## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Mid-Chapter Check Point - Page 381: 11

#### Answer

Domain of $f\left( x \right):\left( -\infty ,\infty \right)$ , Range of $f\left( x \right):\left[ -2,\infty \right)$ #### Work Step by Step

The graph is a parabola. Since $f\left( x \right)$ is a polynomial, it exists for every value of $x$ on the real axis. Therefore, the domain of $f\left( x \right)=3{{x}^{2}}-6x+1$ is $\left( -\infty ,\infty \right)$. Since, \begin{align} & 3{{x}^{2}}-6x+1=3\left( {{x}^{2}}-2x \right)+1 \\ & =3\left( {{x}^{2}}-2x+1-1 \right)+1 \\ & =3\left( {{\left( x-1 \right)}^{2}}-1 \right)+1 \\ & \end{align} Also, \begin{align} & {{\left( x-1 \right)}^{2}}\ge 0 \\ & {{\left( x-1 \right)}^{2}}-1\ge -1 \\ & 3\left( {{\left( x-1 \right)}^{2}}-1 \right)\ge -3 \\ & 3\left( {{\left( x-1 \right)}^{2}}-1 \right)+1\ge -2 \\ \end{align} So, the range of the function is $\left[ -2,\infty \right)$.

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