Solution Bank Rational Exponents & Radicals Target D
Answers to the Practice Problems are below in random order.
\(\begin{align}&\text{No}\ & \ \ \ &\text{Yes}\ & \ \ \ &(x,y) → (y,x)\ \text{No;} f(x)\ \text{does not pass horizontal line test}\\\\
&\text{No}\ & \ \ \ &\text{Yes}\ & \ \ \ &\text{Yes;}\ g(x) \text{passes the horizontal line test}\\\\
&y=\sqrt{x+6}\ & \ \ \ &\text{Yes}\ & \ \ \ &\text{example:} x\ge0\\\\
&y=\pm\sqrt{x+6}\ & \ \ \ &f^{-1}(x)=\frac{3}{4}x+\frac{3}{2}\ & \ \ \ &\text{When domain is restricted, it allows} f(x) \text{to pass the horizontal line test}\\\\
&y=-\sqrt{x+6}\ & \ \ \ &k^{-1}(x)=\sqrt[3]{x-1}+3\ & \ \ \ &f(x) \text{does not pass horizontal line test}\\\\
&y=3x^2-30x+54\ & \ \ \ &l^{-1}(x)=4(x-5)^3\ & \ \ \ &m^{-1}(x)=\sqrt[5]{\frac{x+1}{4}}\\\\
&h^{-1}(x)=\sqrt[4]{\frac{x+8}{3}}\ & \ \ \ &\frac{256}{81}\ & \ \ \ &g^{-1}(x)=-\sqrt{2x-18}\\\\
&\frac{2x^2}{7y^{\frac{17}{6}}}\ & \ \ \ &j^{-1}(x)=\sqrt[3]{x+4}\ & \ \ \ &m\left(m^{-1}(x)\right)=x\ \text{and}\ m^{-1}\left(m(x)\right)=x\\\\
&g^{-1}(x)=-\frac{1}{2}x+\frac{7}{2}\ & \ \ \ &f^{-1}(x)=\frac{1}{3}x+2\end{align}\)
Answers to the Practice Problems are below in random order.
\(\begin{align}&\text{No}\ & \ \ \ &\text{Yes}\ & \ \ \ &(x,y) → (y,x)\ \text{No;} f(x)\ \text{does not pass horizontal line test}\\\\
&\text{No}\ & \ \ \ &\text{Yes}\ & \ \ \ &\text{Yes;}\ g(x) \text{passes the horizontal line test}\\\\
&y=\sqrt{x+6}\ & \ \ \ &\text{Yes}\ & \ \ \ &\text{example:} x\ge0\\\\
&y=\pm\sqrt{x+6}\ & \ \ \ &f^{-1}(x)=\frac{3}{4}x+\frac{3}{2}\ & \ \ \ &\text{When domain is restricted, it allows} f(x) \text{to pass the horizontal line test}\\\\
&y=-\sqrt{x+6}\ & \ \ \ &k^{-1}(x)=\sqrt[3]{x-1}+3\ & \ \ \ &f(x) \text{does not pass horizontal line test}\\\\
&y=3x^2-30x+54\ & \ \ \ &l^{-1}(x)=4(x-5)^3\ & \ \ \ &m^{-1}(x)=\sqrt[5]{\frac{x+1}{4}}\\\\
&h^{-1}(x)=\sqrt[4]{\frac{x+8}{3}}\ & \ \ \ &\frac{256}{81}\ & \ \ \ &g^{-1}(x)=-\sqrt{2x-18}\\\\
&\frac{2x^2}{7y^{\frac{17}{6}}}\ & \ \ \ &j^{-1}(x)=\sqrt[3]{x+4}\ & \ \ \ &m\left(m^{-1}(x)\right)=x\ \text{and}\ m^{-1}\left(m(x)\right)=x\\\\
&g^{-1}(x)=-\frac{1}{2}x+\frac{7}{2}\ & \ \ \ &f^{-1}(x)=\frac{1}{3}x+2\end{align}\)
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