Answer Key 11.3

$\begin{array}{rrl} \\ \\ \\ (g\circ f)(x)&=&-(\sqrt[5]{-x-3})^5-3 \\ &=&-(-x-3)-3 \\ &=&x+3-3 \\ &=&x\hspace{0.75in}\text{Inverse} \end{array}$
$\begin{array}{rrl} \\ \\ \\ (g\circ f)(x)&=&4-\left(\dfrac{4}{x}\right) \\ \\ &=&4-\dfrac{4}{x}\hspace{0.5in}\text{Not inverse} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ (g\circ f)(x)&=&-10\left(\dfrac{x-5}{10}\right)+5 \\ \\ &=& -x+5+5\\ \\ &=&-x+10\hspace{0.75in}\text{Not inverse} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ (f\circ g)(x)&=&\dfrac{(10x+5)-5}{10} \\ \\ &=&\dfrac{10x+5-5}{10} \\ \\ &=&\dfrac{10x}{10} \\ \\ &=&x\hspace{0.75in}\text{Inverse} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (f\circ g)(x)&=&\dfrac{-2}{\dfrac{3x+2}{x+2}+3} \\ \\ &=& \dfrac{-2(x+2)}{3x+2+3(x+2)}\\ \\ &=& \dfrac{-2x-4}{3x+2+3x+6}\\ \\ &=& \dfrac{-2x-4}{6x+8}\\ \\ &=& \dfrac{-x-2}{3x+4}\hspace{0.75in}\text{Not inverse} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (f\circ g)&=&\dfrac{-\left(\dfrac{-2x+1}{-x-1}\right)-1}{\dfrac{-2x+1}{-x-1}-2} \\ \\ &=&\dfrac{-(-2x+1)-1(-x-1)}{-2x+1-2(-x-1)} \\ \\ &=&\dfrac{2x-1+x+1}{-2x+1+2x+2} \\ \\ &=&\dfrac{3x}{3} \\ \\ &=&x\hspace{0.75in}\text{Inverse} \end{array}$
$\begin{array}{rrcrr} \\ \\ \\ \\ \\ \\ y&=&(x-2)^5&+&3 \\ x&=&(y-2)^5&+&3 \\ -3&&&-&3 \\ \midrule x-3&=&(y-2)^5&& \\ \sqrt[5]{x-3}&=&y-2&& \\ \\ y&=&\sqrt[5]{x-3}&+&2 \end{array}$
$\begin{array}{rrcrr} \\ \\ \\ \\ \\ \\ y&=&\sqrt[3]{x+1}&+&2 \\ x&=&\sqrt[3]{y+1}&+&2 \\ -2&&&-&2 \\ \midrule x-2&=&\sqrt[3]{y+1}&& \\ (x-2)^3&=&y+1&& \\ \\ y&=&(x-2)^3&-&1 \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ y&=&\dfrac{4}{x+2} \\ \\ x&=&\dfrac{4}{y+2} \\ \\ y+2&=&\dfrac{4}{x} \\ \\ y&=&\dfrac{4}{x}-2 \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ y&=&\dfrac{3}{x-3} \\ \\ x&=&\dfrac{3}{y-3} \\ \\ y-3&=&\dfrac{3}{x} \\ \\ y&=&\dfrac{3}{x}+3 \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ y&=&\dfrac{-2x-2}{x+2} \\ \\ x&=&\dfrac{-2y-2}{y+2} \\ \\ x(y+2)&=&-2y-2 \\ xy+2x&=&-2y-2 \\ -xy+2&&-xy+2 \\ \midrule 2x+2&=&-2y-xy \\ 2x+2&=&y(-2-x) \\ \\ y&=&\dfrac{2x+2}{-2-x} \\ \\ y&=&-\dfrac{2x+2}{2+x} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ y&=&\dfrac{9+x}{3} \\ \\ x&=&\dfrac{9+y}{3} \\ \\ 3x&=&9+y \\ y&=&3x-9 \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ y&=&\dfrac{10-x}{5} \\ \\ x&=&\dfrac{10-y}{5} \\ \\ 5x&=&10-y \\ y&=&10-5x \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ y&=&\dfrac{5x-15}{2} \\ \\ x&=&\dfrac{5y-15}{2} \\ \\ 5y-15&=&2x \\ \\ 5y&=&2x+15 \\ \\ y&=&\dfrac{2x+15}{5} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ y&=&-(x-1)^3 \\ x&=&-(y-1)^3 \\ \sqrt[3]{x}&=&-(y-1) \\ \sqrt[3]{x}&=&-y+1 \\ -y&=&\sqrt[3]{x}-1 \\ \\ y&=&1-\sqrt[3]{x} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ y&=&\dfrac{12-3x}{4} \\ \\ x&=&\dfrac{12-3y}{4} \\ \\ 4x&=&12-3y \\ \\ 3y&=&12-4x \\ \\ y&=&\dfrac{12-4x}{3} \\ \\ y&=&4-\dfrac{4}{3}x \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ y&=&(x-3)^3 \\ x&=&(y-3)^3 \\ \sqrt[3]{x}&=&y-3 \\ \\ y&=&\sqrt[3]{x}+3 \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ y&=&\sqrt[5]{-x}+2 \\ x&=&\sqrt[5]{-y}+2 \\ x-2&=&\sqrt[5]{-y} \\ (x-2)^5&=&-y \\ \\ y&=&-(x-2)^5 \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ y&=&\dfrac{x}{x-1} \\ \\ x&=&\dfrac{y}{y-1} \\ \\ x(y-1)&=&y \\ \\ xy-x&=&y \\ \\ y-xy&=&-x \\ \\ y(1-x)&=&-x \\ \\ y&=&\dfrac{-x}{1-x} \\ \\ y&=&\dfrac{x}{x-1} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ y&=&\dfrac{-3-2x}{x+3} \\ \\ x&=&\dfrac{-3-2y}{y+3} \\ \\ x(y+3)&=&-3-2y \\ xy+3x&=&-3-2y \\ +2y-3x&&-3x+2y \\ \midrule xy+2y&=&-3-3x \\ y(x+2)&=&-3-3x \\ \\ y&=&\dfrac{-3-3x}{x+2} \\ \\ y&=&-\dfrac{3x+3}{x+2} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ y&=&\dfrac{x-1}{x+1} \\ \\ x&=&\dfrac{y-1}{y+1} \\ \\ x(y+1)&=&y-1 \\ xy+x&=&y-1 \\ xy-y&=&-x-1 \\ y(x-1)&=&-x-1 \\ \\ y&=&\dfrac{-x-1}{x-1} \\ \\ y&=&-\dfrac{x+1}{x-1} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ y&=&\dfrac{x}{x+2} \\ \\ x&=&\dfrac{y}{y+2} \\ \\ x(y+2)&=&y \\ xy+2x&=&y \\ 2x&=&y-xy \\ 2x&=&y(1-x) \\ \\ y&=&\dfrac{2x}{1-x} \end{array}$

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