f \right)\left( x \right)$ for the functions $f\left( x \right)={{x}^{2}}\text{+3 and }g\left( x \right)=4x-1$ is $4{{x}^{2}}+11$. (c) From part (a), $\left( f\circ g \right)\left( x \right)={{\left( 4x-1 \right)}^{2}}+3$. Therefore, \begin{align} & \left( f\circ g \right)\left( 3 \right)={{\left( 4\times 3-1 \right)}^{2}}+3 \\ & ={{\left( 11 \right)}^{2}}+3 \\ & =121+3 \\ & =124 \end{align} Therefore, the value of $\left( f\circ g \right)\left( 3 \right)$ for the functions $f\left( x \right)={{x}^{2}}\text{+3 and }g\left( x \right)=4x-1$ is $124$.

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