Chapter 1 - Review Exercises - Page 300: 24

See the explanation below.

Work Step by Step

(a) Let us consider the provided function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-1}{x-1}\text{ if }x\ne 1 \\ & \text{12 if }x=\text{4} \\ \end{align} \right. To find $f\left( -2 \right)$, put $x=-2$ in function $f\left( x \right)=\frac{{{x}^{2}}-1}{x-1}$. When $x\ne 1$, \begin{align} & f\left( x \right)=\frac{{{x}^{2}}-1}{x-1} \\ & f\left( -2 \right)=\frac{{{\left( -2 \right)}^{2}}-1}{-2-1} \\ & =\frac{4-1}{-3} \\ & =-1. \end{align} (b) Let us consider provided function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-1}{x-1}\text{ if }x\ne 1 \\ & \text{12 if }x=\text{4} \\ \end{align} \right.. To find $f\left( 1 \right)$, put $x=1$ in function $f\left( x \right)=12$. When $x=1$, $f\left( 1 \right)=12$. (c) Let us consider the provided function f\left( x \right)=\left\{ \begin{align} & \frac{{{x}^{2}}-1}{x-1}\text{ if }x\ne 1 \\ & \text{12 if }x=\text{4} \\ \end{align} \right.. To find $f\left( 2 \right)$, put $x=2$ in function $f\left( x \right)=\frac{{{x}^{2}}-1}{x-1}$. When $x\ne 1$, \begin{align} & f\left( x \right)=\frac{{{x}^{2}}-1}{x-1} \\ & f\left( 2 \right)=\frac{{{\left( 2 \right)}^{2}}-1}{2-1} \\ & =\frac{4-1}{1} \\ & =3. \end{align}

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